phy hw

Week2.zip

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Week 2/Week 2/MidTerm2.pdf

MidTerm 2 Name

1. Current enters a resistor at terminal a and leaves at terminal b. a) (2) T F The current at terminal b is less than the current at terminal a. b) (2) T F The electric potential at terminal b is less than the electric potential at terminal a. c) (2) T F The direction of electron flow in the resistor is from a to b.Ω

2. Three resistors RA > RB > RC are wired in series in some circuit. Their respective currents, Ia, Ib, Ic, a. are all equal: Ia = Ib = Ic. b. are related by Ia > Ib > Ic. c. are related by Ia < Ib < Ic.Ω

3. Three resistors RA > RB > RC are wired in parallel in some circuit. Their respective currents, Ia, Ib, Ic, a. are all equal: Ia = Ib = Ic. b. are related by Ia > Ib > Ic. c. are related by Ia < Ib < Ic.Ω

4. T F The sum of the voltage changes around any loop of a circuit is zero. 5. T F The sum of currents entering a node of a circuit equal those leaving that node. 6. T F A real battery’s terminal voltage can be greater than its emf E.Ω 7. Consider the circuit here with resistors RA, RB and RC . Rank the resistors according to the currents IA, IB, IC that flow through them, from largest to smallest (like IC > IB > IA or use “=” sign if two are the same).

8. Consider the circuit here with resistors RA, RB and RC . Rank the resistors according to the voltage drops VA, VB, VC across them, from largest to smallest (like VA > VB > VC or use “=” sign if two are the same).

9. In the circuit shown, the resistor R = 44.0Ω is using 2.50 W of electrical power. The resistance r = 0.320Ω is the internal resistance of the E-volt battery.

a) (4) How large is the current through the battery?Ω

b) How large is the terminal voltage of the battery?

c) How large is the emf of the battery?

10. In the diagram, each resistor is 1.0 kΩ. 11. What pair of resistors is in series?

a. A & B. b. B & C. c. C & D. d. D & E. e. no pair is in series. 12. What pair of resistors is in parallel?

a. A & B. b. B & C. c. C & D. d. D & E. e. no pair is in parallel. 13. Calculate the equivalent resistance between the terminals a and b.Ω

11. A uniform magnetic field points out of the page as shown. Three particles shot into the region curve as shown. For each particle, indicate whether it has negative (-), zero (0), or positive (+) electric charge:

particle a: particle b: particle c:

12. A solenoid carries a current as shown. 13. The direction of its magnetic field at point A is closest toΩ

a. ↑ b. ↓ c. d. → e. ⊗ f. � b) The direction of its magnetic field at point B is closest to

a. ↑ b. ↓ c. d. → e. ⊗ f. � c) The direction of its magnetic field at point C is closest to

a. ↑ b. ↓ c. d. → e. ⊗ f. � 13. A 120.0-meter long straight wire carries a current of 5.00 A towards the north. The Earth’s 0.550-gauss magnetic field points north but 56.0◦ below horizontal.

14. In what direction is the force F~ on the wire? Draw and label F~ on the diagram, or use or← if F~ is out-of-the-page or into-the-page. 15. Calculate the magnitude of the magnetic force on the wire.Ω

14. A bar magnet is set close to a current-carrying coil as shown. The force between bar magnet and coil is

a. zero. b. attractive. c. repulsive.

15. A region has a uniform magnetic field pointing vertically on the page as shown. 16. The north pole of a compass needle placed in this magnetic field will point:Ω

a. ↑ b. ↓ c. d. → e. ⊗ f. � b) The magnetic force on a proton instantaneously moving out of the page points:

a. ↑ b. ↓ c. d. → e. ⊗ f. � c) The magnetic force on a wire carrying a currrent into the page points:

a. ↑ b. ↓ c. d. → e. ⊗ f. � 16. A doubly charged helium ion (an alpha particle, mass m = 6.64 × 10−27 kg) has been accelerated through a potential di erence of -1250 V. It is projected into a region of uniform magnetic field and then performs uniform circular motion with a period of 0.108 µs in a plane perpendicular to the field.

17. How large is its kinetic energy, in eV?Ω

b) Find the speed of the helium ion in m/s.

c) Calculate the strength of the magnetic field it is moving in, in tesla. (Hint: Apply Newton’s 2nd Law.)

17. Two long straight wires separated by 3.00 m carry 25.0 A currents. I1 is out of the page and I2 is into the page. 18. Determine the magnitude of the net magnetic field produced at point P, between the two wires and 2.00 m from I1 and 1.00 m from I2.Ω

b) What is the direction of the net magnetic field at P?

18. Imagine you take the south pole of a bar magnet and move it towards a square loop of wire facing you. The induced current in the wire loop is

a. zero. b. clockwise. c. counterclockwise.Ω

19. A wire loop is near a current as shown. The induced current in the wire loop is

a. zero. b. clockwise. c. counterclockwise.Ω

20. The magnetic field perpendicular to a circular wire loop of diameter 4.0 cm and resistance 0.040 Ωchanges from +0.68 T to −0.32 T in a time of 55 ms, where + means the field points away from the observer and − means towards the observer.

a. clockwise. b. counterclockwise.a) For the observer, in what direction is the induced current? b) Calculate the magnitude of the induced emf, according to Faraday’s Law of Induction.Ω

21. The AC power from a small generator running on gasoline comes out at 50.0 Hz, with 48.0 volts rms. The output of the generator is connected directly to a light bulb whose resistance is 12.0Ω .

22. How large is the rms current through the lightbulb?Ω

b) How large is the peak voltage across the lightbulb?

c) What average electrical power is the lightbulb using?

22. An ideal transformer has 2.00×103 turns on the primary side and 1.00×102 turns on the secondary side. The rms current of the primary side is 1.20 A while connected to 124 VAC, 60.0 Hz power. A load is connected to the secondary side.

a) This transformer is connected as? a. step-up. b. step-down. a) How large is the rms voltage across the secondary side of the transformer?

b)How large is the rms current through the secondary side of the transformer?

