Phys 230 Winter 2010 - Chapter 24

1

Capacitors and DielectricsCapacitors and Capacitance

Capacitors in Series and Parallel

Dielectrics and molecular model of Induced Charge

Energy Storage in Capacitors and Electric Field Energy

/ abC Q V

, 1 2

1 1 1...

eq sC C C

, 1 2 ...eq pC C C

221 1

2 2 2

QU CV QV

C

2

0

1/

2u U Vol E

0 0

0

, ,E VC

K E VC K K

2

Capacitors and Capacitance

Capacitors are an efficient way to separate charge in order to store energy electrically.

A capacitor is any configuration of two conductors separated by an insulator.

Capacitors are one of the three basic electrical elements in any electrical circuit (the other

two are resistors and inductors). Electrical circuits form the basis of the majority of devices

in our modern technological society.

An abstract capacitor

Capacitor symbol:

When we „charge a capacitor‟ (via a battery for example), we transfer charge –Q onto one

conductor, and the other conductor acquires a charge of +Q. However, the net charge on a

capacitor is always zero. When we refer to the charge of a capacitor, we refer to the magnitude

of the charge on either of the two conductors.

Vertical lines represent the conductors.

Horizontal lines represent wires

3

Capacitors and Capacitance

Once the capacitor is charged, a potential difference exists between the two capacitors. If the battery is removed

after the system reaches equilibrium, the potential difference is equal to that of the battery.

1or Coulomb‟s law

Experiments1 show that the quantity of charge Q on a capacitor is linearly proportional to the potential difference

between the capacitors. The constant of proportionality is called the capacitance C:

ab

QC

V

SI Unit: 1 C/V = 1 F(arad)

Surprisingly C depends only on the geometry (shapes, sizes, orientations) of the conductors, and the insulating

material that exists between them. The greater the capacitance, the greater the magnitude of charge Q that can

be put on each conductor for a given voltage.

Thus capacitance is a measure of a capacitor to store energy.

Cylindrical capacitors

capacitance = amount of charge the capacitor can store

per unit of potential difference

4

Example 1: The isolated charged sphere (See YF 24.67 – Capacitance of the Earth)

Although we said that a capacitor consists of two conductors, we can consider the simple case of an isolated

charged sphere and an imaginary spherical shell “at infinity” (playing the role of the second conductor) on which

the field lines from the sphere terminate: the E-field surrounding the real sphere is the same. Thus we can speak

of the capacitance due to the real sphere of radius R.

In Ch. 23 we computed the potential outside a charged spherical shell relative to infinity as:0

1

4r

QV

r

Thus the capacitance is immediately given by: 04R

QC R

V

Notice the capacitance depends only on R, i.e. the geometry.

5

Example 2: Parallel Plate Capacitor

Start with the E-field between the two sheets of charge. Superposition gives us:

0 0 0

ˆ ˆ ˆ( ) ( )2 2

E E E j j j

(Recall Ch 21/22)

The voltage between the plates is thus:

0 0 0

( )b b

aba a

dV E dl dy a b Ed

and the capacitance is 0

0/

Q AC

d d

Notice: The capacitance is proportional to the area, and inversely proportional to the distance between the plates.

This depends only on the geometry (i.e. A and d)./Q A

ˆ ˆl yj dl j dy

(recall: )

In detail:

This simple result holds only for uniform fields!

Notice that [ε0] = Farad/m

6

Example 3: Cylindrical Capacitor

Consider an inner cylindrical conductor of radius ra carrying linear charge

density +λ and an outer cylindrical shell of radius rb, with charge density –λ.

From Ch 21/22:0

ˆ2

E rr

between the cylinders.

Integrating along a radial path , the voltage between the two cylinders is:ˆdl r dr

0 0

ln 02 2

b b

a a

r rb

abr r

a

rdrV E dl

r r

and the capacitance is0

0

2

ln( / ) / 2 ln( / )b a b a

LQC

r r r r

7

Example 4: (Concentric) Spherical Capacitor

With the Gaussian surface as shown,

Gauss‟s law gives us: 2

0

ˆ4

QE r

r for a br r r

Integrating along a radial path , the voltage between the two shells isˆdl r dr

2

0 0 0

( )1 10

4 4 4

b b

a a

r rb a

abr r

b a a b

Q r rQ dr QV E dl

r r r r r

The capacitance is:04 a b

b a

r rC

r r

Note: The geometric mean of two quantities a and b is ab4 a br r is the geometric mean of the areas

2 24 , 4a br r

b ar r is the distance between the spheres.

so we can write the result as 0 /gmC A d

Notice that0

0

4lim 4b

a ba

rb a

r rr

r r

recovering our earlier result.

