PHYS 232 - UNIVERSITY PHYSICS
II - Calculation of electric forces
between point charges
Question Bank - Set 1
Liberty University
Question 1
Question
Three point charges are arranged on the vertices of an equilateral triangle with
side length a, as shown below. The charges are +q,+2q, and −3q. Calculate
the magnitude and direction of the net electric force acting on the charge +2q.
+q
−3q+2q
Solution
Step 1: Calculate the electric force between +qand +2q. The electric force
between two charges q1and q2separated by a distance ris given by Coulomb’s
law:
F=k|q1q2|
r2
where kis the Coulomb constant (8.988 ×109N m2/C2).
In this case, q1= +q,q2= +2q, and r=a(the side length of the equilateral
triangle):
F+q,+2q=k|+q·+2q|
a2=2kq2
a2
Step 2: Calculate the electric force between −3qand +2q. Similarly, the
electric force between −3qand +2qis:
F−3q,+2q=k| − 3q·+2q|
a=6kq2
a2
Step 3: Find the net force. To find the net force on the charge +2q, we need
to consider the vector sum of the forces from +qand −3q. Since the forces act
along a straight line through the charge +2q, we only need to consider their
magnitudes and signs. The net force is:
Fnet =F+q,+2q−F−3q,+2q=2kq2
a2−6kq2
a2=−4kq2
a2
Therefore, the magnitude of the net electric force acting on the charge +2q
is 4kq2
a2, and its direction is attractive (towards the −3qcharge).
Question 2
Question
Three point charges are arranged in an equilateral triangle as shown below.
Calculate the magnitude and direction of the net force on the charge q1=
+2.00 µC.
q2= +4.00 µC
q1= +2.00 µCq3= +6.00 µC
Solution
Step 1: First, let’s calculate the magnitudes of the forces on q1due to q2and
q3using Coulomb’s law: F=k|q1||q2|
r2, where k= 8.99 ×109N m2/C2.
The distance between q1and q2is the length of the side of the equilateral
triangle, r=l= 1 m.
Step 2: Calculating the force on q1due to q2:
F12 =k|q1||q2|
r2
F12 = (8.99 ×109)(2.00 ×10−6)(4.00 ×10−6)
(1)2
F12 = 7.19 ×10−2N
2
Step 3: Calculating the force on q1due to q3:
F13 =k|q1||q3|
r2
Since q3is 2l= 2 m away from q1:
F13 = (8.99 ×109)(2.00 ×10−6)(6.00 ×10−6)
(2)2
F13 = 4.498 ×10−2N
Step 4: Now, let’s find the net force on q1by vectorially adding the forces
F12 and F13. Given the forces are along the same line of the equilateral triangle,
they are in opposite directions and have the same magnitude.
Net force on q1=F12 −F13 = 7.19 ×10−2−4.498 ×10−2N
Net force on q1= 2.692 ×10−2N (in the direction towards q2)
Therefore, the magnitude of the net force on q1is 2.692 ×10−2N and the
direction is towards q2.
Question 3
Question
Two point charges, q1= 3.0µC and q2=−4.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Step 2: Calculate the distance between
the charges. Step 3: Apply Coulomb’s law to find the electric force between the
charges.
Step 1: Convert the charges to coulombs. The charge q1= 3.0µC can be
converted to coulombs as follows:
q1= 3.0×10−6C= 3.0µC
Similarly, the charge q2=−4.0µC can be converted to coulombs as follows:
q2=−4.0×10−6C=−4.0µC
3
Step 2: Calculate the distance between the charges. Given that the two
charges are placed 10.0 cm apart, we convert this distance to meters:
d= 10.0cm = 0.1m
Step 3: Apply Coulomb’s law to find the electric force between the charges.
Coulomb’s law states that the magnitude of the electric force between two
charges is given by:
F=k· |q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F=(8.99 ×109N m2/C2)· |3.0×10−6C· −4.0×10−6C|
(0.1m)2
F=8.99 ×109·12.0×10−12
0.01
F=107.88 ×10−3
0.01 = 10.788 N
Therefore, the magnitude of the electric force between the charges is 10.788
N.
Question 4
Question
Three point charges are arranged in an equilateral triangle. The charge at each
corner of the triangle is +q. Determine the magnitude and direction of the net
force on each charge.
Solution
Let’s denote the charges as q1,q2, and q3, each with magnitude +q.
Step 1: Calculate the magnitude of the force on q1due to q2. The force
F21 on q1due to q2is given by Coulomb’s Law:
F21 =k· |q1|·|q2|
r2
21
Since the charges are +q, the force will be repulsive.
Step 2: Calculate the magnitude of the force on q1due to q3. The force
F31 on q1due to q3is given by Coulomb’s Law:
F31 =k· |q1|·|q3|
r2
31
4
Since the charges are +q, the force will be repulsive.
Step 3: Resolve the forces on q1into components to determine the net force.
Since the forces are acting at 120◦angles to each other, the superposition of the
forces will yield a net force of magnitude:
Fnet1=qF2
21 +F2
31 + 2 ·F21 ·F31 ·cos(120◦)
Step 4: Determine the direction of the net force on q1. The direction of the
net force on q1can be found using trigonometry.
Step 5: Repeat the above steps to determine the net forces on q2and q3.
The net forces on q2and q3can be calculated in a similar manner, accounting
for the forces between all pairs of charges.
By analyzing the symmetry of the problem, one can determine that the net
forces will be along the lines joining the charges, and the magnitudes will be
equal.
Question 5
Question
Consider three point charges placed at the corners of an equilateral triangle
of side length d. The charges are as follows: q1=−3µC,q2= 5µC, and
q3=−2µC. Calculate the net electric force acting on q3due to q1and q2.
Solution
Step 1: Find the distance between q1and q3(and q2and q3). The distance
between any two adjacent charges in an equilateral triangle can be calculated
using the Pythagorean theorem. For an equilateral triangle of side length d,
drawing an altitude from one of the vertices to the opposite side forms a right
triangle with hypotenuse of length d and base of length d/2.
Step 2: Calculate the net force acting on q3due to q1. The force F13 between
q1and q3is given by Coulomb’s law:
F13 =k|q1||q3|
r2
13
where kis the Coulomb constant and r13 is the distance between q1and q3.
Step 3: Calculate the force acting on q3due to q2. The force F23 between
q2and q3is also given by Coulomb’s law:
F23 =k|q2||q3|
r2
23
where r23 is the distance between q2and q3.
Step 4: Determine the net force acting on q3. To find the net force acting on
q3, we need to consider the vector sum of F13 and F23. Remember to account
for the direction of each force.
5
Step 5: Perform the calculations. Substitute the given values for the charges
and distances into the expressions for F13 and F23 , taking into account the signs
of the charges. Then, find the net force by adding the two forces vectorially.
Question 6
Question
Two charges, q1=−3µC and q2= 5µC, are placed 2 meters apart. Calculate
the magnitude of the electric force between these two charges.
Solution
Step 1: Recall that the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into Coulomb’s Law:
F= 8.99 ×109|(−3µC)(5µC)|
(2 m)2
Step 3: Calculate the magnitudes of the charges in coulombs:
q1=−3µC =−3×10−6C
q2= 5µC = 5 ×10−6C
Step 4: Substitute the magnitudes of the charges into the formula:
F= 8.99 ×109|(−3×10−6C)(5 ×10−6C)|
(2 m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×10915 ×10−12C2
4m2
F= 8.99 ×109×3.75 ×10−12 N
F= 33.7125 ×10−3N
F= 3.37125 ×10−2N
Step 6: Thus, the magnitude of the electric force between the two charges is
3.37125 ×10−2N.
6
Question 7
Question
Two point charges, Q1= 5 ×10−6C and Q2=−3×10−6C, are placed 8 meters
apart in a vacuum. Calculate the magnitude of the electric force between the
charges.
Solution
Step 1: Calculate the electric force between the charges using Coulomb’s Law
formula:
F=k
Q1·Q2
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2), Q1and Q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into the formula and solve for the force:
F= 8.99 ×109
(5 ×10−6)·(−3×10−6)
82
F= 8.99 ×109
−15 ×10−12
64
F= 8.99 ×109
−0.23438 ×10−12
F= 8.99 ×109×0.23438 ×10−12
F= 2.11 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.11 ×
10−3N.
Question 8
Question
Two electrons are located 1.5 nm apart. Calculate the electric force between
them. The charge of an electron is −1.6×10−19 C.
Solution
Step 1: Calculate the electric force using Coulomb’s Law: The equation for the
electric force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
7
where Fis the electric force, kis Coulomb’s constant (8.99 ×109N m2/C2),
q1and q2are the charges of the two particles, and ris the separation distance
between the charges.
Step 2: Substitute the given values into the equation: Given that the charge
of an electron is −1.6×10−19 C, the charges of the two electrons are q1=q2=
−1.6×10−19 C. The separation distance ris 1.5 nm = 1.5×10−9m.
Step 3: Calculate the electric force: Plugging in the values, we get:
F= (8.99 ×109)|(−1.6×10−19)(−1.6×10−19)|
(1.5×10−9)2
Step 4: Simplify and calculate the force:
F= (8.99 ×109)(2.56 ×10−38)
2.25 ×10−18
F= (8.99 ×109)×1.138 ×10−20
F= 1.023 ×10−10 N
Therefore, the electric force between the two electrons is 1.023 ×10−10 N.
Question 9
Question
Three point charges are arranged as shown in the diagram below. The charges
are q1=−3µC, q2= 5 µC, and q3=−2µC. The distance rbetween q1and q2
is 4meters, and the distance between q1and q3is 3meters. Calculate the total
electric force on q1due to q2and q3.
q1q2
q3
Solution
Step 1: Calculate the force between q1and q2: The electric force between two
point charges is given by Coulomb’s law:
F12 =k· |q1·q2|
r2
8
where k= 8.99×109N m2/C2is the Coulomb constant, q1=−3µC, q2= 5 µC,
and r= 4 m. Plugging in the values, we get:
F12 =8.99 ×109·|−3·5|
42= 1.1246 ×10−7N
Step 2: Calculate the force between q1and q3: Using the same formula as
above with r= 3 m, we have:
F13 =k· |q1·q3|
r2=8.99 ×109·|−3·(−2)|
32= 1.1988 ×10−7N
Step 3: Calculate the total electric force on q1: The total force on q1is the
vector sum of the forces F12 and F13. Since the forces are in opposite directions,
we subtract them:
Ftotal =F12 −F13 = (1.1246 ×10−7)−(1.1988 ×10−7) = −7.42 ×10−9N
Thus, the total electric force on q1due to q2and q3is −7.42 ×10−9N,
directed towards q3.
Question 10
Question
Three point charges are arranged at the vertices of an equilateral triangle with
sides of length d= 1.00 m. The charges are q1= +5.00µC, q2=−2.00µC, and
q3=−3.00µC. Calculate the magnitude and direction of the net electric force
acting on q1due to the other two charges.
Solution
Step 1: Calculate the electric force on q1due to q2. The electric force between
two point charges q1and q2is given by Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
where - kis the Coulomb constant (8.99 ×109N m2/C2), - |q1|and |q2|are
the magnitudes of the charges, - ris the distance between the charges.
Given q1= +5.00µC, q2=−2.00µC, and d= 1.00 m, the distance between
q1and q2(and also between all charges) is d.
Therefore,
F12 =(8.99 ×109N m2/C2)·(5.00 ×10−6C)·(2.00 ×10−6C)
(1.00 m)2
F12 = 26.97 ×10−3N(repulsive, to the right)
9
Step 2: Calculate the electric force on q1due to q3. Following a similar
calculation as above, the electric force between q1and q3is found to be:
F13 = 44.94×10−3N(at an angle of 120 degrees counterclockwise from horizontal)
Step 3: Find the net force on q1. The net force acting on q1is the vector
sum of the forces due to q2and q3. Using vector addition, we have:
Net force =p(F12 +F13 cos(120◦))2+ (F13 sin(120◦))2
Net force =p(26.97 ×10−3+ 44.94 ×10−3cos(120◦))2+ (44.94 ×10−3sin(120◦))2
Net force = 64.09 ×10−3N
The direction of the net force can be found by calculating the angle it makes
with the horizontal and is given by:
θ= tan−1F13 sin(120◦)
F12 +F13 cos(120◦)
θ=−30◦
Therefore, the magnitude of the net electric force acting on q1is 64.09×10−3
N, and it is directed at an angle of 30◦below the horizontal.
Question 11
Question
Three point charges are arranged in a line as shown below:
q1=−3µC q2= 5 µC q3=−7µC
The charges are located at positions:
x1= 1.0m, x2= 0.0m, x3=−2.0m
Calculate the net electric force on the charge q3due to the other two charges.
Solution
Step 1: Calculate the electric force between q1and q3The electric force between
two charges q1and q3can be calculated using Coulomb’s Law:
F13 =k|q1q3|
r2
13
where k= 8.99 ×109Nm2/C2is the Coulomb’s constant.
10
Substitute the given values:
F13 =(8.99 ×109)(3 ×10−6)(7 ×10−6)
(3)2
F13 =8.99 ×3×7
3= 71.92 ×10−6= 7.192 ×10−5N(to the right)
Step 2: Calculate the electric force between q2and q3Similarly, the electric
force between two charges q2and q3can be calculated using Coulomb’s Law:
F23 =k|q2q3|
r2
23
Substitute the given values:
F23 =(8.99 ×109)(5 ×10−6)(7 ×10−6)
(2)2
F23 =8.99 ×5×7
4= 31.465 ×10−6= 3.1465 ×10−5N(to the left)
Step 3: Find the net electric force on q3To find the net force on q3, we need
to consider the direction and add the two forces:
Fnet =F13 +F23 = 7.192×10−5−3.1465×10−5= 4.0455×10−5N(to the right)
Therefore, the net electric force on q3due to the other two charges is 4.0455×
10−5Nto the right.
Question 12
Question
Two point charges, q1= 3 nC and q2=−5nC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Convert the charges to coulombs.
Given: q1= 3 nC = 3 ×10−9C
q2=−5nC =−5×10−9C
Step 2: Calculate the distance between the charges.
Given: r= 10 cm = 0.1m
Step 3: Calculate the electric force using Coulomb’s Law. Coulomb’s Law
states that the magnitude of the electric force between two point charges is given
by:
F=k|q1q2|
r2
11
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values:
F= (8.99 ×109)|(3 ×10−9)(−5×10−9)|
(0.1)2
Step 4: Calculate the direction of the force.
The direction of the force is along the line joining the two charges. Since the
charges have opposite signs, the force is attractive.
Step 5: Calculate the magnitude of the force.
F= (8.99 ×109)15 ×10−18
0.01
F=134.85 ×10−9
0.01
F= 13.485 ×10−6N= 13.485 µN
Therefore, the magnitude of the electric force that q1exerts on q2is 13.485 µN
and the direction is attractive.
Question 13
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are separated by a distance
of 10.0cm. Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two point charges,
we will use Coulomb’s law:
F=k
q1·q2
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the
charges, and ris the distance between the charges.
Step 1: Convert the given charges to coulombs: Given: q1= 4.0µC =
4.0×10−6Cand q2=−6.0µC =−6.0×10−6C.
Step 2: Determine the distance between the charges in meters: Given:
r= 10.0cm = 10.0×10−2m= 0.10 m.
Step 3: Calculate the electric force between the charges using Coulomb’s
law:
F= 8.99 ×109N·m2/C2×
(4.0×10−6C)·(−6.0×10−6C)
(0.10 m)2
Step 4: Solve for the electric force:
F= 8.99×109×4.0×10−6× −6.0×10−6
0.102= 8.99×109×−24 ×10−12
0.01 = 8.99×109×−2.4×10−9N
12
F=−21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 14
Question
Three point charges are arranged in an equilateral triangle as shown below.
The charges have magnitudes of +2.5µC,−3.0µC, and +4.0µC. Calculate the
electric force on the −3.0µC charge.
+2.5µC
−3.0µC +4.0µC
Solution
Step 1: We first need to calculate the electric forces each charge exerts on the
−3.0µC charge. Let’s denote the distance between the charges as r. The electric
force Fbetween two point charges q1and q2separated by a distance ris given
by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2).
Calculating the forces between the charges:
For +2.5µC and −3.0µC:
F2.5µC,−3.0µC =8.99 ×109·|2.5×10−6·(−3.0×10−6)|
r2
For +4.0µC and −3.0µC:
F4.0µC,−3.0µC =8.99 ×109·|4.0×10−6·(−3.0×10−6)|
r2
Step 2: Next, we need to find the electric force acting on the −3.0µC charge
due to the other two charges. Since the charges are arranged symmetrically in
an equilateral triangle, the magnitudes of F2.5µC,−3.0µC and F4.0µC,−3.0µC are
the same.
Therefore, the total electric force on the −3.0µC charge is:
Ftotal = 2 ·F2.5µC,−3.0µC
Step 3: Finally, calculate the total electric force on the −3.0µC charge.
Perform the necessary calculations to find the numerical value of this force.
13
Question 15
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed at points Aand B
respectively. Point Ais located at coordinates (0,0) in meters and point Bis
at coordinates (3,4). Calculate the magnitude of the electric force experienced
by q2due to q1.
Solution
Step 1: Calculate the distance between the two point charges. Given that point
Ais located at coordinates (0,0) and point Bis located at coordinates (3,4), the
distance rbetween the two point charges can be calculated using the distance
formula:
r=p(x2−x1)2+ (y2−y1)2
where x1= 0,y1= 0,x2= 3, and y2= 4. Therefore,
r=p(3 −0)2+ (4 −0)2=p32+ 42=√9 + 16 = √25 = 5 m
Step 2: Calculate the magnitude of the electric force. The magnitude of
the electric force between two point charges can be calculated using Coulomb’s
Law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant. Substitute the given
values q1=−3.0µC,q2= 5.0µC, and r= 5 m into the formula:
F= (8.99 ×109)| − 3.0×10−6×5.0×10−6|
52
F= 8.99 ×109×15 ×10−12
25 = 8.99 ×109×0.6×10−12 = 5.394 ×10−3N
Therefore, the magnitude of the electric force experienced by q2due to q1is
5.394 ×10−3N.
Question 16
Question
Two point charges, q1= 4.0µC and q2=−8.0µC, are placed 0.20 m apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
14
Solution
Given: q1= 4.0µC= 4.0×10−6Cq2=−8.0µC=−8.0×10−6Cr= 0.20 m
We know that the electric force between two charges q1and q2separated by
a distance ris given by Coulomb’s law:
F=k|q1q2|
r2
Step 1: Calculate the electric force constant k.
k=1
4πε0
The permittivity of free space ε0= 8.85 ×10−12 C2/N·m2Thus,
k=1
4π×8.85 ×10−12
Step 2: Substitute the given values into Coulomb’s law to find the magnitude
of the electric force.
F=1
4π×8.85 ×10−12 ·|4.0×10−6× −8.0×10−6|
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F=1
4π×8.85 ×10−12 ·32 ×10−12
0.04
F=1
4π×8.85 ×10−12 ·800 ×10−12
F=800
4π×8.85
F≈800
35.16
F≈22.74 N
Therefore, the magnitude of the electric force between the charges is approx-
imately 22.74 N.