Prefixes a=10−18, f=10−15, p=10−12, n=10−9, µ = 10−6, m=10−3, c=10−2, k=103, M=106, G=109, T=1012, P=1015

Physical Constants

k = 1/4π�0 = 8.988 GNm2/C2 (Coulomb’s Law) �0 = 1/4πk = 8.854 pF/m (permittivity of space) e = 1.602 × 10−19 C (proton charge) µ0 = 4π × 10−7 T·m/A (permeability of space) me = 9.11 × 10−31 kg (electron mass) mp = 1.67 × 10−27 kg (proton mass)

Units

NA = 6.02 × 1023/mole (Avogadro’s #) 1 u = 1 g/NA = 1.6605 × 10−27 kg (mass unit) 1.0 eV = 1.602 × 10−19 J (electron-volt) 1 V = 1 J/C = 1 volt = 1 joule/coulomb 1 F = 1 C/V = 1 farad = 1 C2/J 1 H = 1 V·s/A = 1 henry = 1 J/A2 1 A = 1 C/s = 1 ampere = 1 coulomb/second 1 Ω = 1 V/A = 1 ohm = 1 J·s/C2 1 T = 1 N/A·m = 1 tesla = 1 newton/ampere·meter 1 G = 10−4 T = 1 gauss = 10−4 tesla

Vectors

Written ~V or V, described by magnitude=V , direction=θ or by components (Vx, Vy). Vx = V cos θ, Vy = V sin θ,

V = √

V 2x + V 2y , tan θ = Vy Vx

. θ is the angle from ~V to +x-axis.

Addition: A + B, head to tail. Subtraction: A − B is A + (−B), −B is B reversed.

Trig summary

sin θ = (opp) (hyp)

, cos θ = (adj) (hyp)

, tan θ = (opp) (adj)

, (opp)2 + (adj)2 = (hyp)2.

sin θ = sin(180◦ − θ), cos θ = cos(−θ), tan θ = tan(180◦ + θ), sin2 θ + cos2 θ = 1.

Chapter 16 Equations

Charges: Q = ±N e, ∆Q1 + ∆Q2 = 0, e = 1.602 × 10−19 C.

Electric Force: F = k Q1Q2

r2 , k = 8.988 × 109 N·m2/C2, F = Q1Q2

4π�0r2 , �0 = 14πk = 8.854 pF/m.

~F = ~F1 + ~F2 + ~F3 + ... superposition of forces. Electric Field:

~E = ~F q , q= test charge. Or: ~F = q ~E.

|~E| = E = k Q r2

= Q 4π�0r2

, due to point charge. Negative Q makes inward ~E, positive Q makes outward ~E. ~E = ~E1 + ~E2 + ~E3 + ... superposition of many electric fields. E = k Q

r2 = electric field around a point charge or outside a spherical charge distribution.

Chapter 17 Equations

Potential Energy and Work: Wba = FE d cos θ = work done by electric force FE on test charge, in displacement d from a to b. Wba = −q∆V = −q(Vb − Va) = work done by electric force on a test charge, moved from a to b. ∆PE = q∆V = q(Vb − Va) = change in electric potential energy of the system. Also: ∆PE = −Wba. ∆KE + ∆P E = 0, or, ∆KE = −∆P E = −q∆V , principle of conservation of mechanical energy.

Potential: ∆V = ∆PE

q = definition of change in electric potential.

∆V = Ed = potential change in a uniform electric field.

Eq.-1

V = k Q r

= potential produced by a point charge or outside a spherical charge distribution. PE = qV = potential energy for a test charge at a point in a field. PE = k Q1Q2

r12 = potential energy of a pair of charges.

Capacitance: Q = CV , C = K�0 Ad = capacitor equations.

U = 1 2 QV = 1

2 CV 2 = 1

2 Q2

C = stored energy.

E = Q/A �0

= electric field strength very near a charged conductor.

Chapter 18 Equations

Electric current and power: I = ∆Q

∆t , ∆Q = I∆t current definition. V = IR, I = V /R Ohm’s law.

R = ρL/A calculation of resistance. ρ = ρ0[1 + α(T − T0)] resistivity changes. P = IV , P = I2R, P = V 2/R. P = instantaneous work/time.

Alternating current: V = V0 sin 2πf t = time-dependent AC voltage. I = I0 sin 2πf t = time-dependent AC current.

Vrms = √

V 2 = V0/ √

2 = root-mean-square voltage. Irms = √

I2 = I0/ √

2 = root-mean-square current. AC power in resistors:

P = 1 2 I20 R =

1 2 V 20 /R =

1 2 I0V0 = average power. P = I2rmsR = V

2 rms/R = IrmsVrms = average power.

Chapter 19 Equations

Resistor Combinations Req = R1 + R2 + R3 + ... (series) 1Req =

1 R1

+ 1 R2

+ 1 R3

+ ... (parallel) Real batteries

Vab = E − Ir (terminal voltage) Vab = IR (connected to load R) Kirchhoff’s Rules∑

∆V = 0 (loop rule, energy conservation) ∑

I = 0 (node rule, charge conservation)

Chapter 20 Equations

Magnetic forces, torque F = IlB sin θ (on a current) F = qvB sin θ (on a moving charge) F/l = µ0

2π I1I2

d (between currents) F = qvB = mv2/r (during cyclotron motion)

τ = N BAI sin θ (torque on a coil) v = ωr = 2πf r = 2πr/T (circular motion) Magnetic Fields

B = µ0 2π

I r

(due to long straight wire) B = µ0IN/l (inside a solenoid)

Right Hand Rules Force (thumb) = [I (4 fingers)] × [magnetic field (palm)] (force on a current) Force (thumb) = [qv (4 fingers)] × [magnetic field (palm)] (force on a moving charge) Current (thumb) ⇐⇒ [magnetic field (4 fingers)] (magnetic field around a wire) Current (4 fingers) ⇐⇒ [magnetic field (thumb)] (magnetic field inside a current loop)

Chapter 21 Equations

Faraday’s Induced EMF ΦB = BA cos θ (magnetic flux) E = −N ∆ΦB∆t (induced emf) E = Blv (moving conductor) E = N BAω sin ωt (AC generator) V −E = IR (motor’s counter-emf) E1 = −M ∆I2∆t (mutual inductance emf) VS /VP = NS /NP (transformer equation) IP VP = IS VS (power in = power out)

Eq.-2

Week 2/Week 2/6.pdf

Electromagnetic Induction

Chapter 21

Faraday’s Law

Part 1

Faraday’s observation

Faraday: If the magnetic field changes, or if the magnet and coil are in relative motion, there will be an induced emf (and therefore current) in the coil.

Electric currents produce magnetic fields. 19th century puzzle: Can magnetic fields produce currents? A static magnet will produce no current in a stationary coil.

Magnetic Flux For a “loop” of wire (not necessarily circular) with area A, in an external magnetic field B, the magnetic flux through the loop is:

SI units of Magnetic Flux: 1 T·m2 = 1 weber = 1 Wb

A = area of loop θ = angle between B and the normal to the loop

ϑcosBA=Φ

Induced emf (Voltage) from changing Magnetic Flux

Faraday: If the magnetic field changes, or if the magnet and coil are in relative motion, there will be an induced emf (and therefore current) in the coil.

Key Concept: The magnetic flux through the coil must change, this will induce an emf e in the coil, which produces a current I = emf/R in the coil.

Such a current is said to be induced by the time varying magnet flux that “links” the coil.