8

YF 24.54 In one type of computer keyboard, each key holds a small metal plate

that serves as one plate of a parallel-plate air-filled capacitor. When the key is

depressed, the plate separation decreases and the capacitance increases. Electronic

circuitry detects the change in capacitance and thus detects that the key has been

pressed. In one particular keyboard the area of each metal plate is 42.0 mm2, and the

separation between the plates is 0.700 mm before the key is depressed.

(a) Calculate the capacitance before the key is depressed.

(b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the

key be depressed before the circuitry detects its depression?

YF24.10 A cylindrical capacitor has an inner conductor of radius 0.250 cm, surrounded by an outer hollow

conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The

capacitance is 36.7 pF. (a) What is the inner radius of hollow tube? (b) When charged to 125 V, what is λ?

9

Capacitors in Circuits: Series Configuration

Capacitors often occur in combination as parts of electrical circuits. One configuration

is a series configuration1 in which the capacitors occur in sequence as we trace along

one path in a circuit.

1Generically, a series configuration is one in which the current flow through

both of the elements in series is the same. However, for „capacitors‟ specifically

this is misleading since no current flows through the capacitor. This interpretation

of the word series is more appropriate for resistors (Ch 25) and inductors (Ch 29/30).

Claim: In series, the charge on each plate of each capacitor has the same magnitude.

The total voltage across both capacitors is the sum of the voltages across each.

Proof: The part of the circuit containing c has zero net charge before the circuit is

connected to a battery. It is isolated and so the net charge cannot change. When we

connect a battery (so Vab > 0), electrons flow from the top plate of C1 and are deposited

on the bottom plate of C2. Eventually equilibrium is reached, and the positive charge +Q left behind on the top plate

of C1 attracts a charge of –Q to the bottom plate of C1, which means that charge +Q must be left at the top of C2.

10

Capacitors in Circuits: Series Configuration

In equilibrium, the configuration is electrostatic and the E-fields inside the conducting

wires are 0, i.e. the potential at pt a is the same as the top plate of C1 (same with pt

b & bottom plate of C2). Thus the total voltage across both capacitors is the same

as that across the battery. Since the bottom plate of C1 and the top plate of C2 are

also at the same potential in equilibrium, the sum of the voltages across both are

ac cb abV V V

The equivalent capacitance of the series combination is the capacitance required of a single capacitor if it was to

replace the two capacitors with the same charge and voltage. Using our two claims,

1 2/ /eq

ab ac cb

Q Q QC

V V V Q C Q C

1 2

1 1 1

eqC C C

Similarly, if we have more than two in series:

, 1 2 31/ 1/ 1/ 1/ ...eq seriesC C C C

Memorize: In series Vtotal = V1 + V2 + ...

11

Capacitors in Circuits: Parallel Configuration

Circuit elements that are in parallel have the same voltage drop across them1.

In other words the upper/lower plates of the two capacitors and the wire that

connects them are equipotential surfaces in equilibrium.

1as opposed to the same current through them

Now the charges on each are (in general) different. Now the equivalent

capacitance is determined from:1 2

1 2eq

ab ab

Q QQC C C

V V

The same argument extends to parallel configurations of more than two capacitors:

, 1 2 3 ...eq parallelC C C C

Memorize: In parallel Qtotal = Q1 + Q2 + ...

12

Capacitor Networks: Circuit Reduction Example (SJ26.3ex)

Find the equivalent capacitance between a and b:

4.0 μF

a b

The key is to reduce the circuit one step at a time...1.0

3.0

6.0

2.0 8.0

4.0 4.0

8.0 8.0

ba a b

2.0

4.0

a b

6.0

Ceq= 1.0 + 3.0 = 4.0 μF

Ceq= 6.0 + 2.0 = 8.0 μF

1/Ceq = 1/4.0 + 1/4.0 = 1/2.0

1/Ceq = 1/8.0 + 1/8.0 = 1/4.0

Ceq = 2.0 + 4.0 = 6.0 μF

13

YF24.59 (What is equivalent capacitance good for?)

In the figure, C1=C5= 8.4 μF and C2=C3=C4= 4.2 μF. The applied potential is

Vab= 220 V. (a) What is the equivalent capacitance of the network between

a and b? (b) Calculate the charge and potential difference across each capacitor.

(The key idea is to work backwards from the reduced circuit applying

the rules for parallel and series that determine charges and voltages.)