Question 17
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 0.10 m apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
15
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k· |q1·q2|
r2
where k≈8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the distance between the charges.
Step 2: Plug in the given values to find the electric force:
F=(8.99 ×109N·m2/C2)·|−4.0×10−6C·8.0×10−6C|
(0.10 m)2
Step 3: Calculate the electric force:
F=8.99 ×109·4.0×10−6·8.0×10−6
0.01
F=287.68
0.01
F= 28768 N
Step 4: Since q1is negative and q2is positive, the force will be attractive.
Therefore, the electric force that q1exerts on q2is 28768 N, directed towards q1.
Question 19
Question
Three point charges are arranged on the x-axis as follows: +qat x= 0,−3qat
x= 4a, and +2qat x= 7a, where qand aare positive constants. Calculate the
net force on the +2qcharge due to the other charges.
Solution
Step 1: Calculate the force on the +2qcharge due to the +qcharge: The force
F1can be calculated using Coulomb’s Law, which states that the magnitude of
the force between two point charges is given by:
F=k·|q1·q2|
r2
where kis Coulomb’s constant, q1and q2are the charges, and ris the distance
between the charges.
The force on the +2qcharge due to the +qcharge is:
F1=k·|(+2q)(q)|
(7a−0)2
16
Step 2: Calculate the force on the +2qcharge due to the −3qcharge: The
force F2can be calculated using Coulomb’s Law as well:
F2=k·|(+2q)(−3q)|
(7a−4a)2
Step 3: Determine the direction of the net force: Since the forces due to the
+qand −3qcharges are in opposite directions, the net force on the +2qcharge
will be the difference between F1and F2:
Fnet =F1−F2
Fnet =k·2q·q
(7a)2−2q·3q
3a)2
Therefore, the net force on the +2qcharge due to the other charges can be
calculated by subtracting the force on the +2qcharge due to the −3qcharge
from the force due to the +qcharge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle with
sides of length d. Charge Qis placed at each vertex. Calculate the magnitude
of the electric force on one of the charges due to the other two charges.
Solution
Let’s denote the three charges as Q1,Q2, and Q3, with Q1being the charge we
want to calculate the force on. The distance between each charge and Q1is d.
Step 1: Calculate the electric force on Q1due to Q2. The electric force
between two charges is given by Coulomb’s Law:
F=k· |Q1|·|Q2|
r2
where kis the electrostatic constant 8.99×109N m2/C2and r=dis the distance
between the charges.
Plugging in the values, the force on Q1due to Q2is:
F1=8.99 ×109·Q·Q
d2=8.99 ×109·Q2
d2
Step 2: Calculate the direction of the force due to Q2. Since Q2is at one
vertex of an equilateral triangle, the force on Q1due to Q2will be along the
line connecting Q1and Q2. This force will have a direction opposite to that of
the line connecting Q1and Q2.
17
Step 3: Calculate the electric force on Q1due to Q3. Similarly, the force
on Q1due to Q3will be along the line connecting Q1and Q3and will have a
direction opposite to that of the line connecting Q1and Q3. Thus, the force
will be:
F2=8.99 ×109·Q2
d2
Step 4: Calculate the total electric force on Q1. Since the forces F1and F2
have the same magnitude and opposite directions, the total electric force on Q1
due to Q2and Q3is:
Ftotal =F1−F2=8.99 ×109·Q2
d2−8.99 ×109·Q2
d2= 0
Thus, the net electric force on Q1due to Q2and Q3is zero.
Question 22
Question
Two point charges, q1= +3.0µC and q2=−6.0µC, are placed 8.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units (Coulombs).
Given: q1= +3.0µC = 3.0×10−6C
q2=−6.0µC =−6.0×10−6C
Step 2: Find the distance between the charges in meters.
Given: r= 8.0cm = 8.0×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law,
F=k|q1q2|
r2.
Where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109)· |3.0×10−6· −6.0×10−6|
(8.0×10−2)2
Step 4: Calculate the magnitude of the electric force.
F=8.99 ×109·18 ×10−12
64 ×10−4
F=161.82 ×10−3
64 ×10−4
F=161.82
64 = 2.53 N
Therefore, the magnitude of the electric force between the charges is 2.53 N.
18
Question 23
Question
Two point charges Q1 = +3.0 µC and Q2 = -5.0 µC are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that Q1 exerts on
Q2.
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Iden-
tify the direction of the force by considering the signs of the charges. Step 3:
Calculate the magnitude of the electric force using Coulomb’s law.
Step 1: Convert the charges from microcoulombs to coulombs: Q1 =
3.0µC = 3.0×10−6Cand Q2 = −5.0µC =−5.0×10−6C.
Step 2: The force will be attractive since Q1 is positive and Q2 is negative.
Step 3: Use Coulomb’s Law to find the magnitude of the electric force:
F=k|Q1·Q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), ris the distance between
the charges (0.1 m), Q1 = 3.0×10−6C, and Q2 = −5.0×10−6C.
Substitute the values into the equation:
F= (8.99 ×109)|3.0×10−6· −5.0×10−6|
(0.1)2
F= (8.99 ×109)15 ×10−12
0.01
F= (8.99 ×109)×1.5×10−9
F= 13.485 ×100
Therefore, F= 13.485N
The magnitude of the electric force that Q1 exerts on Q2 is 13.485 N, directed
towards Q1.
Question 24
Question
Two point charges, q1=−4.00 µC and q2= 3.00 µC, are separated by a distance
of 10.0 cm in a vacuum. Calculate the magnitude of the electric force between
the charges.
19
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Deter-
mine the distance between the charges in meters. Step 3: Use Coulomb’s Law
to calculate the electric force between the charges.
Step 1: Convert the charges from microcoulombs to coulombs. q1=
−4.00 µC =−4.00 ×10−6C q2= 3.00 µC = 3.00 ×10−6C
Step 2: Determine the distance between the charges in meters. Given:
distance r= 10.0cm = 0.10 m
Step 3: Use Coulomb’s Law to calculate the electric force between the
charges. Coulomb’s Law: F=k|q1q2|
r2where k= 8.99 ×109Nm2/C2is the
electrostatic constant.
Substitute the known values into the equation: F=(8.99 ×109)×|−4.00 ×10−6×3.00 ×10−6|
(0.10)2
F=(8.99 ×109)×1.2×10−11
0.01
F=10.788 ×10−2
0.01
F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788
N.
Question 25
Question
Two point charges, q1=−4µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the given charges to coulombs. Recall that 1µC= 10−6C.
⇒q1=−4µC=−4×10−6C=−4×10−6C
⇒q2= 5 µC= 5 ×10−6C= 5 ×10−6C
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The formula for the magnitude of the electric force between two charges q1and
q2separated by a distance ris given by:
F=k
q1·q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Step 3: Substitute the given values into the formula.
F= 8.99 ×109
(−4×10−6)·(5 ×10−6)
(0.10)2
20
F= 8.99 ×109
−20 ×10−12
0.01
F= 8.99 ×109
−2×10−9
F= 8.99 ×109×2×10−9
F= 17.98 ×100
F= 17.98 N
Therefore, the magnitude of the electric force between the charges is 17.98
N.
21
Step 3: Find the net force. To find the net force on the charge +2q, we need
to consider the vector sum of the forces from +qand −3q. Since the forces act
along a straight line through the charge +2q, we only need to consider their
magnitudes and signs. The net force is:
Fnet =F+q,+2q−F−3q,+2q=2kq2
a2−6kq2
a2=−4kq2
a2
Therefore, the magnitude of the net electric force acting on the charge +2q
is 4kq2
a2, and its direction is attractive (towards the −3qcharge).
Question 2
Question
Three point charges are arranged in an equilateral triangle as shown below.
Calculate the magnitude and direction of the net force on the charge q1=
+2.00 µC.
q2= +4.00 µC
q1= +2.00 µCq3= +6.00 µC
Solution
Step 1: First, let’s calculate the magnitudes of the forces on q1due to q2and
q3using Coulomb’s law: F=k|q1||q2|
r2, where k= 8.99 ×109N m2/C2.
The distance between q1and q2is the length of the side of the equilateral
triangle, r=l= 1 m.
Step 2: Calculating the force on q1due to q2:
F12 =k|q1||q2|
r2
F12 = (8.99 ×109)(2.00 ×10−6)(4.00 ×10−6)
(1)2
F12 = 7.19 ×10−2N
2
Step 3: Calculating the force on q1due to q3:
F13 =k|q1||q3|
r2
Since q3is 2l= 2 m away from q1:
F13 = (8.99 ×109)(2.00 ×10−6)(6.00 ×10−6)
(2)2
F13 = 4.498 ×10−2N
Step 4: Now, let’s find the net force on q1by vectorially adding the forces
F12 and F13. Given the forces are along the same line of the equilateral triangle,
they are in opposite directions and have the same magnitude.
Net force on q1=F12 −F13 = 7.19 ×10−2−4.498 ×10−2N
Net force on q1= 2.692 ×10−2N (in the direction towards q2)
Therefore, the magnitude of the net force on q1is 2.692 ×10−2N and the
direction is towards q2.
Question 3
Question
Two point charges, q1= 3.0µC and q2=−4.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Step 2: Calculate the distance between
the charges. Step 3: Apply Coulomb’s law to find the electric force between the
charges.
Step 1: Convert the charges to coulombs. The charge q1= 3.0µC can be
converted to coulombs as follows:
q1= 3.0×10−6C= 3.0µC
Similarly, the charge q2=−4.0µC can be converted to coulombs as follows:
q2=−4.0×10−6C=−4.0µC
3
Step 2: Calculate the distance between the charges. Given that the two
charges are placed 10.0 cm apart, we convert this distance to meters:
d= 10.0cm = 0.1m
Step 3: Apply Coulomb’s law to find the electric force between the charges.
Coulomb’s law states that the magnitude of the electric force between two
charges is given by:
F=k· |q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F=(8.99 ×109N m2/C2)· |3.0×10−6C· −4.0×10−6C|
(0.1m)2
F=8.99 ×109·12.0×10−12
0.01
F=107.88 ×10−3
0.01 = 10.788 N
Therefore, the magnitude of the electric force between the charges is 10.788
N.
Question 4
Question
Three point charges are arranged in an equilateral triangle. The charge at each
corner of the triangle is +q. Determine the magnitude and direction of the net
force on each charge.
Solution
Let’s denote the charges as q1,q2, and q3, each with magnitude +q.
Step 1: Calculate the magnitude of the force on q1due to q2. The force
F21 on q1due to q2is given by Coulomb’s Law:
F21 =k· |q1|·|q2|
r2
21
Since the charges are +q, the force will be repulsive.
Step 2: Calculate the magnitude of the force on q1due to q3. The force
F31 on q1due to q3is given by Coulomb’s Law:
F31 =k· |q1|·|q3|
r2
31
4
Since the charges are +q, the force will be repulsive.
Step 3: Resolve the forces on q1into components to determine the net force.
Since the forces are acting at 120◦angles to each other, the superposition of the
forces will yield a net force of magnitude:
Fnet1=qF2
21 +F2
31 + 2 ·F21 ·F31 ·cos(120◦)
Step 4: Determine the direction of the net force on q1. The direction of the
net force on q1can be found using trigonometry.
Step 5: Repeat the above steps to determine the net forces on q2and q3.
The net forces on q2and q3can be calculated in a similar manner, accounting
for the forces between all pairs of charges.
By analyzing the symmetry of the problem, one can determine that the net
forces will be along the lines joining the charges, and the magnitudes will be
equal.
Question 5
Question
Consider three point charges placed at the corners of an equilateral triangle
of side length d. The charges are as follows: q1=−3µC,q2= 5µC, and
q3=−2µC. Calculate the net electric force acting on q3due to q1and q2.
Solution
Step 1: Find the distance between q1and q3(and q2and q3). The distance
between any two adjacent charges in an equilateral triangle can be calculated
using the Pythagorean theorem. For an equilateral triangle of side length d,
drawing an altitude from one of the vertices to the opposite side forms a right
triangle with hypotenuse of length d and base of length d/2.
Step 2: Calculate the net force acting on q3due to q1. The force F13 between
q1and q3is given by Coulomb’s law:
F13 =k|q1||q3|
r2
13
where kis the Coulomb constant and r13 is the distance between q1and q3.
Step 3: Calculate the force acting on q3due to q2. The force F23 between
q2and q3is also given by Coulomb’s law:
F23 =k|q2||q3|
r2
23
where r23 is the distance between q2and q3.
Step 4: Determine the net force acting on q3. To find the net force acting on
q3, we need to consider the vector sum of F13 and F23. Remember to account
for the direction of each force.
5
Step 5: Perform the calculations. Substitute the given values for the charges
and distances into the expressions for F13 and F23 , taking into account the signs
of the charges. Then, find the net force by adding the two forces vectorially.
Question 6
Question
Two charges, q1=−3µC and q2= 5µC, are placed 2 meters apart. Calculate
the magnitude of the electric force between these two charges.
Solution
Step 1: Recall that the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into Coulomb’s Law:
F= 8.99 ×109|(−3µC)(5µC)|
(2 m)2
Step 3: Calculate the magnitudes of the charges in coulombs:
q1=−3µC =−3×10−6C
q2= 5µC = 5 ×10−6C
Step 4: Substitute the magnitudes of the charges into the formula:
F= 8.99 ×109|(−3×10−6C)(5 ×10−6C)|
(2 m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×10915 ×10−12C2
4m2
F= 8.99 ×109×3.75 ×10−12 N
F= 33.7125 ×10−3N
F= 3.37125 ×10−2N
Step 6: Thus, the magnitude of the electric force between the two charges is
3.37125 ×10−2N.
6
Question 7
Question
Two point charges, Q1= 5 ×10−6C and Q2=−3×10−6C, are placed 8 meters
apart in a vacuum. Calculate the magnitude of the electric force between the
charges.
Solution
Step 1: Calculate the electric force between the charges using Coulomb’s Law
formula:
F=k
Q1·Q2
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2), Q1and Q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into the formula and solve for the force:
F= 8.99 ×109
(5 ×10−6)·(−3×10−6)
82
F= 8.99 ×109
−15 ×10−12
64
F= 8.99 ×109
−0.23438 ×10−12
F= 8.99 ×109×0.23438 ×10−12
F= 2.11 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.11 ×
10−3N.
Question 8
Question
Two electrons are located 1.5 nm apart. Calculate the electric force between
them. The charge of an electron is −1.6×10−19 C.
Solution
Step 1: Calculate the electric force using Coulomb’s Law: The equation for the
electric force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
7
where Fis the electric force, kis Coulomb’s constant (8.99 ×109N m2/C2),
q1and q2are the charges of the two particles, and ris the separation distance
between the charges.
Step 2: Substitute the given values into the equation: Given that the charge
of an electron is −1.6×10−19 C, the charges of the two electrons are q1=q2=
−1.6×10−19 C. The separation distance ris 1.5 nm = 1.5×10−9m.
Step 3: Calculate the electric force: Plugging in the values, we get:
F= (8.99 ×109)|(−1.6×10−19)(−1.6×10−19)|
(1.5×10−9)2
Step 4: Simplify and calculate the force:
F= (8.99 ×109)(2.56 ×10−38)
2.25 ×10−18
F= (8.99 ×109)×1.138 ×10−20
F= 1.023 ×10−10 N
Therefore, the electric force between the two electrons is 1.023 ×10−10 N.
Question 9
Question
Three point charges are arranged as shown in the diagram below. The charges
are q1=−3µC, q2= 5 µC, and q3=−2µC. The distance rbetween q1and q2
is 4meters, and the distance between q1and q3is 3meters. Calculate the total
electric force on q1due to q2and q3.
q1q2
q3
Solution
Step 1: Calculate the force between q1and q2: The electric force between two
point charges is given by Coulomb’s law:
F12 =k· |q1·q2|
r2
8
where k= 8.99×109N m2/C2is the Coulomb constant, q1=−3µC, q2= 5 µC,
and r= 4 m. Plugging in the values, we get:
F12 =8.99 ×109·|−3·5|
42= 1.1246 ×10−7N
Step 2: Calculate the force between q1and q3: Using the same formula as
above with r= 3 m, we have:
F13 =k· |q1·q3|
r2=8.99 ×109·|−3·(−2)|
32= 1.1988 ×10−7N
Step 3: Calculate the total electric force on q1: The total force on q1is the
vector sum of the forces F12 and F13. Since the forces are in opposite directions,
we subtract them:
Ftotal =F12 −F13 = (1.1246 ×10−7)−(1.1988 ×10−7) = −7.42 ×10−9N
Thus, the total electric force on q1due to q2and q3is −7.42 ×10−9N,
directed towards q3.
Question 10
Question
Three point charges are arranged at the vertices of an equilateral triangle with
sides of length d= 1.00 m. The charges are q1= +5.00µC, q2=−2.00µC, and
q3=−3.00µC. Calculate the magnitude and direction of the net electric force
acting on q1due to the other two charges.
Solution
Step 1: Calculate the electric force on q1due to q2. The electric force between
two point charges q1and q2is given by Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
where - kis the Coulomb constant (8.99 ×109N m2/C2), - |q1|and |q2|are
the magnitudes of the charges, - ris the distance between the charges.
Given q1= +5.00µC, q2=−2.00µC, and d= 1.00 m, the distance between
q1and q2(and also between all charges) is d.
Therefore,
F12 =(8.99 ×109N m2/C2)·(5.00 ×10−6C)·(2.00 ×10−6C)
(1.00 m)2
F12 = 26.97 ×10−3N(repulsive, to the right)
9
Step 2: Calculate the electric force on q1due to q3. Following a similar
calculation as above, the electric force between q1and q3is found to be:
F13 = 44.94×10−3N(at an angle of 120 degrees counterclockwise from horizontal)
Step 3: Find the net force on q1. The net force acting on q1is the vector
sum of the forces due to q2and q3. Using vector addition, we have:
Net force =p(F12 +F13 cos(120◦))2+ (F13 sin(120◦))2
Net force =p(26.97 ×10−3+ 44.94 ×10−3cos(120◦))2+ (44.94 ×10−3sin(120◦))2
Net force = 64.09 ×10−3N
The direction of the net force can be found by calculating the angle it makes
with the horizontal and is given by:
θ= tan−1F13 sin(120◦)
F12 +F13 cos(120◦)
θ=−30◦
Therefore, the magnitude of the net electric force acting on q1is 64.09×10−3
N, and it is directed at an angle of 30◦below the horizontal.
Question 11
Question
Three point charges are arranged in a line as shown below:
q1=−3µC q2= 5 µC q3=−7µC
The charges are located at positions:
x1= 1.0m, x2= 0.0m, x3=−2.0m
Calculate the net electric force on the charge q3due to the other two charges.
Solution
Step 1: Calculate the electric force between q1and q3The electric force between
two charges q1and q3can be calculated using Coulomb’s Law:
F13 =k|q1q3|
r2
13
where k= 8.99 ×109Nm2/C2is the Coulomb’s constant.