Problem A magnetic field is oriented at an angle of 32º to the normal of a rectangular area 5.5 cm by 7.2 cm. If the magnetic flux through this surface has a magnitude of 4.8 × 10-5 T·m2, what is the strength of the magnetic field?

cos or cos

4.8 10 14.3 mT

cos 0.055 0.072 cos 32

BA B A

B A

θ θ

−

Φ Φ = =

Φ × = = =

× ×

Faraday’s Law of Induction

Faraday’s Law: The instantaneous EMF (voltage) induced in a circuit (w/ N loops) equals the rate of change of magnetic flux through the circuit:

if tt

N t

N − Φ−Φ

−= Δ ΔΦ

−=ε

The minus sign just indicates the direction of the induced emf. To calculate the magnitude, we will use:

if tt

N t

N − Φ−Φ

= Δ ΔΦ

=ε

Faraday’s Law of Induction

iiifff

tt ABAB

tt − −

−= − Φ−Φ

−= θθ

ε coscos

Induction by relative motion

iiifff

tt ABAB

tt − −

−= − Φ−Φ

−= θθ

ε coscos

tt BB

A − −

−= θε cos

Induction by Rotational Motion

As a coil rotates in a constant magnetic field (uniform or not) the flux through the loop changes, inducing an emf in the coil.

iiifff

tt ABAB

tt − −

−= − Φ−Φ

−= θθ

ε coscos

tt BA

− −

−= θθ

ε coscos

Good to discuss The graph shows the magnitude B of a uniform magnetic field that is perpendicular to the plane of a conducting loop. Rank the five regions indicated on the graph according to the magnitude of the emf induced in the loop, from least to greatest.

Problem This is a plot of the magnetic flux through a coil as a function of time. At what times shown in this plot does (a) the magnetic flux and (b) the induced emf have the greatest magnitude?

Problem A 0.25 T magnetic field is perpendicular to a circular loop of wire with 50 turns and a radius 15 cm. The magnetic field is reduced to zero in 0.12 s. What is the magnitude of the induced emf?

( )( )2 and

0 0.25 0.15 50 50 7.36 volts

0.12 0.12 f i

emf N emf N t t

emf π

ΔΦ ΔΦ = − =

Δ Δ

− × ×Φ − Φ = = =

Lenz’s Law

Part 2

Lenz’s Law

Lenz’s Law: An induced current always flows in a direction that opposes the change that caused it.

In this example the magnetic field in the downward direction through the loop is increasing. So a current is generated in the loop which produces an upward magnetic field inside the loop to oppose the change.

Magnet moving down toward loop

Induced current

Induced B field

Induction by Relative Motion

• When a permanent magnet moves relative to a coil, the magnetic flux through the coil changes (WHY?), inducing an emf in the coil.

• In a) the magnitude of the flux is increasing

• In c) the flux is decreasing in magnitude.

• In a) and c) the induced current is in the opposite direction (Lenz’s law).

Lenz’s Law Good to discuss A square loop of wire lies in the plane of the page. A decreasing magnetic field is directed into the page. The induced current in the loop is: A) counterclockwise B) clockwise C) zero D) depends upon whether or not B is decreasing at a constant rate E) clockwise in two of the loop sides and counterclockwise in the other two

Good to discuss A long, straight wire carries a steady current I. A rectangular conducting loop lies in the same plane as the wire, with two sides parallel to the wire and two sides perpendicular. Suppose the loop is pushed toward the wire as shown. Given the direction of I, the induced current in the loop is 1. clockwise. 2. counterclockwise. 3. need more information

Problem The figure shows a circuit containing a resistor and an uncharged capacitor. Pointing into the plane of the circuit is a uniform magnetic field B. If the magnetic field increases in magnitude with time, which plate of the capacitor (top or bottom) becomes positively charged? The magnetic flux through the circuit is changing so that there are more magnetic field lines into the paper "linking" the circuit. According to Lenz's law the induced cirrent opposes the change. Thus the induced current is counter clock wise. Positive charge builds up on

the bottom plate.

Motional EMF

Part 3

If the moving moving conductor is part of a circuit, then the magnetic flux through the circuit will change with time and a current will be induced (Area of loop = ls):

x x

v l

(1) l s

N B Blv t t

ε ΔΦ Δ

= = = Δ Δ

Problem The figure shows a zero-resistance rod sliding to the right on two zero-resistance rails separated by the distance L = 0.45 m. The rails are connected by a 12.5 Ω resistor, and the entire system is in a uniform magnetic field with a magnitude of 0.75 T.

(a) If the velocity of the bar is 5.0 m/s to the right, what is the current in the circuit? (b) What is the direction of the current in the circuit? (c) What is the magnetic force on the bar? (d) What force must be applied to keep the bar moving at constant velocity?

0.75 0.45 5 a) or 0.14 A

12.5 Blv

emf Blv IR I R

× × = = = = =

b) clockwise

c) to the left .

0.75 0.45 0.14 0.046 N

F BlI

= × × =

d) A force must be applied to the right equal to 0.046 N.

Motional emf An emf will also be produced if a conductor moves through a magnetic field. The emf comes from the motion of charges, which are free to move in the conductor. In this example, why do positive charges collect at the top of the rod?

x x

v L

+ +

Computer Simulation

Generators A generator is a device that converts mechanical energy to electrical energy. Consider a current loop which rotates in a constant magnetic field: The magnetic flux through the loop changes, so an emf is induced. If a loop of area A with N turns rotates with angular speed ω (period of rotation T = 2π/ω) in a constant B field, then the instantaneous induced emf is:

If this loop is part of a circuit, this emf will induce an Alternating Current (AC) in the circuit.

tNBA ωωε sin=

Generator

A coil of wire turns in a magnetic field. The flux in the coil is constantly changing, generating an emf in the coil.

Transformers A transformer is a device used to change the voltage in a circuit. AC currents must be used.

s N N

V V

I I

== p = primary

s = secondary

Step-down transformers 240,000 V in the power lines

120 V in houses

2,400 V local substations

Eddy Currents When a conductor is moved in a magnetic field, there is a force on the electrons (remember they are free to move in a conductor), which then move in the metal. This movement is called an eddy current.

The induced currents produce magnetic fields which tend to oppose the motion of the metal.

××

× ×

× × ×

× ×

Self-inductance

Part 4

Solenoids

If we stack several current loops together we end up with a solenoid. In the limit of a very long solenoid, the magnetic field inside is very uniform:

B=μ0nI n = number of windings per unit length, I = current in windings

Self-Inductance

If you try to change the current instantaneously in a circuit containing a solenoid, the response will instead be gradual. This is because the changing magnetic flux through the solenoid produces a self-induced emf to initially oppose any changes as prescribed by Lenz’s Law. This effect is known as self-induction. Actually all circuit elements have some self-inductance. However, only coils with many turns of wire, such as in a solenoid, have substantial self-inductance.

