Exam One will be handed back at the end of class on W05D1 Mon and Tuesday.

Students can take the exam home and compare with solutions.

Regrading Requests:

Exams must be returned with a request to regrade in class on W05D2 Wed or Thurs

The student must clearly indicate on the cover sheet which problems they would like regraded and briefly why.

Remember regrading can both raise and lower grades.

A sample of exams have been photocopied.

The students must not erase answers, add to answers, or change answers in any way. This is a serious offense which could result in expulsion from MIT.

Exam One Regrade Policy

2

W05D1Conductors and InsulatorsCapacitance & Capacitors

Energy Stored in Capacitors

W05D1 Reading Assignment Course Notes: Sections 3.3, 4.5, 5.1-5.4, 5.6, 5.8

Announcements

No Math Review this week

PS 4 due W05 Tuesday at 9 pm in boxes outside 32-082 or 26-152

W05D2 Reading Assignment

Course Notes: Sections 5.4, 5.6, 5.8-5.9

3

4

Outline

Conductors and Insulators

Conductors as Shields

Capacitance & Capacitors

Energy Stored in Capacitors

5

Conductors and Insulators

Conductor: Charges are free to move

Electrons weakly bound to atoms

Example: metals

Insulator: Charges are NOT free to move

Electrons strongly bound to atoms

Examples: plastic, paper, wood

6

Charge Distribution and Conductors

The Charged Metal Slab Applet: Non-zero charge placed in metal slabhttp://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/chargedmetalslab/chargedmetalslab.htm

Charges move to surface

(move as far apart as possible) Electric field perpendicular to surface, zero inside slab

7

Induced Charge Distribution in External Electric Field

Charging by Induction, Exterior of a Neutral Metallic Boxhttp://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/chargebyinductionBox/chargebyinductionBox.htm

Induced charges move

to surfaceElectric field perpendicular to surface, zero inside slab

8

Conductors in Equilibrium

Conductor Placed in External Electric Field

1)E = 0 inside

2) E perpendicular to surface

3) Induced surface charge distribution

9

Hollow Conductors: Applet

Charge placed OUTSIDE induces charge separation ON OUTSIDE. Electric field is zero inside.

http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm

10

Electric Field on Surface of Conductor

1) E perpendicular to surface

2) Excess charge on surface

Apply Gauss’s Law

EsurfaceA A

0

Esurface 0

11

Conductors are Equipotential Surfaces

1) Conductors are equipotential objects

2) E perpendicular to surface

12

Group Problem: Metal Spheres Connected by a Wire

Two conducting spheres 1 and 2 with radii r1 and r2 are connected by a thin wire. What is the ratio of the charges q1/q2 on the surfaces of the spheres? You may assume that the spheres are very far apart so that the charge distributions on the spheres are uniform.

13

Concept Question: Point Charge in Conductor

A point charge +q is placed inside a hollow cavity of a conductor that carries a net charge +Q. What is the total charge on the outer surface of the conductor?

1. Q.

2. Q + q.

3. q.

4. Q - q.

5. Zero.

14

Concept Q. Ans.: Point Charge in Conductor

Answer 2. Choose Gaussian surface inside conductor. Electric field is zero on Gaussian surface so flux is zero. Therefore charged enclosed is zero. So an induced charge –q appears on cavity surface. Hence an additional charge of +q appears on outer surface giving a total charge of Q + q on outer surface.

15

Hollow Conductors: Applet

Charge placed INSIDE induces balancing charge ON INSIDE. Electric field outside is field of point charge.

http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm

16

Capacitors and Capacitance

Our first of 3 standard electronics devices

(Capacitors, Resistors & Inductors)

17

Capacitors: Store Electric Charge

Capacitor: Two isolated conductors Equal and opposite charges ±QPotential difference between them.

Units: Coulombs/Volt or Farads

C Q

V

C is Always Positive

V

18

Parallel Plate Capacitor; Applet

Oppositely charged plates:Charges move to inner surfaces

http://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/capacitor/capacitor.htm

Electric field perpendicular to surface, zero inside plates

19

Calculating E (Gauss’s Law)

0S

inqd

E A

00

A

QE

E A

Gauss AGauss

0

Note: We only “consider” a single sheet!

Doesn’t the other sheet matter?

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Superposition Principle

0 0 0

ˆ ˆ ˆ2 2

E E E j j j

Between the plates:

Above the plates:

Below plates:

0 0

ˆ ˆ2 2

E E E j j 0

0 0

ˆ ˆ2 2

E E E j j 0

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Parallel Plate Capacitor

C depends only on geometric factors A and d

top

bottom

V d E S

C Q

V

0A

d

Q

A0

dEd

22

Group Problem: Spherical Shells

A spherical conductor of radius a carries a charge +Q. A second thin conducting spherical shell of radius b carries a charge –Q. Calculate the capacitance.