10
Substitute the given values:
F13 =(8.99 ×109)(3 ×10−6)(7 ×10−6)
(3)2
F13 =8.99 ×3×7
3= 71.92 ×10−6= 7.192 ×10−5N(to the right)
Step 2: Calculate the electric force between q2and q3Similarly, the electric
force between two charges q2and q3can be calculated using Coulomb’s Law:
F23 =k|q2q3|
r2
23
Substitute the given values:
F23 =(8.99 ×109)(5 ×10−6)(7 ×10−6)
(2)2
F23 =8.99 ×5×7
4= 31.465 ×10−6= 3.1465 ×10−5N(to the left)
Step 3: Find the net electric force on q3To find the net force on q3, we need
to consider the direction and add the two forces:
Fnet =F13 +F23 = 7.192×10−5−3.1465×10−5= 4.0455×10−5N(to the right)
Therefore, the net electric force on q3due to the other two charges is 4.0455×
10−5Nto the right.
Question 12
Question
Two point charges, q1= 3 nC and q2=−5nC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Convert the charges to coulombs.
Given: q1= 3 nC = 3 ×10−9C
q2=−5nC =−5×10−9C
Step 2: Calculate the distance between the charges.
Given: r= 10 cm = 0.1m
Step 3: Calculate the electric force using Coulomb’s Law. Coulomb’s Law
states that the magnitude of the electric force between two point charges is given
by:
F=k|q1q2|
r2
11
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values:
F= (8.99 ×109)|(3 ×10−9)(−5×10−9)|
(0.1)2
Step 4: Calculate the direction of the force.
The direction of the force is along the line joining the two charges. Since the
charges have opposite signs, the force is attractive.
Step 5: Calculate the magnitude of the force.
F= (8.99 ×109)15 ×10−18
0.01
F=134.85 ×10−9
0.01
F= 13.485 ×10−6N= 13.485 µN
Therefore, the magnitude of the electric force that q1exerts on q2is 13.485 µN
and the direction is attractive.
Question 13
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are separated by a distance
of 10.0cm. Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two point charges,
we will use Coulomb’s law:
F=k
q1·q2
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the
charges, and ris the distance between the charges.
Step 1: Convert the given charges to coulombs: Given: q1= 4.0µC =
4.0×10−6Cand q2=−6.0µC =−6.0×10−6C.
Step 2: Determine the distance between the charges in meters: Given:
r= 10.0cm = 10.0×10−2m= 0.10 m.
Step 3: Calculate the electric force between the charges using Coulomb’s
law:
F= 8.99 ×109N·m2/C2×
(4.0×10−6C)·(−6.0×10−6C)
(0.10 m)2
Step 4: Solve for the electric force:
F= 8.99×109×4.0×10−6× −6.0×10−6
0.102= 8.99×109×−24 ×10−12
0.01 = 8.99×109×−2.4×10−9N
12
F=−21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 14
Question
Three point charges are arranged in an equilateral triangle as shown below.
The charges have magnitudes of +2.5µC,−3.0µC, and +4.0µC. Calculate the
electric force on the −3.0µC charge.
+2.5µC
−3.0µC +4.0µC
Solution
Step 1: We first need to calculate the electric forces each charge exerts on the
−3.0µC charge. Let’s denote the distance between the charges as r. The electric
force Fbetween two point charges q1and q2separated by a distance ris given
by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2).
Calculating the forces between the charges:
For +2.5µC and −3.0µC:
F2.5µC,−3.0µC =8.99 ×109·|2.5×10−6·(−3.0×10−6)|
r2
For +4.0µC and −3.0µC:
F4.0µC,−3.0µC =8.99 ×109·|4.0×10−6·(−3.0×10−6)|
r2
Step 2: Next, we need to find the electric force acting on the −3.0µC charge
due to the other two charges. Since the charges are arranged symmetrically in
an equilateral triangle, the magnitudes of F2.5µC,−3.0µC and F4.0µC,−3.0µC are
the same.
Therefore, the total electric force on the −3.0µC charge is:
Ftotal = 2 ·F2.5µC,−3.0µC
Step 3: Finally, calculate the total electric force on the −3.0µC charge.
Perform the necessary calculations to find the numerical value of this force.
13
Question 15
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed at points Aand B
respectively. Point Ais located at coordinates (0,0) in meters and point Bis
at coordinates (3,4). Calculate the magnitude of the electric force experienced
by q2due to q1.
Solution
Step 1: Calculate the distance between the two point charges. Given that point
Ais located at coordinates (0,0) and point Bis located at coordinates (3,4), the
distance rbetween the two point charges can be calculated using the distance
formula:
r=p(x2−x1)2+ (y2−y1)2
where x1= 0,y1= 0,x2= 3, and y2= 4. Therefore,
r=p(3 −0)2+ (4 −0)2=p32+ 42=√9 + 16 = √25 = 5 m
Step 2: Calculate the magnitude of the electric force. The magnitude of
the electric force between two point charges can be calculated using Coulomb’s
Law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant. Substitute the given
values q1=−3.0µC,q2= 5.0µC, and r= 5 m into the formula:
F= (8.99 ×109)| − 3.0×10−6×5.0×10−6|
52
F= 8.99 ×109×15 ×10−12
25 = 8.99 ×109×0.6×10−12 = 5.394 ×10−3N
Therefore, the magnitude of the electric force experienced by q2due to q1is
5.394 ×10−3N.
Question 16
Question
Two point charges, q1= 4.0µC and q2=−8.0µC, are placed 0.20 m apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
14
Solution
Given: q1= 4.0µC= 4.0×10−6Cq2=−8.0µC=−8.0×10−6Cr= 0.20 m
We know that the electric force between two charges q1and q2separated by
a distance ris given by Coulomb’s law:
F=k|q1q2|
r2
Step 1: Calculate the electric force constant k.
k=1
4πε0
The permittivity of free space ε0= 8.85 ×10−12 C2/N·m2Thus,
k=1
4π×8.85 ×10−12
Step 2: Substitute the given values into Coulomb’s law to find the magnitude
of the electric force.
F=1
4π×8.85 ×10−12 ·|4.0×10−6× −8.0×10−6|
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F=1
4π×8.85 ×10−12 ·32 ×10−12
0.04
F=1
4π×8.85 ×10−12 ·800 ×10−12
F=800
4π×8.85
F≈800
35.16
F≈22.74 N
Therefore, the magnitude of the electric force between the charges is approx-
imately 22.74 N.
Question 17
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 0.10 m apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
15
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k· |q1·q2|
r2
where k≈8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the distance between the charges.
Step 2: Plug in the given values to find the electric force:
F=(8.99 ×109N·m2/C2)·|−4.0×10−6C·8.0×10−6C|
(0.10 m)2
Step 3: Calculate the electric force:
F=8.99 ×109·4.0×10−6·8.0×10−6
0.01
F=287.68
0.01
F= 28768 N
Step 4: Since q1is negative and q2is positive, the force will be attractive.
Therefore, the electric force that q1exerts on q2is 28768 N, directed towards q1.
Question 19
Question
Three point charges are arranged on the x-axis as follows: +qat x= 0,−3qat
x= 4a, and +2qat x= 7a, where qand aare positive constants. Calculate the
net force on the +2qcharge due to the other charges.
Solution
Step 1: Calculate the force on the +2qcharge due to the +qcharge: The force
F1can be calculated using Coulomb’s Law, which states that the magnitude of
the force between two point charges is given by:
F=k·|q1·q2|
r2
where kis Coulomb’s constant, q1and q2are the charges, and ris the distance
between the charges.
The force on the +2qcharge due to the +qcharge is:
F1=k·|(+2q)(q)|
(7a−0)2
16
Step 2: Calculate the force on the +2qcharge due to the −3qcharge: The
force F2can be calculated using Coulomb’s Law as well:
F2=k·|(+2q)(−3q)|
(7a−4a)2
Step 3: Determine the direction of the net force: Since the forces due to the
+qand −3qcharges are in opposite directions, the net force on the +2qcharge
will be the difference between F1and F2:
Fnet =F1−F2
Fnet =k·2q·q
(7a)2−2q·3q
3a)2
Therefore, the net force on the +2qcharge due to the other charges can be
calculated by subtracting the force on the +2qcharge due to the −3qcharge
from the force due to the +qcharge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle with
sides of length d. Charge Qis placed at each vertex. Calculate the magnitude
of the electric force on one of the charges due to the other two charges.
Solution
Let’s denote the three charges as Q1,Q2, and Q3, with Q1being the charge we
want to calculate the force on. The distance between each charge and Q1is d.
Step 1: Calculate the electric force on Q1due to Q2. The electric force
between two charges is given by Coulomb’s Law:
F=k· |Q1|·|Q2|
r2
where kis the electrostatic constant 8.99×109N m2/C2and r=dis the distance
between the charges.
Plugging in the values, the force on Q1due to Q2is:
F1=8.99 ×109·Q·Q
d2=8.99 ×109·Q2
d2
Step 2: Calculate the direction of the force due to Q2. Since Q2is at one
vertex of an equilateral triangle, the force on Q1due to Q2will be along the
line connecting Q1and Q2. This force will have a direction opposite to that of
the line connecting Q1and Q2.
17
Step 3: Calculate the electric force on Q1due to Q3. Similarly, the force
on Q1due to Q3will be along the line connecting Q1and Q3and will have a
direction opposite to that of the line connecting Q1and Q3. Thus, the force
will be:
F2=8.99 ×109·Q2
d2
Step 4: Calculate the total electric force on Q1. Since the forces F1and F2
have the same magnitude and opposite directions, the total electric force on Q1
due to Q2and Q3is:
Ftotal =F1−F2=8.99 ×109·Q2
d2−8.99 ×109·Q2
d2= 0
Thus, the net electric force on Q1due to Q2and Q3is zero.
Question 22
Question
Two point charges, q1= +3.0µC and q2=−6.0µC, are placed 8.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units (Coulombs).
Given: q1= +3.0µC = 3.0×10−6C
q2=−6.0µC =−6.0×10−6C
Step 2: Find the distance between the charges in meters.
Given: r= 8.0cm = 8.0×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law,
F=k|q1q2|
r2.
Where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109)· |3.0×10−6· −6.0×10−6|
(8.0×10−2)2
Step 4: Calculate the magnitude of the electric force.
F=8.99 ×109·18 ×10−12
64 ×10−4
F=161.82 ×10−3
64 ×10−4
F=161.82
64 = 2.53 N
Therefore, the magnitude of the electric force between the charges is 2.53 N.
18
Question 23
Question
Two point charges Q1 = +3.0 µC and Q2 = -5.0 µC are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that Q1 exerts on
Q2.
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Iden-
tify the direction of the force by considering the signs of the charges. Step 3:
Calculate the magnitude of the electric force using Coulomb’s law.
Step 1: Convert the charges from microcoulombs to coulombs: Q1 =
3.0µC = 3.0×10−6Cand Q2 = −5.0µC =−5.0×10−6C.
Step 2: The force will be attractive since Q1 is positive and Q2 is negative.
Step 3: Use Coulomb’s Law to find the magnitude of the electric force:
F=k|Q1·Q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), ris the distance between
the charges (0.1 m), Q1 = 3.0×10−6C, and Q2 = −5.0×10−6C.
Substitute the values into the equation:
F= (8.99 ×109)|3.0×10−6· −5.0×10−6|
(0.1)2
F= (8.99 ×109)15 ×10−12
0.01
F= (8.99 ×109)×1.5×10−9
F= 13.485 ×100
Therefore, F= 13.485N
The magnitude of the electric force that Q1 exerts on Q2 is 13.485 N, directed
towards Q1.
Question 24
Question
Two point charges, q1=−4.00 µC and q2= 3.00 µC, are separated by a distance
of 10.0 cm in a vacuum. Calculate the magnitude of the electric force between
the charges.
19
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Deter-
mine the distance between the charges in meters. Step 3: Use Coulomb’s Law
to calculate the electric force between the charges.
Step 1: Convert the charges from microcoulombs to coulombs. q1=
−4.00 µC =−4.00 ×10−6C q2= 3.00 µC = 3.00 ×10−6C
Step 2: Determine the distance between the charges in meters. Given:
distance r= 10.0cm = 0.10 m
Step 3: Use Coulomb’s Law to calculate the electric force between the
charges. Coulomb’s Law: F=k|q1q2|
r2where k= 8.99 ×109Nm2/C2is the
electrostatic constant.
Substitute the known values into the equation: F=(8.99 ×109)×|−4.00 ×10−6×3.00 ×10−6|
(0.10)2
F=(8.99 ×109)×1.2×10−11
0.01
F=10.788 ×10−2
0.01
F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788
N.
Question 25
Question
Two point charges, q1=−4µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the given charges to coulombs. Recall that 1µC= 10−6C.
⇒q1=−4µC=−4×10−6C=−4×10−6C
⇒q2= 5 µC= 5 ×10−6C= 5 ×10−6C
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The formula for the magnitude of the electric force between two charges q1and
q2separated by a distance ris given by:
F=k
q1·q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Step 3: Substitute the given values into the formula.
F= 8.99 ×109
(−4×10−6)·(5 ×10−6)
(0.10)2
20
F= 8.99 ×109
−20 ×10−12
0.01
F= 8.99 ×109
−2×10−9
F= 8.99 ×109×2×10−9
F= 17.98 ×100
F= 17.98 N
Therefore, the magnitude of the electric force between the charges is 17.98
N.
21
Step 3: Find the net force. To find the net force on the charge +2q, we need
to consider the vector sum of the forces from +qand −3q. Since the forces act
along a straight line through the charge +2q, we only need to consider their
magnitudes and signs. The net force is:
Fnet =F+q,+2q−F−3q,+2q=2kq2
a2−6kq2
a2=−4kq2
a2
Therefore, the magnitude of the net electric force acting on the charge +2q
is 4kq2
a2, and its direction is attractive (towards the −3qcharge).
Question 2
Question
Three point charges are arranged in an equilateral triangle as shown below.
Calculate the magnitude and direction of the net force on the charge q1=
+2.00 µC.
q2= +4.00 µC
q1= +2.00 µCq3= +6.00 µC
Solution
Step 1: First, let’s calculate the magnitudes of the forces on q1due to q2and
q3using Coulomb’s law: F=k|q1||q2|
r2, where k= 8.99 ×109N m2/C2.
The distance between q1and q2is the length of the side of the equilateral
triangle, r=l= 1 m.
Step 2: Calculating the force on q1due to q2:
F12 =k|q1||q2|
r2
F12 = (8.99 ×109)(2.00 ×10−6)(4.00 ×10−6)
(1)2
F12 = 7.19 ×10−2N
2
Step 3: Calculating the force on q1due to q3:
F13 =k|q1||q3|
r2
Since q3is 2l= 2 m away from q1:
F13 = (8.99 ×109)(2.00 ×10−6)(6.00 ×10−6)
(2)2
F13 = 4.498 ×10−2N
Step 4: Now, let’s find the net force on q1by vectorially adding the forces
F12 and F13. Given the forces are along the same line of the equilateral triangle,
they are in opposite directions and have the same magnitude.
Net force on q1=F12 −F13 = 7.19 ×10−2−4.498 ×10−2N
Net force on q1= 2.692 ×10−2N (in the direction towards q2)
Therefore, the magnitude of the net force on q1is 2.692 ×10−2N and the
direction is towards q2.
Question 3
Question
Two point charges, q1= 3.0µC and q2=−4.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Step 2: Calculate the distance between
the charges. Step 3: Apply Coulomb’s law to find the electric force between the
charges.
Step 1: Convert the charges to coulombs. The charge q1= 3.0µC can be
converted to coulombs as follows:
q1= 3.0×10−6C= 3.0µC
Similarly, the charge q2=−4.0µC can be converted to coulombs as follows:
q2=−4.0×10−6C=−4.0µC
3
Step 2: Calculate the distance between the charges. Given that the two
charges are placed 10.0 cm apart, we convert this distance to meters:
d= 10.0cm = 0.1m
Step 3: Apply Coulomb’s law to find the electric force between the charges.
Coulomb’s law states that the magnitude of the electric force between two
charges is given by:
F=k· |q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F=(8.99 ×109N m2/C2)· |3.0×10−6C· −4.0×10−6C|
(0.1m)2
F=8.99 ×109·12.0×10−12
0.01
F=107.88 ×10−3
0.01 = 10.788 N
Therefore, the magnitude of the electric force between the charges is 10.788
N.
Question 4
Question
Three point charges are arranged in an equilateral triangle. The charge at each
corner of the triangle is +q. Determine the magnitude and direction of the net
force on each charge.
Solution
Let’s denote the charges as q1,q2, and q3, each with magnitude +q.
Step 1: Calculate the magnitude of the force on q1due to q2. The force
F21 on q1due to q2is given by Coulomb’s Law:
F21 =k· |q1|·|q2|
r2
21
Since the charges are +q, the force will be repulsive.
Step 2: Calculate the magnitude of the force on q1due to q3. The force
F31 on q1due to q3is given by Coulomb’s Law:
F31 =k· |q1|·|q3|
r2
31
4
Since the charges are +q, the force will be repulsive.
Step 3: Resolve the forces on q1into components to determine the net force.
Since the forces are acting at 120◦angles to each other, the superposition of the
forces will yield a net force of magnitude:
Fnet1=qF2
21 +F2
31 + 2 ·F21 ·F31 ·cos(120◦)
Step 4: Determine the direction of the net force on q1. The direction of the
net force on q1can be found using trigonometry.
Step 5: Repeat the above steps to determine the net forces on q2and q3.
The net forces on q2and q3can be calculated in a similar manner, accounting
for the forces between all pairs of charges.
By analyzing the symmetry of the problem, one can determine that the net
forces will be along the lines joining the charges, and the magnitudes will be
equal.
Question 5
Question
Consider three point charges placed at the corners of an equilateral triangle
of side length d. The charges are as follows: q1=−3µC,q2= 5µC, and
q3=−2µC. Calculate the net electric force acting on q3due to q1and q2.
Solution
Step 1: Find the distance between q1and q3(and q2and q3). The distance
between any two adjacent charges in an equilateral triangle can be calculated
using the Pythagorean theorem. For an equilateral triangle of side length d,
drawing an altitude from one of the vertices to the opposite side forms a right
triangle with hypotenuse of length d and base of length d/2.
Step 2: Calculate the net force acting on q3due to q1. The force F13 between
q1and q3is given by Coulomb’s law:
F13 =k|q1||q3|
r2
13
where kis the Coulomb constant and r13 is the distance between q1and q3.
Step 3: Calculate the force acting on q3due to q2. The force F23 between
q2and q3is also given by Coulomb’s law:
F23 =k|q2||q3|
r2
23
where r23 is the distance between q2and q3.
Step 4: Determine the net force acting on q3. To find the net force acting on
q3, we need to consider the vector sum of F13 and F23. Remember to account
for the direction of each force.
5
Step 5: Perform the calculations. Substitute the given values for the charges
and distances into the expressions for F13 and F23 , taking into account the signs
of the charges. Then, find the net force by adding the two forces vectorially.