Inductance The self-induced emf is given by:

where L is defined as the inductance of the circuit, usually residing in the “inductor. For any geometry of loop, the magnetic flux through the loop, produced by current in the loop is proportional to the current. The inductance L is the constant of proportionality.

The unit of inductance is the Henry 1 H = 1 T·m2/A = 1 (T·m2/s) (s/A) = 1 V·s/A Note that inductance, like capacitance, is purely geometrical.

t I

L t

N Δ Δ

= Δ ΔΦ

=ε

I NLLIN Δ ΔΦ

=⇒=Φ

Inductance of a Solenoid

A solenoid has inductance given by

l l

AnA N

L 20 2

0 μμ =⎟ ⎟ ⎠

⎞ ⎜ ⎜ ⎝

⎛ =

L = inductance of the solenoid N = number of turns in solenoid l = length of solenoid A = cross sectional area of solenoid n = number turns per length

Problem The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross- sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?

( )

272 30

3 3 2 27

4 10 450 a) 7.3 10

0.24 0.24

7.3 10 6.88 10 m 4 10 450

AN A L

πμ

− −

−

× × × = = = ×

= × × = × × ×

( )3 3

b) but or

0 3.2 so 7.3 10 0.425 volts

55 10

emf N L N N L I t I

I emf L

t −

−

ΔΦ ΔΦ = − = ΔΦ = Δ

Δ Δ −Δ

= − = − × × = Δ ×

RL Circuits We can construct a circuit out of inductors and resistors. The circuit will behave similar to an RC circuit, with a time constant given by: τ = L/R

)1()1( // LtRt e R

e R

I −− −=−= εε τ

Week 2/Week 2/5.pdf

Magnetism

Chapter 20 Part 1

The Magnetic Field

What creates magnetic fields?

There is no magnetic charge!

Any permanent magnet has two poles.

What if to cut a permanent magnet in half?

Unlike electrostatics: Magnetic monopoles have never been detected.

Interaction between magnetic poles

Opposite magnetic poles attract each other, and like poles repel each other

Magnetic Field Lines

Assume that there is a magnetic field in some area of space

We can represent magnetic fields with field lines, as we did for electric fields

(1) the direction of the tangent to a magnetic field line at any point gives the direction of at that point

(2) the spacing of the lines represents the magnitude of

B r

Magnetic Field Lines (cont.)

The field lines enter one end of a magnet and exit the other end.

The end of a magnet from which the field lines emerge is called the north pole of the magnet

the other end, where field lines enter the magnet, is called the south pole

The Earth’s Magnetic Field

The spinning iron core of the earth produces a magnetic field.

The magnetic north pole corresponds to the geographic south pole.

South _____

Magnetic Force on a Particle

Part 2

Magnetic force on a charged particle

q – electric charge v – particle velocity B – magnetic field the magnitude

BvqFB rrr

×=

φsinvBqFB =

Right Hand Rule / Teddy Bear Rule / Alligator Rule

Right Hand Rule: Positive and negative particle

BvqFB rrr

×=

The force acting on a charged particle moving through a magnetic field is always perpendicular to the velocity and the field

Magnetic field (definition)

q – electric charge v – particle velocity B – magnetic field

SI unit: Tesla 1 T = newton/(C*m/s) = 1 N/(A*m) 1 tesla = 104 gauss (G)

vq F

B B=

φsinvBqFB =

Notation

To depict a vector oriented perpendicular to the page we use crosses and dots.

A cross indicates a vector going into the page (think of the tail feathers of an arrow disappearing into the page).

A dot indicates a vector coming out of the page (think of the tip of an arrow coming at you out of the page).

B out of the page B into the page

×·

Direction of Magnetic Forces (cont.) The direction of the magnetic force on a moving charge is perpendicular to the plane formed by B and v.

To determine the direction, you must apply the Right Hand Rule (RHR).

Good to discuss

A magnetic field exerts a force on a charged particle: A) always B) never C) if the particle is moving across the field lines D) if the particle is moving along the field lines E) if the particle is at rest

φsinvBqFB =

? Example A charge of 23 mC is moving in the negative x direction at 4 m/s. A magnetic field of 30 T is pointing in the positive y direction. What is the magnitude and direction of the force on the charge? How does your answer change if the charge is –23 mC?

sin 23 10 4 30 sin 90

2.76 10 N

F qvB

θ −

−

= = × × × ×

= ×

o r

Use the right-hand rule!

ˆDirection is , i.e. into the paper

If the charge is negative,

ˆ Direction is .

−

problem

Part 3

The Motion in a Magnetic Field

Motion of Charges in B Fields

If a charged particle is moving in a direction perpendicular to a uniform magnetic field, then its trajectory will be a circle because the force F=qvB is always perpendicular to the velocity, and therefore centripetal.

r mv

maFc 2

== r vm

qvBF 2

qB vm

r =

Recall that so

The radius of the circular trajectory

The period (the time for one full revolution) qB m

v r

T ππ 22

Good to discuss

An electron and a proton are both initially moving with the same speed and in the same direction at 900 to the same uniform magnetic field. They experience magnetic forces, which are initially: A) identical B) equal in magnitude but opposite in direction C) in the same direction and differing in magnitude by a factor of 1840 D) in opposite directions and differing in magnitude by a factor of 1840 E) equal in magnitude but perpendicular to each other

F qvB mv r

= = 2 ?

Good to discuss

An electron and a proton each travel with equal speeds around circular orbits in the same uniform magnetic field, as shown in the diagram (not to scale). The field is into the page on the diagram. (a) Where is the electron (b) What is the direction

qB mv

r =

Good to discuss A uniform magnetic field is directed into the page. A charged particle, moving in the plane of the page, follows a clockwise spiral of decreasing radius as shown. A reasonable explanation is: A) the charge is positive and slowing down B) the charge is negative and slowing down C) the charge is positive and speeding up D) the charge is negative and speeding up E) none of the above

qB mv

r =

? Good to discuss

Cosmic rays (atomic nuclei stripped bare of their electrons) would continuously bombard Earth’s surface if most of them were not deflected by Earth’s magnetic field. Given that Earth is, to an excellent approximation, a magnetic dipole (a bar magnet), the intensity of cosmic rays bombarding its surface is greatest at the 1. poles. 2. mid-latitudes. 3. equator.