This question motivated part of the Fall 2009 Final Exam Question…

14

Energy Storage in Capacitors (A mechanical analogue)

Recall the argument that takes us from Hooke‟s law F=kx to U=1/2 kx2 :

Force applied to stretch a spring from equil. to to x=X: ( )F x kX

Differential work1 done to stretch it from X to X+dX: dW XkX d

dX

1We have a variable force here so your high school definition of

work “W=Fd” only holds infinitesimally, where we can treat F as

constant. The point is F=kX depends on X!

Work done to stretch it from 0 to x is thus:2

0

1

2

x

W dW kXdX kx

The work we do to stretch the spring is stored as potential energy, so21

2U kx

If we agree to define the spring as having no potential energy when it is unstretched, then21

( )2

U x kx

15

Energy Storage in Capacitors

Similarly, suppose we‟ve already placed a charge q on a capacitor of capacitance C.

The voltage across the capacitor (when the charge is q) is then ( )q

V qC

The differential work we do to add another charge dq is: ( )q

dW V q dq dqC

(Since the voltage varies while we‟re adding charge, just as a force on the spring varies

as we‟re adding stretch, we again have to consider the differential work and integrate.)

Thus the work done to charge the capacitor from q=0 to q=Q is:

2

0 2

Q q QW dW dq

C C

If we agree to say that the electrical potential energy of an uncharged capacitor is zero then

2

2

QU

C

In the analogy with the spring,

the charge q plays the role of

stretch x, and the inverse

capacitance (called elastance)

plays the role of the force

constant k.

16

Energy Storage in Capacitors

Alternate forms of this expression:

(using Q=VC)2

21 1

2 2 2

QU CV QV

C

A dramatic application of capacitors

“The Z-Machine”

Electric-Field Energy (NOT a new kind of energy)

We can think of the work done to charge a capacitor as energy stored in

the charge separation/distribution itself (as we did in the last chapter), OR

alternatively, we can think of the energy as being stored in the electric field

between the conductors of the capacitor. They are the SAME energy.

Let’s see how this works for a simple parallel plate capacitor...

17

Electric-Field Energy

For a parallel plate capacitor, the capacitance is given by: 0

AC

d

the charge is given by: Q A

and the E-field mag. is given by: 0/E

Let‟s re-express the potential energy in terms of the electric field:

2 2 22

0

0

1

2 2 2

Q A dU E Ad

C A

More conveniently written in terms of the energy density2

0

1/ / ( )

2u U Vol U Ad E

(i.e. energy per unit volume)

The remarkable thing about this result is that although we derived it for the simple case of a parallel plate

capacitor, it holds for any geometry (in vacuum). In fact, it holds even in empty space away from any charges, i.e.

for any vacuum electric field configuration.

This is a key component of the energy contained in an electromagnetic wave!

Volume for parallel

plate capacitor = Ad

Electric energy

density in vacuum

On your assignment, you will verify this for a cylindrical capacitor.

18

YF24.56/SJEx26.4 Two capacitors C1 =4.0 μF and C2 = 9.0 μF are charged to the same

initial potential difference Vab = V0 = 28 V. The charged capacitors are disconnected from

the source and from each other, and then reconnected to each other with plates of

opposite sign together.

(a) How much does the energy of the system decrease?

(Try for general C1 and C2.)

Be very careful with signs in

charge conservation!

YF24.30 A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is

2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected

from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected

to the potential source so the potential difference between the plates remains constant.

“Potential source” = battery for our purposes here (idealized battery that has constant potential)

Key idea: Q1,f + Q2,f = Q1,0 – Q2,0

19

Dielectrics

A dielectric is a nonconducting material inserted between the two conductors in a capacitor.

Three functions of a dielectric:

1) solves mechanical problem of keeping charged sheets physically apart

2) increases the maximum possible potential difference between sheets (dielectric breakdown)

3) increases the capacitance (charge per unit potential difference) of the capacitor

Q0 , C0

Warning: For problems involving the modification of a capacitor, you must note

whether the modifications are being made while the capacitor is connected to

a battery (so holding V fixed, Q changes), or after the capacitor is disconnected

(so holding Q fixed, V changes).

Inserting a dielectric for a capacitor of charge Q0 (not connected to a battery) and capacitance C0:

a b

Vab

1) decreases the E-field magnitude between the conductors

2) decreases the potential difference between the two conductors

3) increases the capacitance of the conductor

20

Dielectrics: molecular model Why does this happen? With either polar or non-polar

substances, the E-field causes

a re-distribution of charge within

the dielectric. In turn this effectively

creates a formation of bound1

surface charge on each surface of

the dielectric, denoted σi (i = induced).