Question 6
Question
Two charges, q1=−3µC and q2= 5µC, are placed 2 meters apart. Calculate
the magnitude of the electric force between these two charges.
Solution
Step 1: Recall that the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into Coulomb’s Law:
F= 8.99 ×109|(−3µC)(5µC)|
(2 m)2
Step 3: Calculate the magnitudes of the charges in coulombs:
q1=−3µC =−3×10−6C
q2= 5µC = 5 ×10−6C
Step 4: Substitute the magnitudes of the charges into the formula:
F= 8.99 ×109|(−3×10−6C)(5 ×10−6C)|
(2 m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×10915 ×10−12C2
4m2
F= 8.99 ×109×3.75 ×10−12 N
F= 33.7125 ×10−3N
F= 3.37125 ×10−2N
Step 6: Thus, the magnitude of the electric force between the two charges is
3.37125 ×10−2N.
6
Question 7
Question
Two point charges, Q1= 5 ×10−6C and Q2=−3×10−6C, are placed 8 meters
apart in a vacuum. Calculate the magnitude of the electric force between the
charges.
Solution
Step 1: Calculate the electric force between the charges using Coulomb’s Law
formula:
F=k
Q1·Q2
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2), Q1and Q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into the formula and solve for the force:
F= 8.99 ×109
(5 ×10−6)·(−3×10−6)
82
F= 8.99 ×109
−15 ×10−12
64
F= 8.99 ×109
−0.23438 ×10−12
F= 8.99 ×109×0.23438 ×10−12
F= 2.11 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.11 ×
10−3N.
Question 8
Question
Two electrons are located 1.5 nm apart. Calculate the electric force between
them. The charge of an electron is −1.6×10−19 C.
Solution
Step 1: Calculate the electric force using Coulomb’s Law: The equation for the
electric force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
7
where Fis the electric force, kis Coulomb’s constant (8.99 ×109N m2/C2),
q1and q2are the charges of the two particles, and ris the separation distance
between the charges.
Step 2: Substitute the given values into the equation: Given that the charge
of an electron is −1.6×10−19 C, the charges of the two electrons are q1=q2=
−1.6×10−19 C. The separation distance ris 1.5 nm = 1.5×10−9m.
Step 3: Calculate the electric force: Plugging in the values, we get:
F= (8.99 ×109)|(−1.6×10−19)(−1.6×10−19)|
(1.5×10−9)2
Step 4: Simplify and calculate the force:
F= (8.99 ×109)(2.56 ×10−38)
2.25 ×10−18
F= (8.99 ×109)×1.138 ×10−20
F= 1.023 ×10−10 N
Therefore, the electric force between the two electrons is 1.023 ×10−10 N.
Question 9
Question
Three point charges are arranged as shown in the diagram below. The charges
are q1=−3µC, q2= 5 µC, and q3=−2µC. The distance rbetween q1and q2
is 4meters, and the distance between q1and q3is 3meters. Calculate the total
electric force on q1due to q2and q3.
q1q2
q3
Solution
Step 1: Calculate the force between q1and q2: The electric force between two
point charges is given by Coulomb’s law:
F12 =k· |q1·q2|
r2
8
where k= 8.99×109N m2/C2is the Coulomb constant, q1=−3µC, q2= 5 µC,
and r= 4 m. Plugging in the values, we get:
F12 =8.99 ×109·|−3·5|
42= 1.1246 ×10−7N
Step 2: Calculate the force between q1and q3: Using the same formula as
above with r= 3 m, we have:
F13 =k· |q1·q3|
r2=8.99 ×109·|−3·(−2)|
32= 1.1988 ×10−7N
Step 3: Calculate the total electric force on q1: The total force on q1is the
vector sum of the forces F12 and F13. Since the forces are in opposite directions,
we subtract them:
Ftotal =F12 −F13 = (1.1246 ×10−7)−(1.1988 ×10−7) = −7.42 ×10−9N
Thus, the total electric force on q1due to q2and q3is −7.42 ×10−9N,
directed towards q3.
Question 10
Question
Three point charges are arranged at the vertices of an equilateral triangle with
sides of length d= 1.00 m. The charges are q1= +5.00µC, q2=−2.00µC, and
q3=−3.00µC. Calculate the magnitude and direction of the net electric force
acting on q1due to the other two charges.
Solution
Step 1: Calculate the electric force on q1due to q2. The electric force between
two point charges q1and q2is given by Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
where - kis the Coulomb constant (8.99 ×109N m2/C2), - |q1|and |q2|are
the magnitudes of the charges, - ris the distance between the charges.
Given q1= +5.00µC, q2=−2.00µC, and d= 1.00 m, the distance between
q1and q2(and also between all charges) is d.
Therefore,
F12 =(8.99 ×109N m2/C2)·(5.00 ×10−6C)·(2.00 ×10−6C)
(1.00 m)2
F12 = 26.97 ×10−3N(repulsive, to the right)
9
Step 2: Calculate the electric force on q1due to q3. Following a similar
calculation as above, the electric force between q1and q3is found to be:
F13 = 44.94×10−3N(at an angle of 120 degrees counterclockwise from horizontal)
Step 3: Find the net force on q1. The net force acting on q1is the vector
sum of the forces due to q2and q3. Using vector addition, we have:
Net force =p(F12 +F13 cos(120◦))2+ (F13 sin(120◦))2
Net force =p(26.97 ×10−3+ 44.94 ×10−3cos(120◦))2+ (44.94 ×10−3sin(120◦))2
Net force = 64.09 ×10−3N
The direction of the net force can be found by calculating the angle it makes
with the horizontal and is given by:
θ= tan−1F13 sin(120◦)
F12 +F13 cos(120◦)
θ=−30◦
Therefore, the magnitude of the net electric force acting on q1is 64.09×10−3
N, and it is directed at an angle of 30◦below the horizontal.
Question 11
Question
Three point charges are arranged in a line as shown below:
q1=−3µC q2= 5 µC q3=−7µC
The charges are located at positions:
x1= 1.0m, x2= 0.0m, x3=−2.0m
Calculate the net electric force on the charge q3due to the other two charges.
Solution
Step 1: Calculate the electric force between q1and q3The electric force between
two charges q1and q3can be calculated using Coulomb’s Law:
F13 =k|q1q3|
r2
13
where k= 8.99 ×109Nm2/C2is the Coulomb’s constant.
10
Substitute the given values:
F13 =(8.99 ×109)(3 ×10−6)(7 ×10−6)
(3)2
F13 =8.99 ×3×7
3= 71.92 ×10−6= 7.192 ×10−5N(to the right)
Step 2: Calculate the electric force between q2and q3Similarly, the electric
force between two charges q2and q3can be calculated using Coulomb’s Law:
F23 =k|q2q3|
r2
23
Substitute the given values:
F23 =(8.99 ×109)(5 ×10−6)(7 ×10−6)
(2)2
F23 =8.99 ×5×7
4= 31.465 ×10−6= 3.1465 ×10−5N(to the left)
Step 3: Find the net electric force on q3To find the net force on q3, we need
to consider the direction and add the two forces:
Fnet =F13 +F23 = 7.192×10−5−3.1465×10−5= 4.0455×10−5N(to the right)
Therefore, the net electric force on q3due to the other two charges is 4.0455×
10−5Nto the right.
Question 12
Question
Two point charges, q1= 3 nC and q2=−5nC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Convert the charges to coulombs.
Given: q1= 3 nC = 3 ×10−9C
q2=−5nC =−5×10−9C
Step 2: Calculate the distance between the charges.
Given: r= 10 cm = 0.1m
Step 3: Calculate the electric force using Coulomb’s Law. Coulomb’s Law
states that the magnitude of the electric force between two point charges is given
by:
F=k|q1q2|
r2
11
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values:
F= (8.99 ×109)|(3 ×10−9)(−5×10−9)|
(0.1)2
Step 4: Calculate the direction of the force.
The direction of the force is along the line joining the two charges. Since the
charges have opposite signs, the force is attractive.
Step 5: Calculate the magnitude of the force.
F= (8.99 ×109)15 ×10−18
0.01
F=134.85 ×10−9
0.01
F= 13.485 ×10−6N= 13.485 µN
Therefore, the magnitude of the electric force that q1exerts on q2is 13.485 µN
and the direction is attractive.
Question 13
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are separated by a distance
of 10.0cm. Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two point charges,
we will use Coulomb’s law:
F=k
q1·q2
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the
charges, and ris the distance between the charges.
Step 1: Convert the given charges to coulombs: Given: q1= 4.0µC =
4.0×10−6Cand q2=−6.0µC =−6.0×10−6C.
Step 2: Determine the distance between the charges in meters: Given:
r= 10.0cm = 10.0×10−2m= 0.10 m.
Step 3: Calculate the electric force between the charges using Coulomb’s
law:
F= 8.99 ×109N·m2/C2×
(4.0×10−6C)·(−6.0×10−6C)
(0.10 m)2
Step 4: Solve for the electric force:
F= 8.99×109×4.0×10−6× −6.0×10−6
0.102= 8.99×109×−24 ×10−12
0.01 = 8.99×109×−2.4×10−9N
12
F=−21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 14
Question
Three point charges are arranged in an equilateral triangle as shown below.
The charges have magnitudes of +2.5µC,−3.0µC, and +4.0µC. Calculate the
electric force on the −3.0µC charge.
+2.5µC
−3.0µC +4.0µC
Solution
Step 1: We first need to calculate the electric forces each charge exerts on the
−3.0µC charge. Let’s denote the distance between the charges as r. The electric
force Fbetween two point charges q1and q2separated by a distance ris given
by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2).
Calculating the forces between the charges:
For +2.5µC and −3.0µC:
F2.5µC,−3.0µC =8.99 ×109·|2.5×10−6·(−3.0×10−6)|
r2
For +4.0µC and −3.0µC:
F4.0µC,−3.0µC =8.99 ×109·|4.0×10−6·(−3.0×10−6)|
r2
Step 2: Next, we need to find the electric force acting on the −3.0µC charge
due to the other two charges. Since the charges are arranged symmetrically in
an equilateral triangle, the magnitudes of F2.5µC,−3.0µC and F4.0µC,−3.0µC are
the same.
Therefore, the total electric force on the −3.0µC charge is:
Ftotal = 2 ·F2.5µC,−3.0µC
Step 3: Finally, calculate the total electric force on the −3.0µC charge.
Perform the necessary calculations to find the numerical value of this force.
13
Question 15
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed at points Aand B
respectively. Point Ais located at coordinates (0,0) in meters and point Bis
at coordinates (3,4). Calculate the magnitude of the electric force experienced
by q2due to q1.
Solution
Step 1: Calculate the distance between the two point charges. Given that point
Ais located at coordinates (0,0) and point Bis located at coordinates (3,4), the
distance rbetween the two point charges can be calculated using the distance
formula:
r=p(x2−x1)2+ (y2−y1)2
where x1= 0,y1= 0,x2= 3, and y2= 4. Therefore,
r=p(3 −0)2+ (4 −0)2=p32+ 42=√9 + 16 = √25 = 5 m
Step 2: Calculate the magnitude of the electric force. The magnitude of
the electric force between two point charges can be calculated using Coulomb’s
Law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant. Substitute the given
values q1=−3.0µC,q2= 5.0µC, and r= 5 m into the formula:
F= (8.99 ×109)| − 3.0×10−6×5.0×10−6|
52
F= 8.99 ×109×15 ×10−12
25 = 8.99 ×109×0.6×10−12 = 5.394 ×10−3N
Therefore, the magnitude of the electric force experienced by q2due to q1is
5.394 ×10−3N.
Question 16
Question
Two point charges, q1= 4.0µC and q2=−8.0µC, are placed 0.20 m apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
14
Solution
Given: q1= 4.0µC= 4.0×10−6Cq2=−8.0µC=−8.0×10−6Cr= 0.20 m
We know that the electric force between two charges q1and q2separated by
a distance ris given by Coulomb’s law:
F=k|q1q2|
r2
Step 1: Calculate the electric force constant k.
k=1
4πε0
The permittivity of free space ε0= 8.85 ×10−12 C2/N·m2Thus,
k=1
4π×8.85 ×10−12
Step 2: Substitute the given values into Coulomb’s law to find the magnitude
of the electric force.
F=1
4π×8.85 ×10−12 ·|4.0×10−6× −8.0×10−6|
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F=1
4π×8.85 ×10−12 ·32 ×10−12
0.04
F=1
4π×8.85 ×10−12 ·800 ×10−12
F=800
4π×8.85
F≈800
35.16
F≈22.74 N
Therefore, the magnitude of the electric force between the charges is approx-
imately 22.74 N.
Question 17
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 0.10 m apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
15
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k· |q1·q2|
r2
where k≈8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the distance between the charges.
Step 2: Plug in the given values to find the electric force:
F=(8.99 ×109N·m2/C2)·|−4.0×10−6C·8.0×10−6C|
(0.10 m)2
Step 3: Calculate the electric force:
F=8.99 ×109·4.0×10−6·8.0×10−6
0.01
F=287.68
0.01
F= 28768 N
Step 4: Since q1is negative and q2is positive, the force will be attractive.
Therefore, the electric force that q1exerts on q2is 28768 N, directed towards q1.
Question 19
Question
Three point charges are arranged on the x-axis as follows: +qat x= 0,−3qat
x= 4a, and +2qat x= 7a, where qand aare positive constants. Calculate the
net force on the +2qcharge due to the other charges.
Solution
Step 1: Calculate the force on the +2qcharge due to the +qcharge: The force
F1can be calculated using Coulomb’s Law, which states that the magnitude of
the force between two point charges is given by:
F=k·|q1·q2|
r2
where kis Coulomb’s constant, q1and q2are the charges, and ris the distance
between the charges.
The force on the +2qcharge due to the +qcharge is:
F1=k·|(+2q)(q)|
(7a−0)2
16
Step 2: Calculate the force on the +2qcharge due to the −3qcharge: The
force F2can be calculated using Coulomb’s Law as well:
F2=k·|(+2q)(−3q)|
(7a−4a)2
Step 3: Determine the direction of the net force: Since the forces due to the
+qand −3qcharges are in opposite directions, the net force on the +2qcharge
will be the difference between F1and F2:
Fnet =F1−F2
Fnet =k·2q·q
(7a)2−2q·3q
3a)2
Therefore, the net force on the +2qcharge due to the other charges can be
calculated by subtracting the force on the +2qcharge due to the −3qcharge
from the force due to the +qcharge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle with
sides of length d. Charge Qis placed at each vertex. Calculate the magnitude
of the electric force on one of the charges due to the other two charges.
Solution
Let’s denote the three charges as Q1,Q2, and Q3, with Q1being the charge we
want to calculate the force on. The distance between each charge and Q1is d.
Step 1: Calculate the electric force on Q1due to Q2. The electric force
between two charges is given by Coulomb’s Law:
F=k· |Q1|·|Q2|
r2
where kis the electrostatic constant 8.99×109N m2/C2and r=dis the distance
between the charges.
Plugging in the values, the force on Q1due to Q2is:
F1=8.99 ×109·Q·Q
d2=8.99 ×109·Q2
d2
Step 2: Calculate the direction of the force due to Q2. Since Q2is at one
vertex of an equilateral triangle, the force on Q1due to Q2will be along the
line connecting Q1and Q2. This force will have a direction opposite to that of
the line connecting Q1and Q2.
17
Step 3: Calculate the electric force on Q1due to Q3. Similarly, the force
on Q1due to Q3will be along the line connecting Q1and Q3and will have a
direction opposite to that of the line connecting Q1and Q3. Thus, the force
will be:
F2=8.99 ×109·Q2
d2
Step 4: Calculate the total electric force on Q1. Since the forces F1and F2
have the same magnitude and opposite directions, the total electric force on Q1
due to Q2and Q3is:
Ftotal =F1−F2=8.99 ×109·Q2
d2−8.99 ×109·Q2
d2= 0
Thus, the net electric force on Q1due to Q2and Q3is zero.
Question 22
Question
Two point charges, q1= +3.0µC and q2=−6.0µC, are placed 8.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units (Coulombs).
Given: q1= +3.0µC = 3.0×10−6C
q2=−6.0µC =−6.0×10−6C
Step 2: Find the distance between the charges in meters.
Given: r= 8.0cm = 8.0×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law,
F=k|q1q2|
r2.
Where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109)· |3.0×10−6· −6.0×10−6|
(8.0×10−2)2
Step 4: Calculate the magnitude of the electric force.
F=8.99 ×109·18 ×10−12
64 ×10−4
F=161.82 ×10−3
64 ×10−4
F=161.82
64 = 2.53 N
Therefore, the magnitude of the electric force between the charges is 2.53 N.
18
Question 23
Question
Two point charges Q1 = +3.0 µC and Q2 = -5.0 µC are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that Q1 exerts on
Q2.
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Iden-
tify the direction of the force by considering the signs of the charges. Step 3:
Calculate the magnitude of the electric force using Coulomb’s law.
Step 1: Convert the charges from microcoulombs to coulombs: Q1 =
3.0µC = 3.0×10−6Cand Q2 = −5.0µC =−5.0×10−6C.
Step 2: The force will be attractive since Q1 is positive and Q2 is negative.
Step 3: Use Coulomb’s Law to find the magnitude of the electric force:
F=k|Q1·Q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), ris the distance between
the charges (0.1 m), Q1 = 3.0×10−6C, and Q2 = −5.0×10−6C.
Substitute the values into the equation:
F= (8.99 ×109)|3.0×10−6· −5.0×10−6|
(0.1)2
F= (8.99 ×109)15 ×10−12
0.01
F= (8.99 ×109)×1.5×10−9
F= 13.485 ×100
Therefore, F= 13.485N
The magnitude of the electric force that Q1 exerts on Q2 is 13.485 N, directed
towards Q1.
Question 24
Question
Two point charges, q1=−4.00 µC and q2= 3.00 µC, are separated by a distance
of 10.0 cm in a vacuum. Calculate the magnitude of the electric force between
the charges.
19
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Deter-
mine the distance between the charges in meters. Step 3: Use Coulomb’s Law
to calculate the electric force between the charges.
Step 1: Convert the charges from microcoulombs to coulombs. q1=
−4.00 µC =−4.00 ×10−6C q2= 3.00 µC = 3.00 ×10−6C
Step 2: Determine the distance between the charges in meters. Given:
distance r= 10.0cm = 0.10 m
Step 3: Use Coulomb’s Law to calculate the electric force between the
charges. Coulomb’s Law: F=k|q1q2|
r2where k= 8.99 ×109Nm2/C2is the
electrostatic constant.
Substitute the known values into the equation: F=(8.99 ×109)×|−4.00 ×10−6×3.00 ×10−6|
(0.10)2
F=(8.99 ×109)×1.2×10−11
0.01
F=10.788 ×10−2
0.01
F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788
N.
Question 25
Question
Two point charges, q1=−4µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the given charges to coulombs. Recall that 1µC= 10−6C.
⇒q1=−4µC=−4×10−6C=−4×10−6C
⇒q2= 5 µC= 5 ×10−6C= 5 ×10−6C
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The formula for the magnitude of the electric force between two charges q1and
q2separated by a distance ris given by:
F=k
q1·q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Step 3: Substitute the given values into the formula.