Isotope Separation qB mv

r = Crossed E and B fields

Magnetic Force on a Current

Part 4

Force on a Current Carrying Wire Current in a wire is a collection of moving charges; therefore, a current carrying wire in a magnetic field also experiences a force. If a wire of length L, carrying a current I, makes an angle θ with a magnetic field B, then the magnitude of the force on the wire is:

θ Ι

• F

θϑ

ϑϑ

sinsin)(

sin)(sin

ILBvB v L

vBItqvBF

=== θsinILBF =

Force on a Current Carrying Wire θsinILBF =

Force on a Current Carrying Wire θsinILBF = The diagram shows a straight wire carrying a flow of electrons into the page. The wire is between the poles of a permanent magnet. The direction of the magnetic force exerted on the wire is: A) → B) ← C) ↓ D) ↑ E) into the page

Magnetic Levitation (Maglev, etc.)

mgILB=

Magnetic Torque on current loop

In a uniform magnetic field, the net force on a current loop (independent of geometry) is 0. However, there can be a torque

r = distance from axis of rotation to loop segment F = magnetic force on segment θ = angle between vector r and vector F.

Only the vertical segments of the loop experience a force. The torque will rotate the loop so that the plane of the loop is perpendicular to the magnetic field.

θτ sinrF=

)( 22

hwIB w

IhB w

IhB =+=τ

IBA=τ

Magnetic Torque on current loop (cont.)

Torque exerted on a rectangular loop of area A

If the loop has N turns, then

θτ sinIBA=

θτ sinNIBA=

A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic field parallel to the side MK as indicated. The loop will tend to rotate: A) about PQ with KL coming out of the page B) about PQ with KL going into the page C) about RS with MK coming out of the page D) about RS with MK going into the page E) about an axis perpendicular to the page

Magnetic Fields Due to Currents

Part 5

Experimental observation in 1820

Hans Oersted:

Electric currents can create magnetic fields

The magnitude of the field produced at point P

A general equation

for Physics 232

AmT r

ids dB

/1026.1

sin 4

6 0

2 0

⋅×=

−μ

ϑ π μ

A Long Straight Wire

Proof: integration or Ampere’s law

AmT R i

/1026.1 2

6 0

⋅×=

−μ π μ

Conceptual question The equation above is true for an infinitely long, straight conductor carrying a current.

Of course, there is no such thing as an infinitely long anything. How would you decide whether a particular wire is long enough to be considered infinite?

? R i

B π μ 2

0= Problem

Consider the long, straight, current-carrying wires shown in the figure. One wire carries a current of 6.2 A in the negative y direction; the other carries a current of 4.5 A in the positive x direction. Calculate the magnitude and direction of the net magnetic field at points A and B.

problem

Force between two parallel currents

Magnitude of Ba

d i

B aa π μ 2

Force on a length L of wire b

abba LBiF =

d iLi

F baba π μ

2 0=

Each of two parallel wires with currents I1 and I2, experiences a magnetic force given by

L = length of wire d = distance between the two wires

If the currents are parallel, the force is attractive. If the currents are anti-parallel the force is repulsive.

L d II

F π

μ 2

210=

Good to discuss Two long parallel straight wires carry equal currents in opposite directions. At a point midway between the wires, the magnetic field they produce is: A) zero B) non-zero and along a line connecting the wires C) non-zero and parallel to the wires D) non-zero and perpendicular to the plane of the two wires E) none of the above

? Good to discuss

On a computer chip, two conducting strips carry charge from P to Q and from R to S. If the current direction is reversed in both wires, the net magnetic force of strip 1 on strip 2 1. remains the same. 2. reverses. 3. changes in magnitude, but not in direction. 4. changes to some other direction. 5. other

Conceptual question Streams of charged particles emitted from the sun during unusual sunspot activity create a disturbance in the earth’s magnetic field (called a magnetic storm). How can they cause such a disturbance?

? B Fields of Current Distributions By winding wires in various geometries, we can produce different magnetic fields. For example, a current loop (perpendicular to plane, radius R, current emerging from plane at top of loop):

Magnitude of magnetic field at the center of loop:

R IN

B 2

0μ=

N = # of loops of wire (i.e. # turns)

Direction of magnetic field from the RHR.

Solenoids

If we stack several current loops together we end up with a solenoid. In the limit of a very long solenoid, the magnetic field inside is very uniform:

n = number of windings per unit length, I = current in windings

nIB 0μ=

Magnetic Materials

On atomic level – moving electrons (microscopic current loops) create magnetic fields. In many materials, these currents are randomly oriented (net magnetic field is zero). In some materials, the presence of an external magnetic field can cause the loops to become oriented

Paramagnetism – orientation with an external field Diamagnetism – orientation against an external field Ferromagnetism – line up loops (magnetic domains) without an external field

Week 2/Week 2/4.pdf

Electric Current and

Direct Current Circuits

Chapter 19

Electric Current

Resistance and Ohm’s Law

Power in Electric Circuits

Direct Current Circuits

Combination Circuits

Electric Current

Part 1

Real life is mostly dynamic

Chapters 17 and 18: Electrostatics - charges at rest Chapter 19: Charges in motion Examples of electric current Small currents: nerve currents, …, computers (laptops),

Larger currents: microwave, dishwashers, …,

Very large current: lightning strokes, …

Electric Current

Part 1

Electric current is a stream of moving charges

Important to know! A net electric charge moving through a surface is not zero

Examples: • Current: An electric bulb – the net transport of charge

is not zero. • No Current: The flow of water through a garden hose

(water molecules are neutral) . There is NO net transport of charge.

What particles are moving?

In most cases (electric current in a wire) the charge is carried by electrons moving through a metal wire.

In liquids the charge is often carried by positive ions.

Most important applications – electronics (moving electrons)

“Electricity is actually made up of extremely tiny particles called electrons, that you cannot see with the naked eye unless you have been drinking.” Dave Barry

What “pushes” particles? Electric potential – electromotive force External electric field (or electric potential) “pushes” electrons.

In the absence of electric potential the net transport of charge is zero (Just a chaotic motion)

The directions of current

That is just an agreement to draw diagrams

A current is drawn in the direction in which positive charges carries would move, even if the actual charge carries are negative (electrons) and move in the opposite direction

A mathematical definition for electric current

If charger Δq passes through a plane in time Δt, then the current I through that plane is define as

SI unit: coulomb per second = ampere (A) 1 ampere = 1 A = 1 coulomb per second = 1 C/s

t q

I Δ Δ

Conceptual question

Current is a measure of:

A) force that moves a charge past a point B) resistance to the movement of a charge past a point C) energy used to move a charge past a point D) amount of charge that moves past a point per unit time E) speed with which a charge moves past a point

Resistance and Ohm’s Law

Part 2

First Newton’s Law

If there is no net force on a body, the body must remain at rest if it is initially at rest, or move in a straight line at constant speed if it is in motion

However - No electromotive force (or electric potential) and the net transport of charge is zero, or electric current is zero

Is there a contradiction?

