COULOMB’S LAW - NUMERICAL EXPLORATIONS ON
CHARGE MAGNITUDE AND FORCE CALCULATION
1 INTRODUCTION
This document presents a series of numerical explorations related to Coulomb’s law. Each
exploration is accompanied by a detailed analysis and solution.
2 EXPLORATION 1: CHARGE MAGNITUDE
Two point charges are separated by a distance of 0.3 m. The electrostatic force between them
is 1.6 × 10−4 N. If one charge is 2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
1. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
2. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
3. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
3 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
1. Use Coulomb’s law directly.
2. Pay attention to units and scientific notation.
3. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
4 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
1. Start with Coulomb’s law and rearrange to solve for 𝑟.
2. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
5 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
1. Let the initial force be 𝐹1 and the final force be 𝐹2.
2. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
6 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
1. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
2. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
7 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
1. Work done is the change in potential energy.
2. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
8 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
1. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
2. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
9 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
1. Calculate the electric field due to each charge separately.
2. Use the superposition principle to find the net field.
3. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a distance of 0.3 m. The electrostatic force between them
is 1.6 × 10−4 N. If one charge is 2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
4. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
5. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
6. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
10 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
7. Use Coulomb’s law directly.
8. Pay attention to units and scientific notation.
9. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
11 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
10. Start with Coulomb’s law and rearrange to solve for 𝑟.
11. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
12 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
12. Let the initial force be 𝐹1 and the final force be 𝐹2.
13. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
13 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
14. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
15. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
14 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
16. Work done is the change in potential energy.
17. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
15 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
18. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
19. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
16 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
20. Calculate the electric field due to each charge separately.
21. Use the superposition principle to find the net field.
22. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a distance of 0.3 m. The electrostatic force between them
is 1.6 × 10−4 N. If one charge is 2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
23. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
24. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
25. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
17 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
26. Use Coulomb’s law directly.
27. Pay attention to units and scientific notation.
28. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
18 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
29. Start with Coulomb’s law and rearrange to solve for 𝑟.
30. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
19 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
31. Let the initial force be 𝐹1 and the final force be 𝐹2.
32. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
20 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
33. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
34. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
21 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
35. Work done is the change in potential energy.
36. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
22 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
37. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
38. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
23 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
39. Calculate the electric field due to each charge separately.
40. Use the superposition principle to find the net field.
41. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
42. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
43. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
44. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
24 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
45. Use Coulomb’s law directly.
46. Pay attention to units and scientific notation.
47. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
25 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
48. Start with Coulomb’s law and rearrange to solve for 𝑟.
49. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
26 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
50. Let the initial force be 𝐹1 and the final force be 𝐹2.
51. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
27 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
52. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
53. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
28 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
54. Work done is the change in potential energy.
55. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
29 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
56. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
57. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
30 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
58. Calculate the electric field due to each charge separately.
59. Use the superposition principle to find the net field.
60. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
61. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
62. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
63. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
31 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
64. Use Coulomb’s law directly.
65. Pay attention to units and scientific notation.
66. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
32 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
67. Start with Coulomb’s law and rearrange to solve for 𝑟.
68. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
33 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
69. Let the initial force be 𝐹1 and the final force be 𝐹2.
70. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
34 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
71. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
72. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
35 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
73. Work done is the change in potential energy.
74. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
36 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
75. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
76. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
37 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
77. Calculate the electric field due to each charge separately.
78. Use the superposition principle to find the net field.
79. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
80. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
81. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
82. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
38 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
83. Use Coulomb’s law directly.
84. Pay attention to units and scientific notation.
85. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
39 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
86. Start with Coulomb’s law and rearrange to solve for 𝑟.
87. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
40 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
88. Let the initial force be 𝐹1 and the final force be 𝐹2.
89. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
41 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
90. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
91. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
42 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
92. Work done is the change in potential energy.
93. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
43 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
94. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
95. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
44 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
96. Calculate the electric field due to each charge separately.
97. Use the superposition principle to find the net field.
98. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
99. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
100. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
101. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
45 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
102. Use Coulomb’s law directly.
103. Pay attention to units and scientific notation.
104. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
46 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
105. Start with Coulomb’s law and rearrange to solve for 𝑟.
106. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
47 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
107. Let the initial force be 𝐹1 and the final force be 𝐹2.
108. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
48 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
109. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
110. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
49 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
111. Work done is the change in potential energy.
112. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
50 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
113. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
114. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
51 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
115. Calculate the electric field due to each charge separately.
116. Use the superposition principle to find the net field.
117. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a distance of 0.3 m. The electrostatic force between them
is 1.6 × 10−4 N. If one charge is 2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
118. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
119. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
120. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
52 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
121. Use Coulomb’s law directly.
122. Pay attention to units and scientific notation.
123. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
53 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
124. Start with Coulomb’s law and rearrange to solve for 𝑟.
125. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
54 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
126. Let the initial force be 𝐹1 and the final force be 𝐹2.
127. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
55 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
128. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
129. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
56 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
130. Work done is the change in potential energy.
131. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
57 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
132. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
133. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
58 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
134. Calculate the electric field due to each charge separately.
135. Use the superposition principle to find the net field.
136. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
137. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
138. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
139. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
59 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
140. Use Coulomb’s law directly.
141. Pay attention to units and scientific notation.
142. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
60 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
143. Start with Coulomb’s law and rearrange to solve for 𝑟.
144. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
61 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
145. Let the initial force be 𝐹1 and the final force be 𝐹2.
146. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
62 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
147. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
148. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
63 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
149. Work done is the change in potential energy.
150. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
64 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
151. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
152. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
65 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
153. Calculate the electric field due to each charge separately.
154. Use the superposition principle to find the net field.
155. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
156. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
157. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
158. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
66 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
159. Use Coulomb’s law directly.
160. Pay attention to units and scientific notation.
161. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
67 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
162. Start with Coulomb’s law and rearrange to solve for 𝑟.
163. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
68 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
164. Let the initial force be 𝐹1 and the final force be 𝐹2.
165. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
69 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
166. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
167. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
70 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
168. Work done is the change in potential energy.
169. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
71 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
170. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
171. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
72 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
172. Calculate the electric field due to each charge separately.
173. Use the superposition principle to find the net field.
174. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
175. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
176. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
177. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
73 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
178. Use Coulomb’s law directly.
179. Pay attention to units and scientific notation.
180. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
74 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
181. Start with Coulomb’s law and rearrange to solve for 𝑟.
182. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
75 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
183. Let the initial force be 𝐹1 and the final force be 𝐹2.
184. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
76 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
185. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
186. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
77 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
187. Work done is the change in potential energy.
188. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
78 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
189. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
190. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
79 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
191. Calculate the electric field due to each charge separately.
192. Use the superposition principle to find the net field.
193. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
194. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
195. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
196. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
80 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
197. Use Coulomb’s law directly.
198. Pay attention to units and scientific notation.
199. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
81 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
200. Start with Coulomb’s law and rearrange to solve for 𝑟.
201. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
82 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
202. Let the initial force be 𝐹1 and the final force be 𝐹2.
203. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
83 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
204. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
205. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
84 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
206. Work done is the change in potential energy.
207. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
85 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
208. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
209. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
86 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
210. Calculate the electric field due to each charge separately.
211. Use the superposition principle to find the net field.
212. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
Two point charges are separated by a
distance of 0.3 m. The electrostatic force between them is 1.6 × 10−4 N. If one charge is
2 × 10−6 C, what is the magnitude of the other charge?
Analysis:
213. Recall Coulomb’s law: 𝐹=𝑘|𝑞1𝑞2|
𝑟2
214. We know 𝐹, 𝑟, and one charge (𝑞1). We need to find 𝑞2.
215. Rearrange the equation to solve for 𝑞2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
1.6×10−4 =(8.99×109)|(2×10−6)𝑞2|
(0.3)2
𝑞2=1.6×10−4×0.32
8.99×109×2×10−6
𝑞2=8×10−7 C
87 EXPLORATION 2: FORCE CALCULATION
Calculate the electrostatic force between two charges of 3 × 10−6 C and −2 × 10−6 C separated
by a distance of 0.1 m in air.
Analysis:
216. Use Coulomb’s law directly.
217. Pay attention to units and scientific notation.
218. The negative sign in one charge doesn’t affect the magnitude of the force.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
=(8.99×109)|(3×10−6)(−2×10−6)|
(0.1)2
=8.99×109×6×10−12
0.01
=5.394×10−3 N
88 EXPLORATION 3: DISTANCE DETERMINATION
Two point charges, 4 × 10−6 C and −3 × 10−6 C, experience an electrostatic force of 0.054 N.