F= 8.99 ×109
(−4×10−6)·(5 ×10−6)
(0.10)2
20
F= 8.99 ×109
−20 ×10−12
0.01
F= 8.99 ×109
−2×10−9
F= 8.99 ×109×2×10−9
F= 17.98 ×100
F= 17.98 N
Therefore, the magnitude of the electric force between the charges is 17.98
N.
21
Step 3: Find the net force. To find the net force on the charge +2q, we need
to consider the vector sum of the forces from +qand −3q. Since the forces act
along a straight line through the charge +2q, we only need to consider their
magnitudes and signs. The net force is:
Fnet =F+q,+2q−F−3q,+2q=2kq2
a2−6kq2
a2=−4kq2
a2
Therefore, the magnitude of the net electric force acting on the charge +2q
is 4kq2
a2, and its direction is attractive (towards the −3qcharge).
Question 2
Question
Three point charges are arranged in an equilateral triangle as shown below.
Calculate the magnitude and direction of the net force on the charge q1=
+2.00 µC.
q2= +4.00 µC
q1= +2.00 µCq3= +6.00 µC
Solution
Step 1: First, let’s calculate the magnitudes of the forces on q1due to q2and
q3using Coulomb’s law: F=k|q1||q2|
r2, where k= 8.99 ×109N m2/C2.
The distance between q1and q2is the length of the side of the equilateral
triangle, r=l= 1 m.
Step 2: Calculating the force on q1due to q2:
F12 =k|q1||q2|
r2
F12 = (8.99 ×109)(2.00 ×10−6)(4.00 ×10−6)
(1)2
F12 = 7.19 ×10−2N
2
Step 3: Calculating the force on q1due to q3:
F13 =k|q1||q3|
r2
Since q3is 2l= 2 m away from q1:
F13 = (8.99 ×109)(2.00 ×10−6)(6.00 ×10−6)
(2)2
F13 = 4.498 ×10−2N
Step 4: Now, let’s find the net force on q1by vectorially adding the forces
F12 and F13. Given the forces are along the same line of the equilateral triangle,
they are in opposite directions and have the same magnitude.
Net force on q1=F12 −F13 = 7.19 ×10−2−4.498 ×10−2N
Net force on q1= 2.692 ×10−2N (in the direction towards q2)
Therefore, the magnitude of the net force on q1is 2.692 ×10−2N and the
direction is towards q2.
Question 3
Question
Two point charges, q1= 3.0µC and q2=−4.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Step 2: Calculate the distance between
the charges. Step 3: Apply Coulomb’s law to find the electric force between the
charges.
Step 1: Convert the charges to coulombs. The charge q1= 3.0µC can be
converted to coulombs as follows:
q1= 3.0×10−6C= 3.0µC
Similarly, the charge q2=−4.0µC can be converted to coulombs as follows:
q2=−4.0×10−6C=−4.0µC
3
Step 2: Calculate the distance between the charges. Given that the two
charges are placed 10.0 cm apart, we convert this distance to meters:
d= 10.0cm = 0.1m
Step 3: Apply Coulomb’s law to find the electric force between the charges.
Coulomb’s law states that the magnitude of the electric force between two
charges is given by:
F=k· |q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F=(8.99 ×109N m2/C2)· |3.0×10−6C· −4.0×10−6C|
(0.1m)2
F=8.99 ×109·12.0×10−12
0.01
F=107.88 ×10−3
0.01 = 10.788 N
Therefore, the magnitude of the electric force between the charges is 10.788
N.
Question 4
Question
Three point charges are arranged in an equilateral triangle. The charge at each
corner of the triangle is +q. Determine the magnitude and direction of the net
force on each charge.
Solution
Let’s denote the charges as q1,q2, and q3, each with magnitude +q.
Step 1: Calculate the magnitude of the force on q1due to q2. The force
F21 on q1due to q2is given by Coulomb’s Law:
F21 =k· |q1|·|q2|
r2
21
Since the charges are +q, the force will be repulsive.
Step 2: Calculate the magnitude of the force on q1due to q3. The force
F31 on q1due to q3is given by Coulomb’s Law:
F31 =k· |q1|·|q3|
r2
31
4
Since the charges are +q, the force will be repulsive.
Step 3: Resolve the forces on q1into components to determine the net force.
Since the forces are acting at 120◦angles to each other, the superposition of the
forces will yield a net force of magnitude:
Fnet1=qF2
21 +F2
31 + 2 ·F21 ·F31 ·cos(120◦)
Step 4: Determine the direction of the net force on q1. The direction of the
net force on q1can be found using trigonometry.
Step 5: Repeat the above steps to determine the net forces on q2and q3.
The net forces on q2and q3can be calculated in a similar manner, accounting
for the forces between all pairs of charges.
By analyzing the symmetry of the problem, one can determine that the net
forces will be along the lines joining the charges, and the magnitudes will be
equal.
Question 5
Question
Consider three point charges placed at the corners of an equilateral triangle
of side length d. The charges are as follows: q1=−3µC,q2= 5µC, and
q3=−2µC. Calculate the net electric force acting on q3due to q1and q2.
Solution
Step 1: Find the distance between q1and q3(and q2and q3). The distance
between any two adjacent charges in an equilateral triangle can be calculated
using the Pythagorean theorem. For an equilateral triangle of side length d,
drawing an altitude from one of the vertices to the opposite side forms a right
triangle with hypotenuse of length d and base of length d/2.
Step 2: Calculate the net force acting on q3due to q1. The force F13 between
q1and q3is given by Coulomb’s law:
F13 =k|q1||q3|
r2
13
where kis the Coulomb constant and r13 is the distance between q1and q3.
Step 3: Calculate the force acting on q3due to q2. The force F23 between
q2and q3is also given by Coulomb’s law:
F23 =k|q2||q3|
r2
23
where r23 is the distance between q2and q3.
Step 4: Determine the net force acting on q3. To find the net force acting on
q3, we need to consider the vector sum of F13 and F23. Remember to account
for the direction of each force.
5
Step 5: Perform the calculations. Substitute the given values for the charges
and distances into the expressions for F13 and F23 , taking into account the signs
of the charges. Then, find the net force by adding the two forces vectorially.
Question 6
Question
Two charges, q1=−3µC and q2= 5µC, are placed 2 meters apart. Calculate
the magnitude of the electric force between these two charges.
Solution
Step 1: Recall that the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into Coulomb’s Law:
F= 8.99 ×109|(−3µC)(5µC)|
(2 m)2
Step 3: Calculate the magnitudes of the charges in coulombs:
q1=−3µC =−3×10−6C
q2= 5µC = 5 ×10−6C
Step 4: Substitute the magnitudes of the charges into the formula:
F= 8.99 ×109|(−3×10−6C)(5 ×10−6C)|
(2 m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×10915 ×10−12C2
4m2
F= 8.99 ×109×3.75 ×10−12 N
F= 33.7125 ×10−3N
F= 3.37125 ×10−2N
Step 6: Thus, the magnitude of the electric force between the two charges is
3.37125 ×10−2N.
6
Question 7
Question
Two point charges, Q1= 5 ×10−6C and Q2=−3×10−6C, are placed 8 meters
apart in a vacuum. Calculate the magnitude of the electric force between the
charges.
Solution
Step 1: Calculate the electric force between the charges using Coulomb’s Law
formula:
F=k
Q1·Q2
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2), Q1and Q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into the formula and solve for the force:
F= 8.99 ×109
(5 ×10−6)·(−3×10−6)
82
F= 8.99 ×109
−15 ×10−12
64
F= 8.99 ×109
−0.23438 ×10−12
F= 8.99 ×109×0.23438 ×10−12
F= 2.11 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.11 ×
10−3N.
Question 8
Question
Two electrons are located 1.5 nm apart. Calculate the electric force between
them. The charge of an electron is −1.6×10−19 C.
Solution
Step 1: Calculate the electric force using Coulomb’s Law: The equation for the
electric force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
7
where Fis the electric force, kis Coulomb’s constant (8.99 ×109N m2/C2),
q1and q2are the charges of the two particles, and ris the separation distance
between the charges.
Step 2: Substitute the given values into the equation: Given that the charge
of an electron is −1.6×10−19 C, the charges of the two electrons are q1=q2=
−1.6×10−19 C. The separation distance ris 1.5 nm = 1.5×10−9m.
Step 3: Calculate the electric force: Plugging in the values, we get:
F= (8.99 ×109)|(−1.6×10−19)(−1.6×10−19)|
(1.5×10−9)2
Step 4: Simplify and calculate the force:
F= (8.99 ×109)(2.56 ×10−38)
2.25 ×10−18
F= (8.99 ×109)×1.138 ×10−20
F= 1.023 ×10−10 N
Therefore, the electric force between the two electrons is 1.023 ×10−10 N.
Question 9
Question
Three point charges are arranged as shown in the diagram below. The charges
are q1=−3µC, q2= 5 µC, and q3=−2µC. The distance rbetween q1and q2
is 4meters, and the distance between q1and q3is 3meters. Calculate the total
electric force on q1due to q2and q3.
q1q2
q3
Solution
Step 1: Calculate the force between q1and q2: The electric force between two
point charges is given by Coulomb’s law:
F12 =k· |q1·q2|
r2
8
where k= 8.99×109N m2/C2is the Coulomb constant, q1=−3µC, q2= 5 µC,
and r= 4 m. Plugging in the values, we get:
F12 =8.99 ×109·|−3·5|
42= 1.1246 ×10−7N
Step 2: Calculate the force between q1and q3: Using the same formula as
above with r= 3 m, we have:
F13 =k· |q1·q3|
r2=8.99 ×109·|−3·(−2)|
32= 1.1988 ×10−7N
Step 3: Calculate the total electric force on q1: The total force on q1is the
vector sum of the forces F12 and F13. Since the forces are in opposite directions,
we subtract them:
Ftotal =F12 −F13 = (1.1246 ×10−7)−(1.1988 ×10−7) = −7.42 ×10−9N
Thus, the total electric force on q1due to q2and q3is −7.42 ×10−9N,
directed towards q3.
Question 10
Question
Three point charges are arranged at the vertices of an equilateral triangle with
sides of length d= 1.00 m. The charges are q1= +5.00µC, q2=−2.00µC, and
q3=−3.00µC. Calculate the magnitude and direction of the net electric force
acting on q1due to the other two charges.
Solution
Step 1: Calculate the electric force on q1due to q2. The electric force between
two point charges q1and q2is given by Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
where - kis the Coulomb constant (8.99 ×109N m2/C2), - |q1|and |q2|are
the magnitudes of the charges, - ris the distance between the charges.
Given q1= +5.00µC, q2=−2.00µC, and d= 1.00 m, the distance between
q1and q2(and also between all charges) is d.
Therefore,
F12 =(8.99 ×109N m2/C2)·(5.00 ×10−6C)·(2.00 ×10−6C)
(1.00 m)2
F12 = 26.97 ×10−3N(repulsive, to the right)
9
Step 2: Calculate the electric force on q1due to q3. Following a similar
calculation as above, the electric force between q1and q3is found to be:
F13 = 44.94×10−3N(at an angle of 120 degrees counterclockwise from horizontal)
Step 3: Find the net force on q1. The net force acting on q1is the vector
sum of the forces due to q2and q3. Using vector addition, we have:
Net force =p(F12 +F13 cos(120◦))2+ (F13 sin(120◦))2
Net force =p(26.97 ×10−3+ 44.94 ×10−3cos(120◦))2+ (44.94 ×10−3sin(120◦))2
Net force = 64.09 ×10−3N
The direction of the net force can be found by calculating the angle it makes
with the horizontal and is given by:
θ= tan−1F13 sin(120◦)
F12 +F13 cos(120◦)
θ=−30◦
Therefore, the magnitude of the net electric force acting on q1is 64.09×10−3
N, and it is directed at an angle of 30◦below the horizontal.
Question 11
Question
Three point charges are arranged in a line as shown below:
q1=−3µC q2= 5 µC q3=−7µC
The charges are located at positions:
x1= 1.0m, x2= 0.0m, x3=−2.0m
Calculate the net electric force on the charge q3due to the other two charges.
Solution
Step 1: Calculate the electric force between q1and q3The electric force between
two charges q1and q3can be calculated using Coulomb’s Law:
F13 =k|q1q3|
r2
13
where k= 8.99 ×109Nm2/C2is the Coulomb’s constant.
10
Substitute the given values:
F13 =(8.99 ×109)(3 ×10−6)(7 ×10−6)
(3)2
F13 =8.99 ×3×7
3= 71.92 ×10−6= 7.192 ×10−5N(to the right)
Step 2: Calculate the electric force between q2and q3Similarly, the electric
force between two charges q2and q3can be calculated using Coulomb’s Law:
F23 =k|q2q3|
r2
23
Substitute the given values:
F23 =(8.99 ×109)(5 ×10−6)(7 ×10−6)
(2)2
F23 =8.99 ×5×7
4= 31.465 ×10−6= 3.1465 ×10−5N(to the left)
Step 3: Find the net electric force on q3To find the net force on q3, we need
to consider the direction and add the two forces:
Fnet =F13 +F23 = 7.192×10−5−3.1465×10−5= 4.0455×10−5N(to the right)
Therefore, the net electric force on q3due to the other two charges is 4.0455×
10−5Nto the right.
Question 12
Question
Two point charges, q1= 3 nC and q2=−5nC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Convert the charges to coulombs.
Given: q1= 3 nC = 3 ×10−9C
q2=−5nC =−5×10−9C
Step 2: Calculate the distance between the charges.
Given: r= 10 cm = 0.1m
Step 3: Calculate the electric force using Coulomb’s Law. Coulomb’s Law
states that the magnitude of the electric force between two point charges is given
by:
F=k|q1q2|
r2
11
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values:
F= (8.99 ×109)|(3 ×10−9)(−5×10−9)|
(0.1)2
Step 4: Calculate the direction of the force.
The direction of the force is along the line joining the two charges. Since the
charges have opposite signs, the force is attractive.
Step 5: Calculate the magnitude of the force.
F= (8.99 ×109)15 ×10−18
0.01
F=134.85 ×10−9
0.01
F= 13.485 ×10−6N= 13.485 µN
Therefore, the magnitude of the electric force that q1exerts on q2is 13.485 µN
and the direction is attractive.
Question 13
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are separated by a distance
of 10.0cm. Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two point charges,
we will use Coulomb’s law:
F=k
q1·q2
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the
charges, and ris the distance between the charges.
Step 1: Convert the given charges to coulombs: Given: q1= 4.0µC =
4.0×10−6Cand q2=−6.0µC =−6.0×10−6C.
Step 2: Determine the distance between the charges in meters: Given:
r= 10.0cm = 10.0×10−2m= 0.10 m.
Step 3: Calculate the electric force between the charges using Coulomb’s
law:
F= 8.99 ×109N·m2/C2×
(4.0×10−6C)·(−6.0×10−6C)
(0.10 m)2
Step 4: Solve for the electric force:
F= 8.99×109×4.0×10−6× −6.0×10−6
0.102= 8.99×109×−24 ×10−12
0.01 = 8.99×109×−2.4×10−9N
12
F=−21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 14
Question
Three point charges are arranged in an equilateral triangle as shown below.
The charges have magnitudes of +2.5µC,−3.0µC, and +4.0µC. Calculate the
electric force on the −3.0µC charge.
+2.5µC
−3.0µC +4.0µC
Solution
Step 1: We first need to calculate the electric forces each charge exerts on the
−3.0µC charge. Let’s denote the distance between the charges as r. The electric
force Fbetween two point charges q1and q2separated by a distance ris given
by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2).
Calculating the forces between the charges:
For +2.5µC and −3.0µC:
F2.5µC,−3.0µC =8.99 ×109·|2.5×10−6·(−3.0×10−6)|
r2
For +4.0µC and −3.0µC:
F4.0µC,−3.0µC =8.99 ×109·|4.0×10−6·(−3.0×10−6)|
r2
Step 2: Next, we need to find the electric force acting on the −3.0µC charge
due to the other two charges. Since the charges are arranged symmetrically in
an equilateral triangle, the magnitudes of F2.5µC,−3.0µC and F4.0µC,−3.0µC are
the same.
Therefore, the total electric force on the −3.0µC charge is:
Ftotal = 2 ·F2.5µC,−3.0µC
Step 3: Finally, calculate the total electric force on the −3.0µC charge.
Perform the necessary calculations to find the numerical value of this force.
13
Question 15
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed at points Aand B
respectively. Point Ais located at coordinates (0,0) in meters and point Bis
at coordinates (3,4). Calculate the magnitude of the electric force experienced
by q2due to q1.
Solution
Step 1: Calculate the distance between the two point charges. Given that point
Ais located at coordinates (0,0) and point Bis located at coordinates (3,4), the
distance rbetween the two point charges can be calculated using the distance
formula:
r=p(x2−x1)2+ (y2−y1)2
where x1= 0,y1= 0,x2= 3, and y2= 4. Therefore,
r=p(3 −0)2+ (4 −0)2=p32+ 42=√9 + 16 = √25 = 5 m
Step 2: Calculate the magnitude of the electric force. The magnitude of
the electric force between two point charges can be calculated using Coulomb’s
Law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant. Substitute the given
values q1=−3.0µC,q2= 5.0µC, and r= 5 m into the formula:
F= (8.99 ×109)| − 3.0×10−6×5.0×10−6|
52
F= 8.99 ×109×15 ×10−12
25 = 8.99 ×109×0.6×10−12 = 5.394 ×10−3N
Therefore, the magnitude of the electric force experienced by q2due to q1is
5.394 ×10−3N.
Question 16
Question
Two point charges, q1= 4.0µC and q2=−8.0µC, are placed 0.20 m apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
14
Solution
Given: q1= 4.0µC= 4.0×10−6Cq2=−8.0µC=−8.0×10−6Cr= 0.20 m
We know that the electric force between two charges q1and q2separated by
a distance ris given by Coulomb’s law:
F=k|q1q2|
r2
Step 1: Calculate the electric force constant k.
k=1
4πε0
The permittivity of free space ε0= 8.85 ×10−12 C2/N·m2Thus,
k=1
4π×8.85 ×10−12
Step 2: Substitute the given values into Coulomb’s law to find the magnitude
of the electric force.
F=1
4π×8.85 ×10−12 ·|4.0×10−6× −8.0×10−6|
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F=1
4π×8.85 ×10−12 ·32 ×10−12
0.04
F=1
4π×8.85 ×10−12 ·800 ×10−12
F=800
4π×8.85
F≈800
35.16
F≈22.74 N
Therefore, the magnitude of the electric force between the charges is approx-
imately 22.74 N.
Question 17
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 0.10 m apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
15
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k· |q1·q2|
r2
where k≈8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the distance between the charges.
Step 2: Plug in the given values to find the electric force:
F=(8.99 ×109N·m2/C2)·|−4.0×10−6C·8.0×10−6C|
(0.10 m)2
Step 3: Calculate the electric force:
F=8.99 ×109·4.0×10−6·8.0×10−6
0.01
F=287.68
0.01
F= 28768 N
Step 4: Since q1is negative and q2is positive, the force will be attractive.