Ideal wires vs. real wires

Motion of electrons (or other charged particles) in real wires is similar to motion with air resistance (frictional force!)

Example: a car

Atomic structure of real wires

Interactive simulations from the Physics Education Technology project at the University of Colorado http://www.colorado.edu/physics/phet/

Resistance

In order for a current I to flow through a material there must be a potential difference, or voltage V, between the two ends of the material. We define the resistance, R, of a material to be:

SI unit: volt per ampere 1 ohm = 1 Ω = 1 volt per ampere = 1 V/A

Ohm’s Law: The electric current is always directly proportional to the potential difference

I V

R =

R V

I =

Resistance and Resistivity

Resistance (R) is a property of an object Resistivity (ρ) is a property of a material

Calculating Resistance from Resistivity

L – length of a wire A – the cross sectional area of a wire

It does not depends on the shape of the area A!

A L

R ρ=

Resistance and Resistivity

An object which provides “resistance” to current flow is called a resistor.

The symbol for a resistor is

Variation with temperature

The values of most physical properties vary with temperature, and resistivity is no exception.

Superconductivity – resistivity drops to ZERO at very small temperatures

Drift speed

No electric field – random motion (but very fast) v ~106 m/s

Electric current – still random motion but with a drift speed vd~10-4 m/s

Conceptual question

If the potential difference across a resistor is doubled:

A) only the resistance is doubled B) only the current is halved C) only the current is doubled D) only the resistance is halved E) both the current and resistance are doubled

Conceptual question

A cylindrical copper rod has resistance R. It is reformed to twice its original length with no change of volume. Its new resistance is:

A) R B) 2R C) 4R D) 8R E) R/2

Power in Electric Circuits

Part 3

Energy and Power

When a charge Δq moves across a potential difference V, its electrical potential energy U changes by the amount

Power is the rate of electric energy transfer

SI unit: watt, W 1 V·A = (1J/C)(1C/s) = 1 J/s = 1 W

qVU Δ=Δ

IV t

qV t U

P = Δ Δ

= Δ Δ

Resistive Dissipation

Since

The rate of electric energy dissipation due to resistance

R V

RIP 2

2 ==

R V

I =

Caution

There is a difference between two equations for power

applies to electric energy transfer of all kinds

apply only to the electric energy transfer to thermal energy in a device with resistanceR

V RIP

2 2 ==

IVP =

Power It is better to send 10,000 kW of electric power long distances at 10,000 V rather than at 120 V because:

A) the insulation is more effective at high voltages B) the resistance of the wires is less at high voltages C) more current is transmitted at high voltages D) there is less heating in the transmission wires E) the iR drop along the wires is greater at high voltage

More efficient power lines? A) USA B) Europe

Power You buy a "75 W" light bulb. The label means that: A) no matter how you use the bulb, the power will be 75 W B) the bulb was filled with 75 W at the factory C) the actual power dissipated will be much higher than 75 W since

most of the power appears as heat D) the bulb is expected to "burn out" after you use up its 75 watts E) none of the above

? Energy usage

1 kilowatt-hour = (1000 W)(3600 s) = = (1000 J/s)(3600 s) = 3.6×106 J

Buying electricity

What do you buy from the power company?

A) Only energy B) Electrons and energy C) Only electrons

? Simple problem

Suppose the electric company charges 10 cents per kW×h. How much does it cost to watch a TV (250 watt) 4 hours a day for 30 days?

A) $1.50 B) $3.0 C) $30.0 D) $150.0 E) none of these

problem

Problem problem

Direct Current Circuits

Part 4

Direct Current (DC) Circuits A circuit is a loop comprised of elements like resistors and capacitors around which current may flow.

For current to continue flowing in a circuit with non-zero resistance, there must be an energy source. This source is often a battery.

The light bulb in this circuit is the resistor. The wires that connect the battery to other circuit elements are assumed to have zero resistance.

battery

A single-loop circuit

a simple equation

where is electromotive force (emf) The potential difference between the terminals of an ideal emf device is equal to the emf of the device

R i

ε =

A single-loop rules

Loop rule: The algebraic sum of all changes in potential encountered in a complete traversal of any loop of a circuit must be zero (Kirhhoff’s loop rule). Resistance Rule: For a move through a resistance in the direction of the current, the change in potential is -iR; in the opposite direction +iR. EMF Rule: For a move through an ideal emf device in the direction of the emf arrow, the change on the potential is +ε; in the opposite direction is -ε

A single-loop rules (cont.)

EMF Rule + Resistance Rule + Loop Rule

0=−εiR

Clockwise

Counterclockwise

0=− iRε

Electric hazard in heart surgery

A a patient is undergoing open-heart surgery. A sustained current as small as 25 μm (25*10-6 A) passing through the hart can be fatal. Assume that the heart has a constant resistance of 250 Ω; determine the minimum voltage that posses a danger to the patient.

problem

V = IR = 6.25 mV

Resistors in series

Part 4a

A single-loop with resistors in series

When a potential difference ε is applied across resistances connected in series, the resistance have identical currents i.

The sum of the potential differences across the resistances is equal to the potential difference ε

Resistors in series

with a single equivalent resistance

0321 =−−− iRiRiRε

321 RRR i

++ =

eqR i

ε =

321 RRRReq ++=

Resistors in parallel

Part 4a

Resistances in Parallel

Multiloop Circuits

Junction rule: The sum of the current entering any junction must be equal to the sum of the currents leaving that junction (Kirhhoff’s junction rule)

210 iii +=

Resistances in Parallel

321 321 R

V R V

R V

iiii ++=++=

eqR V

i =

321

1111 RRRReq

++=

when a potential difference V is applied across resistances connected in parallel, the resistances all have that same potential difference

Resistors in series and parallel (reference set)

321

RRRR IIII VVV

++= === ++=ε

321

1111 RRRR

IIII VVV

++=

++= ===ε

Resistance

Give an example how four resistors of resistance R can be combined to produce an equivalent resistance of R.

problem

Electric current Two 110-V light bulbs, one "25 W" and the other "100 W", are connected in series to a 110 V source. Then:

A) the current in the 100-W bulb is greater than that in the 25-W bulb

B) the current in the 100-W bulb is less than that in the 25-W bulb

C) both bulbs will light with equal brightness D) each bulb will have a potential difference of 55 V E) none of the above

? Consider the circuit shown in the figure, in which three lights, each with a resistance R, are connected in parallel. What happens to the intensity of light 3 when the switch is closed? What happens to the intensities of lights 1 and 2?

Conceptual Question ?

