Lecture 8-Lecture 8-11High Electric Field at Sharp Tips

Two conducting spheres are connected by a long conducting wire. The total charge on them is Q = Q1+Q2.

Potential is the same: 1 2

1 2

kQ kQ

R R 1 1

2 2

Q R

Q R

1 2

2 1

E R

E R

The smaller the radius of curvature, the larger the electric field.

21

11 R

kQE

22

22 R

kQE

With same potential, sphere with smaller radius carry smaller amount of charge

Lecture 8-Lecture 8-22Potential and Conductors

• Entire conductor including its surface(s) has uniform V.

•E an equipotential surface everywhere

•Equipotential surfaces are drawn at constant intervals of V

•Potential difference between nearby equipotentials is approximately equal to E times the separation distance.

Equipotential surfaces

Lecture 8-Lecture 8-33Again, Electrostatic Potential Energy

The electrostatic potential energy of a system (relative to ∞) is the (external) work needed to bring the charges from an infinite separation to their final positions.

r12r13

r23

q1

q2 q3

1 2 1 3 2 3

12 13 23

kq q kq q kq qU W

r r r

2 3 3 1 1 21 2 3

12 13 23 12 13 23

1 1 2 2 3 3

1

2

1

2

kq kq kq kq kq kqq q q

r r r r r r

qV q V q V

Generally for n charges:1

1

2

n

i ii

U qV

Lecture 8-Lecture 8-44Electrostatic Potential Energy of Conductors

bring dq in from ∞

q+dq,v+dv vkq

U W dq dqR

Spherical conductor

q,v

R

Total work to build charge from 0 up to Q:

2

0

1

2 2

Q kq kQU U dq QV

R R

Lecture 8-Lecture 8-55Capacitors

• A capacitor is a device that is capable of storing electric charges and thus electric potential energy. => charging• Its purpose is to release them later in a controlled way. => discharging• Capacitors are used in vast majority of electrical and electronic devices.

+Q -Q

Q

Q

Q

Typically made of two conductors and, when charged, each holds equal and opposite charges.

Lecture 8-Lecture 8-66

Capacitance

• Capacitor plates hold charge Q The two conductors hold charge +Q and –Q, respectively.

• The capacitance C of a capacitor is a measure of how much charge Q it can store for a given potential difference V between the plates.

Q V Expect Q

CLetV

Q Coulomb

C FV Volt

farad

(Often we use V to mean V.)

Lecture 8-Lecture 8-77

Steps to calculate capacitance C

1. Put charges Q and -Q on the two conductors, respectively.

2. Calculate the electric field E between the plates due to the charges Q and -Q, e.g., by using Gauss’s law.

3. Calculate the potential difference V between the plates due to the electric field E by

4. Calculate the capacitance of the capacitor by dividing the charge by the potential difference, i.e., C = Q/V.

b

ba aV E dl

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Lecture 8-Lecture 8-88 Example: (Ideal) Parallel-Plate Capacitor

b

b a

a

V V V E dl Ed

1. Put charges +q and –q on plates of area A and separation d.

2. Calculate E by Gauss’s Law

0

qE A

3. Calculate V by

4. Divide q by V

a

b

0

0

q q q AC

V Ed dqd

A

• q is indeed prop. to V

• C is prop. to A

• C is inversely prop. to d

Lecture 8-Lecture 8-99Ideal vs Real Parallel-Plate Capacitors

Ideal Real

• Uniform E between plates

• No fields outside

• Often described as “thin”

• Non-uniform E, particularly at the edges

• Fields leak outside

Lecture 8-Lecture 8-1010Physics 241 –warm-up quiz

Two identical ideal parallel plate capacitor. Both initially carry charge Q. One is always connected with an ideal battery while the other is not. After increase the distance between the plates by a factor of two, which of the following statements is true.

a) E field is not changed in both A and B . b) E field in A is half of the original value; E field in B

is not changed. c) E field in A is not changed; E field in B is half of the

original value. d) E field is half of the original value in both A and B . e) One need to know the voltage of the battery to give

the answer

A B

Lecture 8-Lecture 8-1111Numerical magnitudes

Let’s say:

Area A= 1 cm² , separation d=1 mm

Then0

12 2 2 4 2

3

13 13

8.85 10 ( / ) 10 ( )

10 ( )

8.85 10 ( / ) 8.85 10 ( )

AC

d

C Nm m

m

C N m F

This is on the order of 1 pF (pico farad). Generally the values of typical capacitors are more conveniently measured in µF or pF.

Lecture 8-Lecture 8-1212Long Cylindrical Capacitor

0 0

22

q qE rL E

rL

0 0

ln /2 2

a b

b a

q qV E dl dr b a

rL L

0

0

2

ln /ln /2

q q LC

qV b ab aL

1. Put charges +q on inner cylinder of radius a, -q on outer cylindrical shell of inner radius b.

2. Calculate E by Gauss’ Law

ab

+q -q

3. Calculate V

4. Divide q by V •

•

• C dep. log. on a, b

q VC L

L

Lecture 8-Lecture 8-1313Spherical Capacitor

204

qE

r

04q ab

CV b a

20 0

1 1

4 4

a

a b

b

q dr qV V V

r a b

04C a as b

1. Put charges +q on inner sphere of radius a, -q on outer shell of inner radius b.

2. Calculate E by Gauss’s Law

3. Calculate V from E

4. Divide q by V • q is proportional to V

• C only depends on a,b

(isolated sphere)

Lecture 8-Lecture 8-1414 Energy of a charged capacitor

How much energy is stored in a charged capacitor?

Calculate the work required (usually provided by a battery) to charge a capacitor to Q

2

0

1 1

2

QQ

U dW qdqC C

Total work is then2

21

2 2 2

QV QU CV

C

( ) /dW V q dq q C dq

Calculate incremental work dW needed to move charge dq from negative plate to the positive plate at voltage V.

Lecture 8-Lecture 8-1515

Where is the energy stored?

• Energy is stored in the electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the electric field.

• The electric field is given by:

0 0

QE

A

20

1

2U E Ad

• The energy density u in the field is given by:

20

1

2

U U

volume Adu E 3/J mUnits:

This is the energy density, u, of the electric field….

• To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor, where

2 2

0

1 1

2 2 ( / )

Q QU

C A d

++++++++ +++++++

- - - - - - - - - - - - - -- Q

+Q

Lecture 8-Lecture 8-1616Physics 241 –Quiz 6a

If we are to make an ideal parallel plate capacitor of capacitance C = 1 F by using square plates with a spacing of 1 mm, what would the edge length of the plates be? Pick the closest value.

a) 10 mb) 1 cmc) 100 md) 1 mme) 1 m

12 2 20 8.85 10 ( / )C N m

Lecture 8-Lecture 8-1717Physics 241 –Quiz 6b

We have an ideal parallel plate capacitor of capacitance C=1 nF that is made of two square plates with a separation of 2 m. A battery keeps a

potential difference of 15 V between the plates. What is the electric field strength E between the plates?

a) 30 N/Cb) 5 x 109 V/mc) 15,000 V/md) 7.5 x 106 N/Ce) Not enough information to tell

12 2 20 8.85 10 ( / )C N m

Lecture 8-Lecture 8-1818Physics 241 –Quiz 6c

We have am ideal parallel plate capacitor of capacitance C=1 F that is made of two square plates. We could double the capacitance by

a) doubling the separation db) doubling charges on the plates Qc) halving potential difference between the plates Vd) increasing the edge length a of the plates by a factor of 1.4e) halving the area of the plates A

12 2 20 8.85 10 ( / )C N m

a

dAa Q

Q