Therefore, the electric force that q1exerts on q2is 28768 N, directed towards q1.
Question 19
Question
Three point charges are arranged on the x-axis as follows: +qat x= 0,−3qat
x= 4a, and +2qat x= 7a, where qand aare positive constants. Calculate the
net force on the +2qcharge due to the other charges.
Solution
Step 1: Calculate the force on the +2qcharge due to the +qcharge: The force
F1can be calculated using Coulomb’s Law, which states that the magnitude of
the force between two point charges is given by:
F=k·|q1·q2|
r2
where kis Coulomb’s constant, q1and q2are the charges, and ris the distance
between the charges.
The force on the +2qcharge due to the +qcharge is:
F1=k·|(+2q)(q)|
(7a−0)2
16
Step 2: Calculate the force on the +2qcharge due to the −3qcharge: The
force F2can be calculated using Coulomb’s Law as well:
F2=k·|(+2q)(−3q)|
(7a−4a)2
Step 3: Determine the direction of the net force: Since the forces due to the
+qand −3qcharges are in opposite directions, the net force on the +2qcharge
will be the difference between F1and F2:
Fnet =F1−F2
Fnet =k·2q·q
(7a)2−2q·3q
3a)2
Therefore, the net force on the +2qcharge due to the other charges can be
calculated by subtracting the force on the +2qcharge due to the −3qcharge
from the force due to the +qcharge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle with
sides of length d. Charge Qis placed at each vertex. Calculate the magnitude
of the electric force on one of the charges due to the other two charges.
Solution
Let’s denote the three charges as Q1,Q2, and Q3, with Q1being the charge we
want to calculate the force on. The distance between each charge and Q1is d.
Step 1: Calculate the electric force on Q1due to Q2. The electric force
between two charges is given by Coulomb’s Law:
F=k· |Q1|·|Q2|
r2
where kis the electrostatic constant 8.99×109N m2/C2and r=dis the distance
between the charges.
Plugging in the values, the force on Q1due to Q2is:
F1=8.99 ×109·Q·Q
d2=8.99 ×109·Q2
d2
Step 2: Calculate the direction of the force due to Q2. Since Q2is at one
vertex of an equilateral triangle, the force on Q1due to Q2will be along the
line connecting Q1and Q2. This force will have a direction opposite to that of
the line connecting Q1and Q2.
17
Step 3: Calculate the electric force on Q1due to Q3. Similarly, the force
on Q1due to Q3will be along the line connecting Q1and Q3and will have a
direction opposite to that of the line connecting Q1and Q3. Thus, the force
will be:
F2=8.99 ×109·Q2
d2
Step 4: Calculate the total electric force on Q1. Since the forces F1and F2
have the same magnitude and opposite directions, the total electric force on Q1
due to Q2and Q3is:
Ftotal =F1−F2=8.99 ×109·Q2
d2−8.99 ×109·Q2
d2= 0
Thus, the net electric force on Q1due to Q2and Q3is zero.
Question 22
Question
Two point charges, q1= +3.0µC and q2=−6.0µC, are placed 8.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units (Coulombs).
Given: q1= +3.0µC = 3.0×10−6C
q2=−6.0µC =−6.0×10−6C
Step 2: Find the distance between the charges in meters.
Given: r= 8.0cm = 8.0×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law,
F=k|q1q2|
r2.
Where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109)· |3.0×10−6· −6.0×10−6|
(8.0×10−2)2
Step 4: Calculate the magnitude of the electric force.
F=8.99 ×109·18 ×10−12
64 ×10−4
F=161.82 ×10−3
64 ×10−4
F=161.82
64 = 2.53 N
Therefore, the magnitude of the electric force between the charges is 2.53 N.
18
Question 23
Question
Two point charges Q1 = +3.0 µC and Q2 = -5.0 µC are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that Q1 exerts on
Q2.
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Iden-
tify the direction of the force by considering the signs of the charges. Step 3:
Calculate the magnitude of the electric force using Coulomb’s law.
Step 1: Convert the charges from microcoulombs to coulombs: Q1 =
3.0µC = 3.0×10−6Cand Q2 = −5.0µC =−5.0×10−6C.
Step 2: The force will be attractive since Q1 is positive and Q2 is negative.
Step 3: Use Coulomb’s Law to find the magnitude of the electric force:
F=k|Q1·Q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), ris the distance between
the charges (0.1 m), Q1 = 3.0×10−6C, and Q2 = −5.0×10−6C.
Substitute the values into the equation:
F= (8.99 ×109)|3.0×10−6· −5.0×10−6|
(0.1)2
F= (8.99 ×109)15 ×10−12
0.01
F= (8.99 ×109)×1.5×10−9
F= 13.485 ×100
Therefore, F= 13.485N
The magnitude of the electric force that Q1 exerts on Q2 is 13.485 N, directed
towards Q1.
Question 24
Question
Two point charges, q1=−4.00 µC and q2= 3.00 µC, are separated by a distance
of 10.0 cm in a vacuum. Calculate the magnitude of the electric force between
the charges.
19
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Deter-
mine the distance between the charges in meters. Step 3: Use Coulomb’s Law
to calculate the electric force between the charges.
Step 1: Convert the charges from microcoulombs to coulombs. q1=
−4.00 µC =−4.00 ×10−6C q2= 3.00 µC = 3.00 ×10−6C
Step 2: Determine the distance between the charges in meters. Given:
distance r= 10.0cm = 0.10 m
Step 3: Use Coulomb’s Law to calculate the electric force between the
charges. Coulomb’s Law: F=k|q1q2|
r2where k= 8.99 ×109Nm2/C2is the
electrostatic constant.
Substitute the known values into the equation: F=(8.99 ×109)×|−4.00 ×10−6×3.00 ×10−6|
(0.10)2
F=(8.99 ×109)×1.2×10−11
0.01
F=10.788 ×10−2
0.01
F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788
N.
Question 25
Question
Two point charges, q1=−4µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the given charges to coulombs. Recall that 1µC= 10−6C.
⇒q1=−4µC=−4×10−6C=−4×10−6C
⇒q2= 5 µC= 5 ×10−6C= 5 ×10−6C
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The formula for the magnitude of the electric force between two charges q1and
q2separated by a distance ris given by:
F=k
q1·q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Step 3: Substitute the given values into the formula.
F= 8.99 ×109
(−4×10−6)·(5 ×10−6)
(0.10)2
20
F= 8.99 ×109
−20 ×10−12
0.01
F= 8.99 ×109
−2×10−9
F= 8.99 ×109×2×10−9
F= 17.98 ×100
F= 17.98 N
Therefore, the magnitude of the electric force between the charges is 17.98
N.
21
Step 3: Find the net force. To find the net force on the charge +2q, we need
to consider the vector sum of the forces from +qand −3q. Since the forces act
along a straight line through the charge +2q, we only need to consider their
magnitudes and signs. The net force is:
Fnet =F+q,+2q−F−3q,+2q=2kq2
a2−6kq2
a2=−4kq2
a2
Therefore, the magnitude of the net electric force acting on the charge +2q
is 4kq2
a2, and its direction is attractive (towards the −3qcharge).
Question 2
Question
Three point charges are arranged in an equilateral triangle as shown below.
Calculate the magnitude and direction of the net force on the charge q1=
+2.00 µC.
q2= +4.00 µC
q1= +2.00 µCq3= +6.00 µC
Solution
Step 1: First, let’s calculate the magnitudes of the forces on q1due to q2and
q3using Coulomb’s law: F=k|q1||q2|
r2, where k= 8.99 ×109N m2/C2.
The distance between q1and q2is the length of the side of the equilateral
triangle, r=l= 1 m.
Step 2: Calculating the force on q1due to q2:
F12 =k|q1||q2|
r2
F12 = (8.99 ×109)(2.00 ×10−6)(4.00 ×10−6)
(1)2
F12 = 7.19 ×10−2N
2
Step 3: Calculating the force on q1due to q3:
F13 =k|q1||q3|
r2
Since q3is 2l= 2 m away from q1:
F13 = (8.99 ×109)(2.00 ×10−6)(6.00 ×10−6)
(2)2
F13 = 4.498 ×10−2N
Step 4: Now, let’s find the net force on q1by vectorially adding the forces
F12 and F13. Given the forces are along the same line of the equilateral triangle,
they are in opposite directions and have the same magnitude.
Net force on q1=F12 −F13 = 7.19 ×10−2−4.498 ×10−2N
Net force on q1= 2.692 ×10−2N (in the direction towards q2)
Therefore, the magnitude of the net force on q1is 2.692 ×10−2N and the
direction is towards q2.
Question 3
Question
Two point charges, q1= 3.0µC and q2=−4.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Step 2: Calculate the distance between
the charges. Step 3: Apply Coulomb’s law to find the electric force between the
charges.
Step 1: Convert the charges to coulombs. The charge q1= 3.0µC can be
converted to coulombs as follows:
q1= 3.0×10−6C= 3.0µC
Similarly, the charge q2=−4.0µC can be converted to coulombs as follows:
q2=−4.0×10−6C=−4.0µC
3
Step 2: Calculate the distance between the charges. Given that the two
charges are placed 10.0 cm apart, we convert this distance to meters:
d= 10.0cm = 0.1m
Step 3: Apply Coulomb’s law to find the electric force between the charges.
Coulomb’s law states that the magnitude of the electric force between two
charges is given by:
F=k· |q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F=(8.99 ×109N m2/C2)· |3.0×10−6C· −4.0×10−6C|
(0.1m)2
F=8.99 ×109·12.0×10−12
0.01
F=107.88 ×10−3
0.01 = 10.788 N
Therefore, the magnitude of the electric force between the charges is 10.788
N.
Question 4
Question
Three point charges are arranged in an equilateral triangle. The charge at each
corner of the triangle is +q. Determine the magnitude and direction of the net
force on each charge.
Solution
Let’s denote the charges as q1,q2, and q3, each with magnitude +q.
Step 1: Calculate the magnitude of the force on q1due to q2. The force
F21 on q1due to q2is given by Coulomb’s Law:
F21 =k· |q1|·|q2|
r2
21
Since the charges are +q, the force will be repulsive.
Step 2: Calculate the magnitude of the force on q1due to q3. The force
F31 on q1due to q3is given by Coulomb’s Law:
F31 =k· |q1|·|q3|
r2
31
4
Since the charges are +q, the force will be repulsive.
Step 3: Resolve the forces on q1into components to determine the net force.
Since the forces are acting at 120◦angles to each other, the superposition of the
forces will yield a net force of magnitude:
Fnet1=qF2
21 +F2
31 + 2 ·F21 ·F31 ·cos(120◦)
Step 4: Determine the direction of the net force on q1. The direction of the
net force on q1can be found using trigonometry.
Step 5: Repeat the above steps to determine the net forces on q2and q3.
The net forces on q2and q3can be calculated in a similar manner, accounting
for the forces between all pairs of charges.
By analyzing the symmetry of the problem, one can determine that the net
forces will be along the lines joining the charges, and the magnitudes will be
equal.
Question 5
Question
Consider three point charges placed at the corners of an equilateral triangle
of side length d. The charges are as follows: q1=−3µC,q2= 5µC, and
q3=−2µC. Calculate the net electric force acting on q3due to q1and q2.
Solution
Step 1: Find the distance between q1and q3(and q2and q3). The distance
between any two adjacent charges in an equilateral triangle can be calculated
using the Pythagorean theorem. For an equilateral triangle of side length d,
drawing an altitude from one of the vertices to the opposite side forms a right
triangle with hypotenuse of length d and base of length d/2.
Step 2: Calculate the net force acting on q3due to q1. The force F13 between
q1and q3is given by Coulomb’s law:
F13 =k|q1||q3|
r2
13
where kis the Coulomb constant and r13 is the distance between q1and q3.
Step 3: Calculate the force acting on q3due to q2. The force F23 between
q2and q3is also given by Coulomb’s law:
F23 =k|q2||q3|
r2
23
where r23 is the distance between q2and q3.
Step 4: Determine the net force acting on q3. To find the net force acting on
q3, we need to consider the vector sum of F13 and F23. Remember to account
for the direction of each force.
5
Step 5: Perform the calculations. Substitute the given values for the charges
and distances into the expressions for F13 and F23 , taking into account the signs
of the charges. Then, find the net force by adding the two forces vectorially.
Question 6
Question
Two charges, q1=−3µC and q2= 5µC, are placed 2 meters apart. Calculate
the magnitude of the electric force between these two charges.
Solution
Step 1: Recall that the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into Coulomb’s Law:
F= 8.99 ×109|(−3µC)(5µC)|
(2 m)2
Step 3: Calculate the magnitudes of the charges in coulombs:
q1=−3µC =−3×10−6C
q2= 5µC = 5 ×10−6C
Step 4: Substitute the magnitudes of the charges into the formula:
F= 8.99 ×109|(−3×10−6C)(5 ×10−6C)|
(2 m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×10915 ×10−12C2
4m2
F= 8.99 ×109×3.75 ×10−12 N
F= 33.7125 ×10−3N
F= 3.37125 ×10−2N
Step 6: Thus, the magnitude of the electric force between the two charges is
3.37125 ×10−2N.
6
Question 7
Question
Two point charges, Q1= 5 ×10−6C and Q2=−3×10−6C, are placed 8 meters
apart in a vacuum. Calculate the magnitude of the electric force between the
charges.
Solution
Step 1: Calculate the electric force between the charges using Coulomb’s Law
formula:
F=k
Q1·Q2
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2), Q1and Q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into the formula and solve for the force:
F= 8.99 ×109
(5 ×10−6)·(−3×10−6)
82
F= 8.99 ×109
−15 ×10−12
64
F= 8.99 ×109
−0.23438 ×10−12
F= 8.99 ×109×0.23438 ×10−12
F= 2.11 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.11 ×
10−3N.
Question 8
Question
Two electrons are located 1.5 nm apart. Calculate the electric force between
them. The charge of an electron is −1.6×10−19 C.
Solution
Step 1: Calculate the electric force using Coulomb’s Law: The equation for the
electric force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
7
where Fis the electric force, kis Coulomb’s constant (8.99 ×109N m2/C2),
q1and q2are the charges of the two particles, and ris the separation distance
between the charges.
Step 2: Substitute the given values into the equation: Given that the charge
of an electron is −1.6×10−19 C, the charges of the two electrons are q1=q2=
−1.6×10−19 C. The separation distance ris 1.5 nm = 1.5×10−9m.
Step 3: Calculate the electric force: Plugging in the values, we get:
F= (8.99 ×109)|(−1.6×10−19)(−1.6×10−19)|
(1.5×10−9)2
Step 4: Simplify and calculate the force:
F= (8.99 ×109)(2.56 ×10−38)
2.25 ×10−18
F= (8.99 ×109)×1.138 ×10−20
F= 1.023 ×10−10 N
Therefore, the electric force between the two electrons is 1.023 ×10−10 N.
Question 9
Question
Three point charges are arranged as shown in the diagram below. The charges
are q1=−3µC, q2= 5 µC, and q3=−2µC. The distance rbetween q1and q2
is 4meters, and the distance between q1and q3is 3meters. Calculate the total
electric force on q1due to q2and q3.
q1q2
q3
Solution
Step 1: Calculate the force between q1and q2: The electric force between two
point charges is given by Coulomb’s law:
F12 =k· |q1·q2|
r2
8
where k= 8.99×109N m2/C2is the Coulomb constant, q1=−3µC, q2= 5 µC,
and r= 4 m. Plugging in the values, we get:
F12 =8.99 ×109·|−3·5|
42= 1.1246 ×10−7N
Step 2: Calculate the force between q1and q3: Using the same formula as
above with r= 3 m, we have:
F13 =k· |q1·q3|
r2=8.99 ×109·|−3·(−2)|
32= 1.1988 ×10−7N
Step 3: Calculate the total electric force on q1: The total force on q1is the
vector sum of the forces F12 and F13. Since the forces are in opposite directions,
we subtract them:
Ftotal =F12 −F13 = (1.1246 ×10−7)−(1.1988 ×10−7) = −7.42 ×10−9N
Thus, the total electric force on q1due to q2and q3is −7.42 ×10−9N,
directed towards q3.
Question 10
Question
Three point charges are arranged at the vertices of an equilateral triangle with
sides of length d= 1.00 m. The charges are q1= +5.00µC, q2=−2.00µC, and
q3=−3.00µC. Calculate the magnitude and direction of the net electric force
acting on q1due to the other two charges.
Solution
Step 1: Calculate the electric force on q1due to q2. The electric force between
two point charges q1and q2is given by Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
where - kis the Coulomb constant (8.99 ×109N m2/C2), - |q1|and |q2|are
the magnitudes of the charges, - ris the distance between the charges.
Given q1= +5.00µC, q2=−2.00µC, and d= 1.00 m, the distance between
q1and q2(and also between all charges) is d.
Therefore,
F12 =(8.99 ×109N m2/C2)·(5.00 ×10−6C)·(2.00 ×10−6C)
(1.00 m)2
F12 = 26.97 ×10−3N(repulsive, to the right)
9
Step 2: Calculate the electric force on q1due to q3. Following a similar
calculation as above, the electric force between q1and q3is found to be:
F13 = 44.94×10−3N(at an angle of 120 degrees counterclockwise from horizontal)
Step 3: Find the net force on q1. The net force acting on q1is the vector
sum of the forces due to q2and q3. Using vector addition, we have:
Net force =p(F12 +F13 cos(120◦))2+ (F13 sin(120◦))2
Net force =p(26.97 ×10−3+ 44.94 ×10−3cos(120◦))2+ (44.94 ×10−3sin(120◦))2
Net force = 64.09 ×10−3N
The direction of the net force can be found by calculating the angle it makes
with the horizontal and is given by:
θ= tan−1F13 sin(120◦)
F12 +F13 cos(120◦)
θ=−30◦
Therefore, the magnitude of the net electric force acting on q1is 64.09×10−3
N, and it is directed at an angle of 30◦below the horizontal.
Question 11
Question
Three point charges are arranged in a line as shown below:
q1=−3µC q2= 5 µC q3=−7µC
The charges are located at positions:
x1= 1.0m, x2= 0.0m, x3=−2.0m
Calculate the net electric force on the charge q3due to the other two charges.
Solution
Step 1: Calculate the electric force between q1and q3The electric force between
two charges q1and q3can be calculated using Coulomb’s Law:
F13 =k|q1q3|
r2
13
where k= 8.99 ×109Nm2/C2is the Coulomb’s constant.
10
Substitute the given values:
F13 =(8.99 ×109)(3 ×10−6)(7 ×10−6)
(3)2
F13 =8.99 ×3×7
3= 71.92 ×10−6= 7.192 ×10−5N(to the right)
Step 2: Calculate the electric force between q2and q3Similarly, the electric
force between two charges q2and q3can be calculated using Coulomb’s Law:
F23 =k|q2q3|
r2
23
Substitute the given values:
F23 =(8.99 ×109)(5 ×10−6)(7 ×10−6)
(2)2
F23 =8.99 ×5×7
4= 31.465 ×10−6= 3.1465 ×10−5N(to the left)
Step 3: Find the net electric force on q3To find the net force on q3, we need
to consider the direction and add the two forces:
Fnet =F13 +F23 = 7.192×10−5−3.1465×10−5= 4.0455×10−5N(to the right)
Therefore, the net electric force on q3due to the other two charges is 4.0455×
10−5Nto the right.