Electric current Two 110-V light bulbs, one "25 W" and the other "100 W", are connected in series to a 110 V source. Then:

A) the current in the 100-W bulb is greater than that in the 25-W bulb

B) the current in the 100-W bulb is less than that in the 25-W bulb

C) both bulbs will light with equal brightness D) each bulb will have a potential difference of 55 V E) none of the above

The resistance of resistor 1 is twice the resistance of resistor 2. The two are connected in parallel and a potential difference is maintained across the combination. Then:

A) the current in 1 is twice that in 2 B) the current in 1 is half that in 2 C) the potential difference across 1 is twice that across 2 D) the potential difference across 1 is half that across 2 E) none of the above are true

Electric current A 120-V power line is protected by a 15-A fuse. What is the maximum number of "120 V, 500 W" light bulbs that can be operated at full brightness from this line?

A) 1 B) 2 C) 3 D) 4

? Electric current Two light bulbs A and B are connected in series to a constant voltage source. When a wire is connected across B as shown, bulb A

1. burns more brightly. 2. burns as brightly. 3. burns more dimly. 4. goes out.

problem

Electric current If the four light bulbs in the figure are identical, which circuit puts out more light?

1. I. 2. The two emit the same amount of light. 3. II

problem

Old-time Christmas tree lights had the property that, when bulb burned out, all the lights were out.

How were these lights connected?

How could you rewire them to prevent all the lights from going out when one of them burned out?

Christmas tree ?

Combination Circuits

Part 4c

Combination Circuits

Equivalent resistance

Part 4c (1)

Resistors in series and parallel (reference set)

321

RRRR IIII VVV

++= === ++=ε

321

1111 RRRR

IIII VVV

++=

++= ===ε

1. Select a group of resistors connected in either series or parallel.

2. Calculate the equivalent resistance for the group. 3. Go to the first step - Select a group of resistors

connected in either series or parallel in the new loop 4. Keep going till you get what you want

A key idea for solving combination circuits (equivalent resistance)

Simple analysis: all resistors are the same

(a) The two vertical resistors are in parallel with one another, hence they can be replaced with their equivalent resistance, R/2.

(b) Now, the circuit consists of three resistors in series. The equivalent resistance of these three resistors is 2.5 R.

Find the equivalent resistance of each combination

problem Simple problem

The equivalent resistance between points A and B of the circuit shown is:

problem

Single battery case

Part 4c (2)

Key ideas for solving combination circuits

1. When a potential difference is applied across resistances connected in series, the resistance have identical currents i.

2. When a potential difference V is applied across resistances connected in parallel, the resistances all have that same potential difference

3. Junction rule: The sum of the current entering any junction must be equal to the sum of the currents leaving that junction

Three identical resistors with resistance of 6 Ω are connected to a battery with an emf of 18 V

a) find the equivalent resistance of the resistor network

b) Find the current in each resistor

A resistor network problem problem

Part (a)

Part (b)

The current in the 13.8 Ω resistor is 0.750 A. Find the current in the other resistors in the circuit.

problem problem

( )

8 4

0.75 A given

13.8 0.75 13.8 10.35

17.2 10.35

0.6 A

8.45 4.11 10.35

0.824 A

drop

V I

I +

= × = × =

= × =

= × + =

13 17 8 4

0.75 0.6 0.824

2.174 A

total

I I I I I

+= + + = + +

problem

Electric current

(a) What is the equivalent resistance of the network shown? Put R1=10.0 Ω, R2=R3=50.0 Ω, R4=25.0 Ω, and E=6.0 V; (assume the battery is ideal.) (b) What is the current in resistors R1 and R4?

problem

Networks that can not be reduced to simple series-parallel combinations of resistors

Part 4c (2)

For solving any combination circuit …

1. If a circuit can be simplified by replacing resistors in series or parallel with their equivalents, do so Case One or “work forward”: reduce the circuit to a single loop Case Two or “work backward”: undoing the resistor simplification processes to find current or potential difference for a particular resistor

2. If a circuit cannot be to a single loop, use Kirchhoff’s rule (the junction rule and the loop rule) to write a set of simultaneous equations. You need have only as many independent equations as there are unknowns

Kirhhoff’s rules

1. Junction rule: The sum of the current entering any junction must be equal to the sum of the currents leaving that junction

2. Loop rule: The algebraic sum of the potential differences in any loop must equal zero

Complex networks problem

A two-cell flash light In a two-cell flash light , the batteries are normally connected in series. Why not connect them in parallel?

Week 2/Week 2/3.pdf

Electric Potential, Electric Potential Energy

and Capacitance

Chapter 18

Electric Potential Energy

Conservation of Energy

Potential of Point Charges

Equipotential Surfaces

Capacitance & Capacitors

Electric Potential Energy

Part 1

Energy: Definitions

Webster’s dictionary:

Energy – the capacity to do work

Work – the transfer of energy

Richard Feynman – Nobel Prize in physics (1965)

The Feynman Lectures on Physics. “...in physics today, we have no knowledge of what energy is.”

We know how to calculate its value for a great variety of situations, but beyond that it’s just an abstract thing which has only one really important property - conservation

Physics 111: Potential Energy

Potential energy U is energy that can be associated with the configuration of a system of objects that exert forces on one another. If the configuration of the system changes, then the potential energy of the system can also change

Potential energy can be defined for conservative forces only

Examples: • gravitational potential energy • spring elastic potential energy

Connection between energy and force - hint

Left side – the kinetic energy has been changed

Right side – the change is equal to Fx·d

2 0

2 mvmv K −=Δ )(2 0

2 0

2 xxavv −=−

)()( 2 1

0 2 0

2 xxmavvmK −=−=Δ

)( 2 1

2 1

0 2 0

2 xxFmvmv −=−

)( 0xxFW −=

Similarities between Coulomb’s law and Newton’s gravitational law

Equations are similar

Both forces are conservative ones. Conservative force: Definition 1. A conservative force does zero total work on any closed path Definition 2. The work done by a conservative force in going from an arbitrary point A to an arbitrary point B is independent of the path from A to B

2 21

r qq

kF r mm

GF cg ==

Connection between a conservative force and potential energy

The change in potential energy dues to the change in configuration Calculus based physics

Algebra based physics (gravitational and spring forces)

∫−=Δ f

dxxFU )(

kxFkxU

mgFmghU

−=−=Δ

==Δ

2 1

Definition for electric potential energy

Electric potential energy can be defined similar to definition of the gravitational potential energy Potential energy change

EdqUFdUU bba 0+=+=

d is the distance between points a and b Definition is true for a constant electric force working along the path of the motion.

a b

EqF 0=

Definition for electric potential

Electric potential is not the same as Electric potential energy but the connection is very simple For applications it is very useful to introduce the potential energy per unit charge

Units: J/C (the joule per coulomb), 1 volt = 1 joule per coulomb with this definition E = N/C = V/m Electric potential is a scalar, not a vector

0q U

V Δ

=Δ

Confusion with definitions for “change in electric potential”, and “change in electric potential energy”

There are two definitions for a “change of something”

In the textbook the change is defined as Therefore

iffi xxxandxxx −=Δ−=Δ

EdVVV EdqUUU

−=−=Δ −=−=Δ 0

a b

EqF 0=

Helpful approach (less confusion)

1. Always draw a diagram with fields and forces 2. Mark initial and final points 3. Watch the field direction and the sign of the charge

+ -

EdqUU EdqUU

−

−= +=

in most textbooks

EdqUUU ab +−=−=Δ EqF +=

EqF −=

Four possible combinations

1. A positive particle moves in the direction of the electric field Potential energy decreases

2. A positive particle moves opposite to the direction of the electric field Potential energy increases

3. A negative particle moves in the direction of the electric field Potential energy increases

4. A positive particle moves opposite to the direction of the electric field Potential energy decreases

blue field red force green displacement

Change in electric potential energy. (notice the direction of the blue arrow.)

