Physics 2112Unit 7: Conductors and Capacitance

Today’s Concept: Conductors Capacitance

Electricity & Magnetism Lecture 7, Slide 1

THERE ARE ONLY THREE THINGS YOU NEED TO KNOW TO DO ALL OF HOMEWORK

1. E = 0 within the material of a conductor:

Charges move inside a conductor in order to cancel out the fields that would be there in the absence of the conductor. This principle determines the induced charge densities on the surfaces of conductors.

3. Definition of Potential: =

b

a

baba ldE

qUV

CONCEPTS DETERMINE THE CALCULATION !

Comments

2. Gauss’ Law: If charge distributions have sufficient symmetry (spherical,

cylindrical, planar), then Gauss’ law can be used to determine the electric field everywhere.

=0

enclosedQAdE

Electricity & Magnetism Lecture 7, Slide 2

Conductors

Charges free to move E = 0 in a conductor Surface = Equipotential E at surface perpendicular to

surface

The Main Points

Electricity & Magnetism Lecture 7, Slide 3

CheckPoint: Two Spherical Conductors 1

Electricity & Magnetism Lecture 7, Slide 4

Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B.

Compare the potential at the surface of conductor A with the potential at the surface of conductor B.

A. VA > VB

B. VA = VB

C. VA < VB

CheckPoint: Two Spherical Conductors 2

Electricity & Magnetism Lecture 7, Slide 5

Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B.

The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now?

A. VA > VB

B. VA = VB

C. VA < VB

CheckPoint: Two Spherical Conductors 3

Electricity & Magnetism Lecture 7, Slide 6

Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B.

What happens to the charge on conductor A after it is connected to conductor B by the wire?

A. QA increasesB. QA decreasesC. QA does not

change

Electric Circuit Element

Capacitor

Unit 7, Slide 7

Same uses as spring in mechanical system• smooth out rough spots• store energy• cause controlled oscillations

Simplest Example: Parallel plate capacitor

Q

E

Capacitor (II)

Unit 7, Slide 8

QCV

d

+Q

|| Fkx

Define capacitance, C, such that:

How easy is it to stuff charge on the plates

How easy is it to stuff charge on the plates

How difficult is it to change the length of the springQ

VUnits of Farad, F = Coulomb/Volt

Key Points

Unit 7, Slide 9

• +Q and –Q always have same magnitude

• Charges don’t more directly from one plate to the other

• Charged from the outside

First determine E field produced by charged conductors:

Second, integrate E to find the potential difference V

x

y

As promised, V is proportional to Q !• Method good for

all cases• Formula good for

parallel plate only

Review of Capacitance Example

o

E

=

=d

ydEV0

What is ?

A = area of plate

AQ

=

+Q

Qd E

dA

QdyEEdyVo

d d

===

0 0

)(

dAC 0=

AQdQ

VQC

o/=

Unit 7, Slide 10

Example 7.1 (Capacitor)

Unit 7, Slide 11

A flat plate capacitor has a capacitance of C = 10pF and an area of A=1cm2. What is the distance between the plates?

CheckPoint Results: Charged Parallel Plates 1

Electricity & Magnetism Lecture 7, Slide 12

Two parallel plates of equal area carry equal and opposite charge Q0. The potential difference between the two plates is measured to be V0. An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1 such that the potential difference between the plates remains the same.

Compare Q1 and Q0.A. Q1 < Q0

B. Q1 = Q0

C. Q1 > Q0

CheckPoint Results: Charged Parallel Plates 1

Electricity & Magnetism Lecture 7, Slide 13

An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value Q1 such that the potential difference between the plates remains the same.

Compare the capacitance of the two configurations in the above problem.

A. C1 > C0

B. C1 = C0

C. C1 < C0

Conceptual Idea:

Find V in terms of some general Q and divide Q out.

Example 7.2 (Linear Capacitor)

metal

metal

a1

a2

a3a4

cross-section A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai).

What is the capacitance C of this capacitor ?

Electricity & Magnetism Lecture 7, Slide 14

VQC

Plan:• Put +Q on outer shell and Q on inner shell• Cylindrical symmetry: Use Gauss’ Law to calculate E everywhere • Integrate E to get V• Take ratio Q/V (should get expression only using geometric parameters (ai, L))

Limiting Case:• L gets bigger, C gets bigger• a2 –> a3, C gets bigger

Example 7.2 (Linear Capacitor)

metal

metal

a1

a2

a3a4

cross-section A capacitor is constructed from two conducting cylindrical shells of radii a1, a2, a3, and a4 and length L (L >> ai).

What is the capacitance C of this capacitor ?

Unit 7, Slide 15

Do Limiting Cases Work?• L gets bigger, C gets bigger• a2 –> a3, C gets bigger

Energy in Capacitors

Electricity & Magnetism Lecture 7, Slide 16

= VdqU = dqCq

U is equal to the amount of work took to put all the charge on the two plates:

Example 7.3 (Energy in Capacitor)

Unit 7, Slide 17

A 8uF parallel plate capacitor is has a potential different of 120V between its two sides. The distance between the plates is d=1mm.

What is the potential stored in the capacitor?

What is the energy density of the capacitor?