Question 12
Question
Two point charges, q1= 3 nC and q2=−5nC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Convert the charges to coulombs.
Given: q1= 3 nC = 3 ×10−9C
q2=−5nC =−5×10−9C
Step 2: Calculate the distance between the charges.
Given: r= 10 cm = 0.1m
Step 3: Calculate the electric force using Coulomb’s Law. Coulomb’s Law
states that the magnitude of the electric force between two point charges is given
by:
F=k|q1q2|
r2
11
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values:
F= (8.99 ×109)|(3 ×10−9)(−5×10−9)|
(0.1)2
Step 4: Calculate the direction of the force.
The direction of the force is along the line joining the two charges. Since the
charges have opposite signs, the force is attractive.
Step 5: Calculate the magnitude of the force.
F= (8.99 ×109)15 ×10−18
0.01
F=134.85 ×10−9
0.01
F= 13.485 ×10−6N= 13.485 µN
Therefore, the magnitude of the electric force that q1exerts on q2is 13.485 µN
and the direction is attractive.
Question 13
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are separated by a distance
of 10.0cm. Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two point charges,
we will use Coulomb’s law:
F=k
q1·q2
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the
charges, and ris the distance between the charges.
Step 1: Convert the given charges to coulombs: Given: q1= 4.0µC =
4.0×10−6Cand q2=−6.0µC =−6.0×10−6C.
Step 2: Determine the distance between the charges in meters: Given:
r= 10.0cm = 10.0×10−2m= 0.10 m.
Step 3: Calculate the electric force between the charges using Coulomb’s
law:
F= 8.99 ×109N·m2/C2×
(4.0×10−6C)·(−6.0×10−6C)
(0.10 m)2
Step 4: Solve for the electric force:
F= 8.99×109×4.0×10−6× −6.0×10−6
0.102= 8.99×109×−24 ×10−12
0.01 = 8.99×109×−2.4×10−9N
12
F=−21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 14
Question
Three point charges are arranged in an equilateral triangle as shown below.
The charges have magnitudes of +2.5µC,−3.0µC, and +4.0µC. Calculate the
electric force on the −3.0µC charge.
+2.5µC
−3.0µC +4.0µC
Solution
Step 1: We first need to calculate the electric forces each charge exerts on the
−3.0µC charge. Let’s denote the distance between the charges as r. The electric
force Fbetween two point charges q1and q2separated by a distance ris given
by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2).
Calculating the forces between the charges:
For +2.5µC and −3.0µC:
F2.5µC,−3.0µC =8.99 ×109·|2.5×10−6·(−3.0×10−6)|
r2
For +4.0µC and −3.0µC:
F4.0µC,−3.0µC =8.99 ×109·|4.0×10−6·(−3.0×10−6)|
r2
Step 2: Next, we need to find the electric force acting on the −3.0µC charge
due to the other two charges. Since the charges are arranged symmetrically in
an equilateral triangle, the magnitudes of F2.5µC,−3.0µC and F4.0µC,−3.0µC are
the same.
Therefore, the total electric force on the −3.0µC charge is:
Ftotal = 2 ·F2.5µC,−3.0µC
Step 3: Finally, calculate the total electric force on the −3.0µC charge.
Perform the necessary calculations to find the numerical value of this force.
13
Question 15
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed at points Aand B
respectively. Point Ais located at coordinates (0,0) in meters and point Bis
at coordinates (3,4). Calculate the magnitude of the electric force experienced
by q2due to q1.
Solution
Step 1: Calculate the distance between the two point charges. Given that point
Ais located at coordinates (0,0) and point Bis located at coordinates (3,4), the
distance rbetween the two point charges can be calculated using the distance
formula:
r=p(x2−x1)2+ (y2−y1)2
where x1= 0,y1= 0,x2= 3, and y2= 4. Therefore,
r=p(3 −0)2+ (4 −0)2=p32+ 42=√9 + 16 = √25 = 5 m
Step 2: Calculate the magnitude of the electric force. The magnitude of
the electric force between two point charges can be calculated using Coulomb’s
Law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant. Substitute the given
values q1=−3.0µC,q2= 5.0µC, and r= 5 m into the formula:
F= (8.99 ×109)| − 3.0×10−6×5.0×10−6|
52
F= 8.99 ×109×15 ×10−12
25 = 8.99 ×109×0.6×10−12 = 5.394 ×10−3N
Therefore, the magnitude of the electric force experienced by q2due to q1is
5.394 ×10−3N.
Question 16
Question
Two point charges, q1= 4.0µC and q2=−8.0µC, are placed 0.20 m apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
14
Solution
Given: q1= 4.0µC= 4.0×10−6Cq2=−8.0µC=−8.0×10−6Cr= 0.20 m
We know that the electric force between two charges q1and q2separated by
a distance ris given by Coulomb’s law:
F=k|q1q2|
r2
Step 1: Calculate the electric force constant k.
k=1
4πε0
The permittivity of free space ε0= 8.85 ×10−12 C2/N·m2Thus,
k=1
4π×8.85 ×10−12
Step 2: Substitute the given values into Coulomb’s law to find the magnitude
of the electric force.
F=1
4π×8.85 ×10−12 ·|4.0×10−6× −8.0×10−6|
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F=1
4π×8.85 ×10−12 ·32 ×10−12
0.04
F=1
4π×8.85 ×10−12 ·800 ×10−12
F=800
4π×8.85
F≈800
35.16
F≈22.74 N
Therefore, the magnitude of the electric force between the charges is approx-
imately 22.74 N.
Question 17
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 0.10 m apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
15
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k· |q1·q2|
r2
where k≈8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the distance between the charges.
Step 2: Plug in the given values to find the electric force:
F=(8.99 ×109N·m2/C2)·|−4.0×10−6C·8.0×10−6C|
(0.10 m)2
Step 3: Calculate the electric force:
F=8.99 ×109·4.0×10−6·8.0×10−6
0.01
F=287.68
0.01
F= 28768 N
Step 4: Since q1is negative and q2is positive, the force will be attractive.
Therefore, the electric force that q1exerts on q2is 28768 N, directed towards q1.
Question 19
Question
Three point charges are arranged on the x-axis as follows: +qat x= 0,−3qat
x= 4a, and +2qat x= 7a, where qand aare positive constants. Calculate the
net force on the +2qcharge due to the other charges.
Solution
Step 1: Calculate the force on the +2qcharge due to the +qcharge: The force
F1can be calculated using Coulomb’s Law, which states that the magnitude of
the force between two point charges is given by:
F=k·|q1·q2|
r2
where kis Coulomb’s constant, q1and q2are the charges, and ris the distance
between the charges.
The force on the +2qcharge due to the +qcharge is:
F1=k·|(+2q)(q)|
(7a−0)2
16
Step 2: Calculate the force on the +2qcharge due to the −3qcharge: The
force F2can be calculated using Coulomb’s Law as well:
F2=k·|(+2q)(−3q)|
(7a−4a)2
Step 3: Determine the direction of the net force: Since the forces due to the
+qand −3qcharges are in opposite directions, the net force on the +2qcharge
will be the difference between F1and F2:
Fnet =F1−F2
Fnet =k·2q·q
(7a)2−2q·3q
3a)2
Therefore, the net force on the +2qcharge due to the other charges can be
calculated by subtracting the force on the +2qcharge due to the −3qcharge
from the force due to the +qcharge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle with
sides of length d. Charge Qis placed at each vertex. Calculate the magnitude
of the electric force on one of the charges due to the other two charges.
Solution
Let’s denote the three charges as Q1,Q2, and Q3, with Q1being the charge we
want to calculate the force on. The distance between each charge and Q1is d.
Step 1: Calculate the electric force on Q1due to Q2. The electric force
between two charges is given by Coulomb’s Law:
F=k· |Q1|·|Q2|
r2
where kis the electrostatic constant 8.99×109N m2/C2and r=dis the distance
between the charges.
Plugging in the values, the force on Q1due to Q2is:
F1=8.99 ×109·Q·Q
d2=8.99 ×109·Q2
d2
Step 2: Calculate the direction of the force due to Q2. Since Q2is at one
vertex of an equilateral triangle, the force on Q1due to Q2will be along the
line connecting Q1and Q2. This force will have a direction opposite to that of
the line connecting Q1and Q2.
17
Step 3: Calculate the electric force on Q1due to Q3. Similarly, the force
on Q1due to Q3will be along the line connecting Q1and Q3and will have a
direction opposite to that of the line connecting Q1and Q3. Thus, the force
will be:
F2=8.99 ×109·Q2
d2
Step 4: Calculate the total electric force on Q1. Since the forces F1and F2
have the same magnitude and opposite directions, the total electric force on Q1
due to Q2and Q3is:
Ftotal =F1−F2=8.99 ×109·Q2
d2−8.99 ×109·Q2
d2= 0
Thus, the net electric force on Q1due to Q2and Q3is zero.
Question 22
Question
Two point charges, q1= +3.0µC and q2=−6.0µC, are placed 8.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units (Coulombs).
Given: q1= +3.0µC = 3.0×10−6C
q2=−6.0µC =−6.0×10−6C
Step 2: Find the distance between the charges in meters.
Given: r= 8.0cm = 8.0×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law,
F=k|q1q2|
r2.
Where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109)· |3.0×10−6· −6.0×10−6|
(8.0×10−2)2
Step 4: Calculate the magnitude of the electric force.
F=8.99 ×109·18 ×10−12
64 ×10−4
F=161.82 ×10−3
64 ×10−4
F=161.82
64 = 2.53 N
Therefore, the magnitude of the electric force between the charges is 2.53 N.
18
Question 23
Question
Two point charges Q1 = +3.0 µC and Q2 = -5.0 µC are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that Q1 exerts on
Q2.
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Iden-
tify the direction of the force by considering the signs of the charges. Step 3:
Calculate the magnitude of the electric force using Coulomb’s law.
Step 1: Convert the charges from microcoulombs to coulombs: Q1 =
3.0µC = 3.0×10−6Cand Q2 = −5.0µC =−5.0×10−6C.
Step 2: The force will be attractive since Q1 is positive and Q2 is negative.
Step 3: Use Coulomb’s Law to find the magnitude of the electric force:
F=k|Q1·Q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), ris the distance between
the charges (0.1 m), Q1 = 3.0×10−6C, and Q2 = −5.0×10−6C.
Substitute the values into the equation:
F= (8.99 ×109)|3.0×10−6· −5.0×10−6|
(0.1)2
F= (8.99 ×109)15 ×10−12
0.01
F= (8.99 ×109)×1.5×10−9
F= 13.485 ×100
Therefore, F= 13.485N
The magnitude of the electric force that Q1 exerts on Q2 is 13.485 N, directed
towards Q1.
Question 24
Question
Two point charges, q1=−4.00 µC and q2= 3.00 µC, are separated by a distance
of 10.0 cm in a vacuum. Calculate the magnitude of the electric force between
the charges.
19
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Deter-
mine the distance between the charges in meters. Step 3: Use Coulomb’s Law
to calculate the electric force between the charges.
Step 1: Convert the charges from microcoulombs to coulombs. q1=
−4.00 µC =−4.00 ×10−6C q2= 3.00 µC = 3.00 ×10−6C
Step 2: Determine the distance between the charges in meters. Given:
distance r= 10.0cm = 0.10 m
Step 3: Use Coulomb’s Law to calculate the electric force between the
charges. Coulomb’s Law: F=k|q1q2|
r2where k= 8.99 ×109Nm2/C2is the
electrostatic constant.
Substitute the known values into the equation: F=(8.99 ×109)×|−4.00 ×10−6×3.00 ×10−6|
(0.10)2
F=(8.99 ×109)×1.2×10−11
0.01
F=10.788 ×10−2
0.01
F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788
N.
Question 25
Question
Two point charges, q1=−4µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the given charges to coulombs. Recall that 1µC= 10−6C.
⇒q1=−4µC=−4×10−6C=−4×10−6C
⇒q2= 5 µC= 5 ×10−6C= 5 ×10−6C
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The formula for the magnitude of the electric force between two charges q1and
q2separated by a distance ris given by:
F=k
q1·q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Step 3: Substitute the given values into the formula.
F= 8.99 ×109
(−4×10−6)·(5 ×10−6)
(0.10)2
20
F= 8.99 ×109
−20 ×10−12
0.01
F= 8.99 ×109
−2×10−9
F= 8.99 ×109×2×10−9
F= 17.98 ×100
F= 17.98 N
Therefore, the magnitude of the electric force between the charges is 17.98
N.
21
Step 3: Find the net force. To find the net force on the charge +2q, we need
to consider the vector sum of the forces from +qand −3q. Since the forces act
along a straight line through the charge +2q, we only need to consider their
magnitudes and signs. The net force is:
Fnet =F+q,+2q−F−3q,+2q=2kq2
a2−6kq2
a2=−4kq2
a2
Therefore, the magnitude of the net electric force acting on the charge +2q
is 4kq2
a2, and its direction is attractive (towards the −3qcharge).
Question 2
Question
Three point charges are arranged in an equilateral triangle as shown below.
Calculate the magnitude and direction of the net force on the charge q1=
+2.00 µC.
q2= +4.00 µC
q1= +2.00 µCq3= +6.00 µC
Solution
Step 1: First, let’s calculate the magnitudes of the forces on q1due to q2and
q3using Coulomb’s law: F=k|q1||q2|
r2, where k= 8.99 ×109N m2/C2.
The distance between q1and q2is the length of the side of the equilateral
triangle, r=l= 1 m.
Step 2: Calculating the force on q1due to q2:
F12 =k|q1||q2|
r2
F12 = (8.99 ×109)(2.00 ×10−6)(4.00 ×10−6)
(1)2
F12 = 7.19 ×10−2N
2
Step 3: Calculating the force on q1due to q3:
F13 =k|q1||q3|
r2
Since q3is 2l= 2 m away from q1:
F13 = (8.99 ×109)(2.00 ×10−6)(6.00 ×10−6)
(2)2
F13 = 4.498 ×10−2N
Step 4: Now, let’s find the net force on q1by vectorially adding the forces
F12 and F13. Given the forces are along the same line of the equilateral triangle,
they are in opposite directions and have the same magnitude.
Net force on q1=F12 −F13 = 7.19 ×10−2−4.498 ×10−2N
Net force on q1= 2.692 ×10−2N (in the direction towards q2)
Therefore, the magnitude of the net force on q1is 2.692 ×10−2N and the
direction is towards q2.
Question 3
Question
Two point charges, q1= 3.0µC and q2=−4.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Step 2: Calculate the distance between
the charges. Step 3: Apply Coulomb’s law to find the electric force between the
charges.
Step 1: Convert the charges to coulombs. The charge q1= 3.0µC can be
converted to coulombs as follows:
q1= 3.0×10−6C= 3.0µC
Similarly, the charge q2=−4.0µC can be converted to coulombs as follows:
q2=−4.0×10−6C=−4.0µC
3
Step 2: Calculate the distance between the charges. Given that the two
charges are placed 10.0 cm apart, we convert this distance to meters:
d= 10.0cm = 0.1m
Step 3: Apply Coulomb’s law to find the electric force between the charges.
Coulomb’s law states that the magnitude of the electric force between two
charges is given by:
F=k· |q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F=(8.99 ×109N m2/C2)· |3.0×10−6C· −4.0×10−6C|
(0.1m)2
F=8.99 ×109·12.0×10−12
0.01
F=107.88 ×10−3
0.01 = 10.788 N
Therefore, the magnitude of the electric force between the charges is 10.788
N.
Question 4
Question
Three point charges are arranged in an equilateral triangle. The charge at each
corner of the triangle is +q. Determine the magnitude and direction of the net
force on each charge.
Solution
Let’s denote the charges as q1,q2, and q3, each with magnitude +q.
Step 1: Calculate the magnitude of the force on q1due to q2. The force
F21 on q1due to q2is given by Coulomb’s Law:
F21 =k· |q1|·|q2|
r2
21
Since the charges are +q, the force will be repulsive.
Step 2: Calculate the magnitude of the force on q1due to q3. The force
F31 on q1due to q3is given by Coulomb’s Law:
F31 =k· |q1|·|q3|
r2
31
4
Since the charges are +q, the force will be repulsive.
Step 3: Resolve the forces on q1into components to determine the net force.
Since the forces are acting at 120◦angles to each other, the superposition of the
forces will yield a net force of magnitude:
Fnet1=qF2
21 +F2
31 + 2 ·F21 ·F31 ·cos(120◦)
Step 4: Determine the direction of the net force on q1. The direction of the
net force on q1can be found using trigonometry.
Step 5: Repeat the above steps to determine the net forces on q2and q3.
The net forces on q2and q3can be calculated in a similar manner, accounting
for the forces between all pairs of charges.
By analyzing the symmetry of the problem, one can determine that the net
forces will be along the lines joining the charges, and the magnitudes will be
equal.
Question 5
Question
Consider three point charges placed at the corners of an equilateral triangle
of side length d. The charges are as follows: q1=−3µC,q2= 5µC, and
q3=−2µC. Calculate the net electric force acting on q3due to q1and q2.
Solution
Step 1: Find the distance between q1and q3(and q2and q3). The distance
between any two adjacent charges in an equilateral triangle can be calculated
using the Pythagorean theorem. For an equilateral triangle of side length d,
drawing an altitude from one of the vertices to the opposite side forms a right
triangle with hypotenuse of length d and base of length d/2.
Step 2: Calculate the net force acting on q3due to q1. The force F13 between
q1and q3is given by Coulomb’s law:
F13 =k|q1||q3|
r2
13
where kis the Coulomb constant and r13 is the distance between q1and q3.
Step 3: Calculate the force acting on q3due to q2. The force F23 between
q2and q3is also given by Coulomb’s law:
F23 =k|q2||q3|
r2
23
where r23 is the distance between q2and q3.
Step 4: Determine the net force acting on q3. To find the net force acting on
q3, we need to consider the vector sum of F13 and F23. Remember to account
for the direction of each force.
5
Step 5: Perform the calculations. Substitute the given values for the charges
and distances into the expressions for F13 and F23 , taking into account the signs
of the charges. Then, find the net force by adding the two forces vectorially.
Question 6
Question
Two charges, q1=−3µC and q2= 5µC, are placed 2 meters apart. Calculate
the magnitude of the electric force between these two charges.
Solution
Step 1: Recall that the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into Coulomb’s Law:
F= 8.99 ×109|(−3µC)(5µC)|
(2 m)2
Step 3: Calculate the magnitudes of the charges in coulombs:
q1=−3µC =−3×10−6C
q2= 5µC = 5 ×10−6C
Step 4: Substitute the magnitudes of the charges into the formula:
F= 8.99 ×109|(−3×10−6C)(5 ×10−6C)|
(2 m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×10915 ×10−12C2
4m2
F= 8.99 ×109×3.75 ×10−12 N
F= 33.7125 ×10−3N
F= 3.37125 ×10−2N
Step 6: Thus, the magnitude of the electric force between the two charges is
3.37125 ×10−2N.