For a uniform field (E=const) and 1D case

More terminology

Electric potential energy (U):

potential energy

electrostatic potential energy

Electric potential (V):

potential

potential difference

voltage (difference)

electrostatic potential

The zero of electrical potential

For calculating physical quantities it is the difference in potential which has significance, not the potential itself. Therefore, we are free to choose as having zero potential any arbitrary point which is convenient.

Typical choices are: • the earth • infinity

Conservation of Energy

Part 2

Conservation of energy

A consequence of the fact that electric force is conservative is that the total energy of an object is conserved (as long as non-conservative forces like friction can be ignored)

bbaa UKUK +=+

bbaa qVmvqVmv +=+ 22

2 1

Problem problem

Net Potential Three possible configurations for an electron e and a proton p are shown below. Take the zero of potential to be at infinity and rank the three configurations according to the potential at S, from most negative to most positive. A) 1, 2, 3 B) 3, 2, 1 C) 2, 3, 1 D) 1 and 2 tie, then 3 E) 1 and 3 tie, then 2

A charge q (q = 6.24 μC) is released from rest at the positive plate and reaches the negative plate with a speed of 3.4 m/s. The plates are connected to a 12-V battery Calculate: (a) the mass of the charge. (b) its kinetic energy at point A and at point B. (c) If d = 2cm, what is the electric field between the plates?

Problem problem

Problem A uniform electric field with a magnitude of 1200 N/C points in the negative x direction. (a) What is the difference in electric potential ΔV=Vb-Va between points a and b (b) What is the difference ΔV=Vb-Vc ? (c) What is the difference ΔV=Vc-Va ? (d) If a particle with mass of 3.5 g and a charge +0.045μC is released from rest at point A, in what direction it will move? (e) What speed will be after moving through a distance of 5 cm?

Let’s discuss this!

problem

The Electric Potential of Point Charges

Part 3

The electric potential of point charges

The difference in electric potential energy and electric potential between two points can be written as

Since the potential can be set to zero at any location, we choose the electric potential to be zero infinitively far from a given origin (Vb→0 as rb → ∞) Thus

ba ba r

qq k

r qq

kUU 00 −=− ba

ba r q

k r q

kVV −=−

Vq r qq

kU 0 0 ==

r q

kV = 26

For a point electric charge

Many Charges and Superposition If we wish to know the potential at a given point in space which results from all surrounding charges, we simply add up the potential at that point due to each charge:

Note that because potential is a scalar, the summation is not difficult. We just need to insert the sign of each charge.

... 3

1 +++= r q

k r q

kV A

How many grains form a pile of sand? 28

One proton

A proton is released from rest in a region of space with non-zero electric field. As the proton moves, does the electric potential energy of the proton increase, decrease or stay the same? Explain.

Two protons

Two protons are released from rest when they are D nm apart. After being released

(a) their kinetic energies gradually decrease to zero as they move apart

(b) their kinetic energies increase as they move apart

(d) their electric potential energy increase as they move apart

Two protons (more)

Two protons are released from rest when they are D nm apart. (a) What is the maximum speed they will reach? (b) What is the maximum acceleration they will achieve? (c) When does this acceleration occur? (d) Will the answers to questions a-c be different if we consider two electrons? (e) What if we have an electron and a proton?

The Electron Volt It is often convenient to work with a unit of energy called the electron volt.

One electron volt is defined as the amount of energy an electron (with charge e) gains when accelerated through a potential difference of -1 V: 1 eV = (1.6 x 10-19 C)V= 1.6 x 10-19 J

Equipotential Surfaces

Part 4

Equipotential Surfaces

A surface in space for which the potential is the same everywhere (like the surface of a conductor) is called an equipotential surface.

Since no work is required to move a charge on an equipotential surface, the electric field at every point on an equipotential surface is perpendicular to the surface.

Equipotentials for a point charge

Equipotential surfaces for a uniform electric field

Equipotential surfaces for two point charges

Capacitance & Capacitors

Part 5

Capacitance and Capacitors

Storing energy as potential energy: stretching a spring lifting a book pulling a bowstring We can also store energy as potential energy in an electric field. Capacitor is a device that is used to do that.

A Simple Capacitor

Two charges (with equal but opposite charges of magnitude q)

Why is that?

Connection: energy -> potential difference

+ -

r q

k r q

kVV ba

ba 2 21 =−=−

Definition for Capacitance

The charge and the potential difference V for a capacitor are proportional to each other

The proportional constant C is called the capacitance of the capacitor. Its value depends only on the geometry of the plates and not on their charge or potential difference SI unit: coulomb/volt = farad, F

CVq =

Charging a capacitor

Two conductors (conducting plates) connected to a battery.

Conductors: negative charge can move rather freely.

Capacitance for different geometries

A parallel plate capacitor

A – the plate area d – the plate separation ε0 – the permittivity constant ε0 = 8.85*10-12F/m = 8.85*10-12C2/N·m2

There are definitions for other geometries Gauss’s law is the toll to calculate.

d A

C 0 ε

Question

A capacitor C "has a charge Q". The actual charges on its plates are: A) Q, Q B) Q/2, Q/2 C) Q, –Q D) Q/2, –Q/2 E) Q, 0

Energy stored in an electric field

Electric potential energy of a charged capacitor

The potential energy of a charged capacitor may be viewed as being stored in the electric field between its plates

2 2

2 1

22 1

CV C

q qVU ===

Capacitor with dielectric

Dielectric: insulating material (plastic, paper …)

k is the dielectric constant air 1.00054 paper 3.5 Silicon 12.0 Water 80.4 Titania ceramic 130.0

0CC κ=

Question

A parallel-plate capacitor is attached to a battery that maintains a constant potential difference V between the plates. While the battery is still connected, a glass slab is inserted so as to just fill the space between the plates. The stored energy 1. increases. 2. decreases. 3. remains the same.