What is the distance between them?
Analysis:
219. Start with Coulomb’s law and rearrange to solve for 𝑟.
220. Use the square root to find 𝑟 from 𝑟2.
Solution:
𝐹 =𝑘|𝑞1𝑞2|
𝑟2
0.054 =(8.99×109)|(4×10−6)(−3×10−6)|
𝑟2
𝑟2=8.99×109×12×10−12
0.054
𝑟 =√8.99×109×12×10−12
0.054
𝑟 =0.2 m
89 EXPLORATION 4: CHARGE RATIO
Two point charges 𝑞1 and 𝑞2 are separated by a fixed distance. If 𝑞1 is doubled and 𝑞2 is
halved, how does the electrostatic force change?
Analysis:
221. Let the initial force be 𝐹1 and the final force be 𝐹2.
222. Express 𝐹2 in terms of 𝐹1 using the charge changes.
Solution:
𝐹1=𝑘|𝑞1𝑞2|
𝑟2
𝐹2=𝑘|(2𝑞1)(0.5𝑞2)|
𝑟2
=𝑘|𝑞1𝑞2|
𝑟2=𝐹1
The force remains unchanged.
90 EXPLORATION 5: SYSTEM OF CHARGES
Three charges are placed on a straight line. Charges of 2 × 10−6 C and −3 × 10−6 C are placed
0.1 m apart. Where should a third charge of 4 × 10−6 C be placed so that the net force on it is
zero?
Analysis:
223. Let the distance of the third charge from the 2 × 10−6 C charge be 𝑥.
224. Set up an equation where the forces from the two charges on the third charge are equal
and opposite.
Solution:
𝑘(2×10−6)(4×10−6)
𝑥2=𝑘(3×10−6)(4×10−6)
(0.1−𝑥)2
2
𝑥2=3
(0.1−𝑥)2
√2(0.1−𝑥)=√3𝑥
0.1√2 =𝑥(√2+√3)
𝑥 = 0.1√2
√2+√3≈0.0414 m
91 EXPLORATION 6: WORK DONE
Calculate the work done in moving a charge of 5 × 10−6 C from a distance of 0.2 m to 0.5 m
from a fixed charge of −3 × 10−6 C.
Analysis:
225. Work done is the change in potential energy.
226. Use the electric potential energy formula: 𝑈=𝑘𝑞1𝑞2
𝑟
Solution:
𝑊 =𝑈𝑓−𝑈𝑖=𝑘𝑞1𝑞2
𝑟𝑓−𝑘𝑞1𝑞2
𝑟𝑖
=(8.99×109)×(−3×10−6)×(5×10−6)×(1
0.5−1
0.2)
=−134.85×10−3+337.125×10−3
=202.275×10−3 J=0.202275 J
92 EXPLORATION 7: ELECTRIC FIELD
At what distance from a point charge of 2 × 10−6 C is the electric field strength 100 N C−1?
Analysis:
227. Use the electric field formula: 𝐸=𝑘𝑞
𝑟2
228. Solve for 𝑟.
Solution:
𝐸 =𝑘𝑞
𝑟2
100 =(8.99×109)2×10−6
𝑟2
𝑟2=8.99×109×2×10−6
100
𝑟 =√8.99×109×2×10−6
100
𝑟 =0.4242 m
93 EXPLORATION 8: SUPERPOSITION
Two point charges, 3 × 10−6 C and −2 × 10−6 C, are placed 0.1 m apart. Calculate the electric
field at a point 0.05 m from the positive charge on the line joining the charges.
Analysis:
229. Calculate the electric field due to each charge separately.
230. Use the superposition principle to find the net field.
231. Pay attention to the direction of each field.
Solution:
𝐸1=𝑘𝑞1
𝑟12=(8.99×109)3×10−6
(0.05)2=10788 N/C (right)
𝐸2=𝑘𝑞2
𝑟22=(8.99×109)2×10−6
(0.05)2=7192 N/C (left)
𝐸net =𝐸1−𝐸2=10788−7192=3596 N/C (right)