6
Question 7
Question
Two point charges, Q1= 5 ×10−6C and Q2=−3×10−6C, are placed 8 meters
apart in a vacuum. Calculate the magnitude of the electric force between the
charges.
Solution
Step 1: Calculate the electric force between the charges using Coulomb’s Law
formula:
F=k
Q1·Q2
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2), Q1and Q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Substitute the given values into the formula and solve for the force:
F= 8.99 ×109
(5 ×10−6)·(−3×10−6)
82
F= 8.99 ×109
−15 ×10−12
64
F= 8.99 ×109
−0.23438 ×10−12
F= 8.99 ×109×0.23438 ×10−12
F= 2.11 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.11 ×
10−3N.
Question 8
Question
Two electrons are located 1.5 nm apart. Calculate the electric force between
them. The charge of an electron is −1.6×10−19 C.
Solution
Step 1: Calculate the electric force using Coulomb’s Law: The equation for the
electric force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
7
where Fis the electric force, kis Coulomb’s constant (8.99 ×109N m2/C2),
q1and q2are the charges of the two particles, and ris the separation distance
between the charges.
Step 2: Substitute the given values into the equation: Given that the charge
of an electron is −1.6×10−19 C, the charges of the two electrons are q1=q2=
−1.6×10−19 C. The separation distance ris 1.5 nm = 1.5×10−9m.
Step 3: Calculate the electric force: Plugging in the values, we get:
F= (8.99 ×109)|(−1.6×10−19)(−1.6×10−19)|
(1.5×10−9)2
Step 4: Simplify and calculate the force:
F= (8.99 ×109)(2.56 ×10−38)
2.25 ×10−18
F= (8.99 ×109)×1.138 ×10−20
F= 1.023 ×10−10 N
Therefore, the electric force between the two electrons is 1.023 ×10−10 N.
Question 9
Question
Three point charges are arranged as shown in the diagram below. The charges
are q1=−3µC, q2= 5 µC, and q3=−2µC. The distance rbetween q1and q2
is 4meters, and the distance between q1and q3is 3meters. Calculate the total
electric force on q1due to q2and q3.
q1q2
q3
Solution
Step 1: Calculate the force between q1and q2: The electric force between two
point charges is given by Coulomb’s law:
F12 =k· |q1·q2|
r2
8
where k= 8.99×109N m2/C2is the Coulomb constant, q1=−3µC, q2= 5 µC,
and r= 4 m. Plugging in the values, we get:
F12 =8.99 ×109·|−3·5|
42= 1.1246 ×10−7N
Step 2: Calculate the force between q1and q3: Using the same formula as
above with r= 3 m, we have:
F13 =k· |q1·q3|
r2=8.99 ×109·|−3·(−2)|
32= 1.1988 ×10−7N
Step 3: Calculate the total electric force on q1: The total force on q1is the
vector sum of the forces F12 and F13. Since the forces are in opposite directions,
we subtract them:
Ftotal =F12 −F13 = (1.1246 ×10−7)−(1.1988 ×10−7) = −7.42 ×10−9N
Thus, the total electric force on q1due to q2and q3is −7.42 ×10−9N,
directed towards q3.
Question 10
Question
Three point charges are arranged at the vertices of an equilateral triangle with
sides of length d= 1.00 m. The charges are q1= +5.00µC, q2=−2.00µC, and
q3=−3.00µC. Calculate the magnitude and direction of the net electric force
acting on q1due to the other two charges.
Solution
Step 1: Calculate the electric force on q1due to q2. The electric force between
two point charges q1and q2is given by Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
where - kis the Coulomb constant (8.99 ×109N m2/C2), - |q1|and |q2|are
the magnitudes of the charges, - ris the distance between the charges.
Given q1= +5.00µC, q2=−2.00µC, and d= 1.00 m, the distance between
q1and q2(and also between all charges) is d.
Therefore,
F12 =(8.99 ×109N m2/C2)·(5.00 ×10−6C)·(2.00 ×10−6C)
(1.00 m)2
F12 = 26.97 ×10−3N(repulsive, to the right)
9
Step 2: Calculate the electric force on q1due to q3. Following a similar
calculation as above, the electric force between q1and q3is found to be:
F13 = 44.94×10−3N(at an angle of 120 degrees counterclockwise from horizontal)
Step 3: Find the net force on q1. The net force acting on q1is the vector
sum of the forces due to q2and q3. Using vector addition, we have:
Net force =p(F12 +F13 cos(120◦))2+ (F13 sin(120◦))2
Net force =p(26.97 ×10−3+ 44.94 ×10−3cos(120◦))2+ (44.94 ×10−3sin(120◦))2
Net force = 64.09 ×10−3N
The direction of the net force can be found by calculating the angle it makes
with the horizontal and is given by:
θ= tan−1F13 sin(120◦)
F12 +F13 cos(120◦)
θ=−30◦
Therefore, the magnitude of the net electric force acting on q1is 64.09×10−3
N, and it is directed at an angle of 30◦below the horizontal.
Question 11
Question
Three point charges are arranged in a line as shown below:
q1=−3µC q2= 5 µC q3=−7µC
The charges are located at positions:
x1= 1.0m, x2= 0.0m, x3=−2.0m
Calculate the net electric force on the charge q3due to the other two charges.
Solution
Step 1: Calculate the electric force between q1and q3The electric force between
two charges q1and q3can be calculated using Coulomb’s Law:
F13 =k|q1q3|
r2
13
where k= 8.99 ×109Nm2/C2is the Coulomb’s constant.
10
Substitute the given values:
F13 =(8.99 ×109)(3 ×10−6)(7 ×10−6)
(3)2
F13 =8.99 ×3×7
3= 71.92 ×10−6= 7.192 ×10−5N(to the right)
Step 2: Calculate the electric force between q2and q3Similarly, the electric
force between two charges q2and q3can be calculated using Coulomb’s Law:
F23 =k|q2q3|
r2
23
Substitute the given values:
F23 =(8.99 ×109)(5 ×10−6)(7 ×10−6)
(2)2
F23 =8.99 ×5×7
4= 31.465 ×10−6= 3.1465 ×10−5N(to the left)
Step 3: Find the net electric force on q3To find the net force on q3, we need
to consider the direction and add the two forces:
Fnet =F13 +F23 = 7.192×10−5−3.1465×10−5= 4.0455×10−5N(to the right)
Therefore, the net electric force on q3due to the other two charges is 4.0455×
10−5Nto the right.
Question 12
Question
Two point charges, q1= 3 nC and q2=−5nC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Convert the charges to coulombs.
Given: q1= 3 nC = 3 ×10−9C
q2=−5nC =−5×10−9C
Step 2: Calculate the distance between the charges.
Given: r= 10 cm = 0.1m
Step 3: Calculate the electric force using Coulomb’s Law. Coulomb’s Law
states that the magnitude of the electric force between two point charges is given
by:
F=k|q1q2|
r2
11
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values:
F= (8.99 ×109)|(3 ×10−9)(−5×10−9)|
(0.1)2
Step 4: Calculate the direction of the force.
The direction of the force is along the line joining the two charges. Since the
charges have opposite signs, the force is attractive.
Step 5: Calculate the magnitude of the force.
F= (8.99 ×109)15 ×10−18
0.01
F=134.85 ×10−9
0.01
F= 13.485 ×10−6N= 13.485 µN
Therefore, the magnitude of the electric force that q1exerts on q2is 13.485 µN
and the direction is attractive.
Question 13
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are separated by a distance
of 10.0cm. Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two point charges,
we will use Coulomb’s law:
F=k
q1·q2
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the
charges, and ris the distance between the charges.
Step 1: Convert the given charges to coulombs: Given: q1= 4.0µC =
4.0×10−6Cand q2=−6.0µC =−6.0×10−6C.
Step 2: Determine the distance between the charges in meters: Given:
r= 10.0cm = 10.0×10−2m= 0.10 m.
Step 3: Calculate the electric force between the charges using Coulomb’s
law:
F= 8.99 ×109N·m2/C2×
(4.0×10−6C)·(−6.0×10−6C)
(0.10 m)2
Step 4: Solve for the electric force:
F= 8.99×109×4.0×10−6× −6.0×10−6
0.102= 8.99×109×−24 ×10−12
0.01 = 8.99×109×−2.4×10−9N
12
F=−21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 14
Question
Three point charges are arranged in an equilateral triangle as shown below.
The charges have magnitudes of +2.5µC,−3.0µC, and +4.0µC. Calculate the
electric force on the −3.0µC charge.
+2.5µC
−3.0µC +4.0µC
Solution
Step 1: We first need to calculate the electric forces each charge exerts on the
−3.0µC charge. Let’s denote the distance between the charges as r. The electric
force Fbetween two point charges q1and q2separated by a distance ris given
by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2).
Calculating the forces between the charges:
For +2.5µC and −3.0µC:
F2.5µC,−3.0µC =8.99 ×109·|2.5×10−6·(−3.0×10−6)|
r2
For +4.0µC and −3.0µC:
F4.0µC,−3.0µC =8.99 ×109·|4.0×10−6·(−3.0×10−6)|
r2
Step 2: Next, we need to find the electric force acting on the −3.0µC charge
due to the other two charges. Since the charges are arranged symmetrically in
an equilateral triangle, the magnitudes of F2.5µC,−3.0µC and F4.0µC,−3.0µC are
the same.
Therefore, the total electric force on the −3.0µC charge is:
Ftotal = 2 ·F2.5µC,−3.0µC
Step 3: Finally, calculate the total electric force on the −3.0µC charge.
Perform the necessary calculations to find the numerical value of this force.
13
Question 15
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed at points Aand B
respectively. Point Ais located at coordinates (0,0) in meters and point Bis
at coordinates (3,4). Calculate the magnitude of the electric force experienced
by q2due to q1.
Solution
Step 1: Calculate the distance between the two point charges. Given that point
Ais located at coordinates (0,0) and point Bis located at coordinates (3,4), the
distance rbetween the two point charges can be calculated using the distance
formula:
r=p(x2−x1)2+ (y2−y1)2
where x1= 0,y1= 0,x2= 3, and y2= 4. Therefore,
r=p(3 −0)2+ (4 −0)2=p32+ 42=√9 + 16 = √25 = 5 m
Step 2: Calculate the magnitude of the electric force. The magnitude of
the electric force between two point charges can be calculated using Coulomb’s
Law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant. Substitute the given
values q1=−3.0µC,q2= 5.0µC, and r= 5 m into the formula:
F= (8.99 ×109)| − 3.0×10−6×5.0×10−6|
52
F= 8.99 ×109×15 ×10−12
25 = 8.99 ×109×0.6×10−12 = 5.394 ×10−3N
Therefore, the magnitude of the electric force experienced by q2due to q1is
5.394 ×10−3N.
Question 16
Question
Two point charges, q1= 4.0µC and q2=−8.0µC, are placed 0.20 m apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
14
Solution
Given: q1= 4.0µC= 4.0×10−6Cq2=−8.0µC=−8.0×10−6Cr= 0.20 m
We know that the electric force between two charges q1and q2separated by
a distance ris given by Coulomb’s law:
F=k|q1q2|
r2
Step 1: Calculate the electric force constant k.
k=1
4πε0
The permittivity of free space ε0= 8.85 ×10−12 C2/N·m2Thus,
k=1
4π×8.85 ×10−12
Step 2: Substitute the given values into Coulomb’s law to find the magnitude
of the electric force.
F=1
4π×8.85 ×10−12 ·|4.0×10−6× −8.0×10−6|
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F=1
4π×8.85 ×10−12 ·32 ×10−12
0.04
F=1
4π×8.85 ×10−12 ·800 ×10−12
F=800
4π×8.85
F≈800
35.16
F≈22.74 N
Therefore, the magnitude of the electric force between the charges is approx-
imately 22.74 N.
Question 17
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 0.10 m apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
15
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k· |q1·q2|
r2
where k≈8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the distance between the charges.
Step 2: Plug in the given values to find the electric force:
F=(8.99 ×109N·m2/C2)·|−4.0×10−6C·8.0×10−6C|
(0.10 m)2
Step 3: Calculate the electric force:
F=8.99 ×109·4.0×10−6·8.0×10−6
0.01
F=287.68
0.01
F= 28768 N
Step 4: Since q1is negative and q2is positive, the force will be attractive.
Therefore, the electric force that q1exerts on q2is 28768 N, directed towards q1.
Question 19
Question
Three point charges are arranged on the x-axis as follows: +qat x= 0,−3qat
x= 4a, and +2qat x= 7a, where qand aare positive constants. Calculate the
net force on the +2qcharge due to the other charges.
Solution
Step 1: Calculate the force on the +2qcharge due to the +qcharge: The force
F1can be calculated using Coulomb’s Law, which states that the magnitude of
the force between two point charges is given by:
F=k·|q1·q2|
r2
where kis Coulomb’s constant, q1and q2are the charges, and ris the distance
between the charges.
The force on the +2qcharge due to the +qcharge is:
F1=k·|(+2q)(q)|
(7a−0)2
16
Step 2: Calculate the force on the +2qcharge due to the −3qcharge: The
force F2can be calculated using Coulomb’s Law as well:
F2=k·|(+2q)(−3q)|
(7a−4a)2
Step 3: Determine the direction of the net force: Since the forces due to the
+qand −3qcharges are in opposite directions, the net force on the +2qcharge
will be the difference between F1and F2:
Fnet =F1−F2
Fnet =k·2q·q
(7a)2−2q·3q
3a)2
Therefore, the net force on the +2qcharge due to the other charges can be
calculated by subtracting the force on the +2qcharge due to the −3qcharge
from the force due to the +qcharge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle with
sides of length d. Charge Qis placed at each vertex. Calculate the magnitude
of the electric force on one of the charges due to the other two charges.
Solution
Let’s denote the three charges as Q1,Q2, and Q3, with Q1being the charge we
want to calculate the force on. The distance between each charge and Q1is d.
Step 1: Calculate the electric force on Q1due to Q2. The electric force
between two charges is given by Coulomb’s Law:
F=k· |Q1|·|Q2|
r2
where kis the electrostatic constant 8.99×109N m2/C2and r=dis the distance
between the charges.
Plugging in the values, the force on Q1due to Q2is:
F1=8.99 ×109·Q·Q
d2=8.99 ×109·Q2
d2
Step 2: Calculate the direction of the force due to Q2. Since Q2is at one
vertex of an equilateral triangle, the force on Q1due to Q2will be along the
line connecting Q1and Q2. This force will have a direction opposite to that of
the line connecting Q1and Q2.
17
Step 3: Calculate the electric force on Q1due to Q3. Similarly, the force
on Q1due to Q3will be along the line connecting Q1and Q3and will have a
direction opposite to that of the line connecting Q1and Q3. Thus, the force
will be:
F2=8.99 ×109·Q2
d2
Step 4: Calculate the total electric force on Q1. Since the forces F1and F2
have the same magnitude and opposite directions, the total electric force on Q1
due to Q2and Q3is:
Ftotal =F1−F2=8.99 ×109·Q2
d2−8.99 ×109·Q2
d2= 0
Thus, the net electric force on Q1due to Q2and Q3is zero.
Question 22
Question
Two point charges, q1= +3.0µC and q2=−6.0µC, are placed 8.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units (Coulombs).
Given: q1= +3.0µC = 3.0×10−6C
q2=−6.0µC =−6.0×10−6C
Step 2: Find the distance between the charges in meters.
Given: r= 8.0cm = 8.0×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law,
F=k|q1q2|
r2.
Where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109)· |3.0×10−6· −6.0×10−6|
(8.0×10−2)2
Step 4: Calculate the magnitude of the electric force.
F=8.99 ×109·18 ×10−12
64 ×10−4
F=161.82 ×10−3
64 ×10−4
F=161.82
64 = 2.53 N
Therefore, the magnitude of the electric force between the charges is 2.53 N.
18
Question 23
Question
Two point charges Q1 = +3.0 µC and Q2 = -5.0 µC are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that Q1 exerts on
Q2.
Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Iden-
tify the direction of the force by considering the signs of the charges. Step 3:
Calculate the magnitude of the electric force using Coulomb’s law.
Step 1: Convert the charges from microcoulombs to coulombs: Q1 =
3.0µC = 3.0×10−6Cand Q2 = −5.0µC =−5.0×10−6C.
Step 2: The force will be attractive since Q1 is positive and Q2 is negative.
Step 3: Use Coulomb’s Law to find the magnitude of the electric force:
F=k|Q1·Q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), ris the distance between
the charges (0.1 m), Q1 = 3.0×10−6C, and Q2 = −5.0×10−6C.
Substitute the values into the equation:
F= (8.99 ×109)|3.0×10−6· −5.0×10−6|
(0.1)2
F= (8.99 ×109)15 ×10−12
0.01
F= (8.99 ×109)×1.5×10−9
F= 13.485 ×100
Therefore, F= 13.485N
The magnitude of the electric force that Q1 exerts on Q2 is 13.485 N, directed
towards Q1.
Question 24
Question
Two point charges, q1=−4.00 µC and q2= 3.00 µC, are separated by a distance
of 10.0 cm in a vacuum. Calculate the magnitude of the electric force between
the charges.
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Solution
Step 1: Convert the charges from microcoulombs to coulombs. Step 2: Deter-
mine the distance between the charges in meters. Step 3: Use Coulomb’s Law
to calculate the electric force between the charges.
Step 1: Convert the charges from microcoulombs to coulombs. q1=
−4.00 µC =−4.00 ×10−6C q2= 3.00 µC = 3.00 ×10−6C
Step 2: Determine the distance between the charges in meters. Given:
distance r= 10.0cm = 0.10 m
Step 3: Use Coulomb’s Law to calculate the electric force between the
charges. Coulomb’s Law: F=k|q1q2|
r2where k= 8.99 ×109Nm2/C2is the
electrostatic constant.
Substitute the known values into the equation: F=(8.99 ×109)×|−4.00 ×10−6×3.00 ×10−6|
(0.10)2
F=(8.99 ×109)×1.2×10−11
0.01
F=10.788 ×10−2
0.01
F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788
N.
Question 25
Question
Two point charges, q1=−4µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the given charges to coulombs. Recall that 1µC= 10−6C.
⇒q1=−4µC=−4×10−6C=−4×10−6C
⇒q2= 5 µC= 5 ×10−6C= 5 ×10−6C
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The formula for the magnitude of the electric force between two charges q1and
q2separated by a distance ris given by:
F=k
q1·q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Step 3: Substitute the given values into the formula.
F= 8.99 ×109
(−4×10−6)·(5 ×10−6)
(0.10)2
20
F= 8.99 ×109
−20 ×10−12
0.01
F= 8.99 ×109
−2×10−9
F= 8.99 ×109×2×10−9
F= 17.98 ×100
F= 17.98 N
Therefore, the magnitude of the electric force between the charges is 17.98
N.
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