23

Concept Question: Isolated Spherical Conductor

What is the capacitance of an isolated spherical conductor of radius a?

1.Capacitance is not well defined.

2.Capacitance is .

3.Capacitance is infinite.

4.Capacitance is zero.

40a

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Concept Q. Ans. Isolated Spherical Conductor

Answer 2. Capacitance is Other equipotential surface (second conducting surface) is located at infinity. In previous calculation set .

40a

C Q

V

40

1

a

1

b

4

0

1

a

1

40a

b

25

Capacitance of Earth

For an isolated spherical conductor of radius a:

aC 04

mF1085.8 120

m104.6 6a

mF7.0F107 4 C

A Farad is REALLY BIG! We usually use pF (10-12) or nF (10-9)

26

Energy To Charge Capacitor

1. Capacitor starts uncharged.2. Carry +dq from bottom to top.

Now top has charge q = +dq, bottom -dq3. Repeat4. Finish when top has charge q = +Q, bottom -Q

+q

-q

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Stored Energy in Charging Capacitor

At some point top plate has +q, bottom has –qPotential difference is V = q / CChange in stored energy done lifting another dq is dU = dq V

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So change in stored energy to move dq is:

Total energy to charge to Q

Stored Energy in Charging Capacitor

dU dqV dq

q

C

1

Cq dq

U dU

1

Cq dq

0

Q

1

C

Q2

2

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Energy Stored in Capacitor

C Q

VSince

22

2

1

2

1

2VCVQ

C

QU

Where is the energy stored???

30

Energy Stored in Capacitor

u

E

oE 2

2

Parallel-plate capacitor: C

oA

dand V Ed

Energy stored in the E field!

U

1

2CV 2

1

2

oA

dEd 2

oE 2

2( Ad ) u

E(volume)

Energy density [J/m3]

31

Demonstration:

Changing Distance Between Circular Capacitor Plates E4

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%204&show=0

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Concept Question: Changing DimensionsA parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d, and then disconnected from the charging source (battery). The plates are pulled apart to a distance D > d. What happens to the magnitude of the potential difference V and charge Q?

1. V, Q increases.

2. V increases, Q is the same.

3. V increases, Q decreases.

4. V is the same, Q increases.

5. V is the same, Q is the same.

6. V is the same, Q decreases.

7. V decreases, Q increases.

8. V decreases, Q is the same.

9. V decreases, Q decreases.

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Concept Q. Answer: Changing Dimensions

With no battery connected to the plates the charge on them has no possibility of changing.

In this situation, the electric field doesn’t change when you change the distance between the plates, so:

V = E d

As d increases, V increases.

Answer: 2. V increases, Q is the same

34

Concept Question: Changing DimensionsA parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d. While still connected to the charging source, the plates are pulled apart to a distance D > d. What happens to the magnitude of the potential difference V and charge Q?

1. V, Q increases.

2. V increases, Q is the same.

3. V increases, Q decreases.

4. V is the same, Q increases.

5. V is the same, Q is the same.

6. V is the same, Q decreases.

7. V decreases, Q increases.

8. V decreases, Q is the same.

9. V decreases, Q decreases.

35

Concept Q. Answer: Changing Dimensions

With a charging source (battery) connected to the plates the potential V between them is held constant

In this situation, since

V = E d

As d increases, E must decrease.

Since the electric field is proportional to the charge on the plates, Q must decrease as well.

Answer: 6. V is the same, Q decreases

36

Concept Question: Changing Dimensions

A parallel-plate capacitor, disconnected from a battery, has plates with equal and opposite charges, separated by a distance d.

Suppose the plates are pulled apart until separated by a distance D > d.

How does the final electrostatic energy stored in the capacitor compare to the initial energy?

1. The final stored energy is smaller

2. The final stored energy is larger

3. Stored energy does not change.

37

Concept Q. Answer: Changing Dimensions

As you pull apart the capacitor plates you increase the amount of space in which the E field is non-zero and hence increase the stored energy. Where does the extra energy come from? From the work you do pulling the plates apart.

Answer: 2. The stored energy increases

38

U

1

2CV 2

1

2(110 4 F)(4 103 V)2 800 J

A 100 microfarad oil-filled capacitor is charged to 4 KV and discharged through a wire

Stored Energy:

Demonstration: Charging Up a Capacitor

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%206&show=0