1as opposed to the free charges found in a conductor.

Net charge in dielectric remains

zero, but the material has become

polarized.

21

Dielectrics: molecular model The effect of the polarization of

the dielectric, and the formation of

opposite surface charges σi, is to

partially cancel the „original‟

electric field.

With no dielectric: 0

0

E

With dielectric:

(parallel plate of course)

0

iE

„0‟ subscripts denote quantities

in vacuum in this section.

Note: E < E0

22

Dielectric Constant

Since we have no way (in this course) of computing the bound surface charges σi on the surface of the dielectric

from first principles (it is a material dependent question), it is very convenient to introduce an experimental

parameter K called the dielectric constant, that captures the same idea: 0 1E

KE

(with Q =const)

Since any re-scaling of E, rescales V by the same amount, i.e. b

aba

V E dl 0 0E KE V KV

If the charge is held fixed on the capacitor (i.e. we disconnect a charge source such as a battery before inserting

the dielectric), then:0

0 0/

Q Q QC K KC

V V K V i.e. the effect of the dielectric is to increase the capacitance.

Here the subscript 0 refers to no dielectric.

If we are given K, we can compute the bound surface charge density: 0

0 0

, iE E

11i

K

(for a parallel plate capacitor)

23

Permittivity & Dielectric Breakdown

Permittivity is defined as , and since K = 1 for a vacuum is called the “permittivity of free space”1. 0K 0

This allows us to write the earlier expressions in terms of instead of both K and : 0

E

0

AC KC

d

(parallel plate capacitance

with dielectric)

If a dielectric is subject to sufficiently strong electric fields, electrons can be torn

from the lattice of the dielectric, and what was formerly an insulator can now act

as a conductor (quite „violently‟ in fact, as lightning demonstrates). This situation

is known as dielectric breakdown. The maximum E-field a dielectric can sustain

is known as its dielectric strength. Dry air has a dielectric strength of 3×106 V/m.

1Many standards organizations now consider

this terminology misleading, and refer to it

simply as the „electric constant‟. This is related

to issues of defining “the vacuum” in a quantum

mechanical world.

24

YF24.47 A 12.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V

between the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates,

completely filling the space between them. (a) How much energy is stored in the capacitor before and after the

dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

Why?

Contrast this example with Example YF 24.11 where the charge (not the voltage) is held fixed.

If a battery remains connected to a capacitor while we change the dielectric configuration, the voltage across it will

remain constant, and the charge will change.

If a battery is disconnected from a capacitor, and the capacitor remains isolated while we change the dielectric

configuration, the charge will remain constant and the potential difference will change.

Summary of Changes in Capacitor state variables under a change in dielectric

25

YF24.78 A fuel gauge uses a capacitor to determine the height of the fuel in a tank.

The effective dielectric constant Keff changes from a value of 1 when the tank is empty

to a value of K, the dielectric constant of the fuel when the tank is full. The appropriate

electronic circuitry can determine the effective dielectric constant of the combined air

and fuel between the capacitor plates. Each of the two rectangular plates has a

width w and a length L. The height of the fuel between the plates is h. You can ignore

any fringing effects. (a) Derive an expression for Keff as a function of h. (b) What

is the effective dielectric constant for a tank ¼, ½ and ¾ full if the fuel is gasoline

(K=1.95)? (c) methanol (K=33.0)? (d) For which fuel is this fuel gauge more practical?

(See also YF24.65, 24.66, 24.71, 24.72, 24.75, 24.76.)This particular system is used on aircraft rather than on cars (which

use a variable resistance) instead of a variable capacitor..

26

Phys 230 Final Exam – Fall 2009 – Chapter Summary Question

9. Consider the capacitive circuit shown below. The battery is a 9.0 V battery, C2 = 2.0 pF and C3 = 3.0 pF. C1 is a parallel-

plate capacitor. Each plate in C1 has an area of 22.6 cm2, and the plates are (initially) separated by 4.0 mm, and a space-

filling dielectric of dielectric constant 2.0 sits between the plates.

(1 pF=10-12 F, 1 cm2=10-4 m, 1 mm = 10-3 m.)

+9 V

C1

C2

C3

(a) [4] What is the charge and potential energy stored on each of the capacitors? (If you don’t get charges that are integer multiples

of 1 pC go back and check your calculation for C1: it should be the same order of magnitude as C2 and C3.)

(b) [3] If we remove the battery from the circuit, reverse the polarity of C1 and then re-connect it to C2 and C3, what is the

new charge on stored on (one plate of) C1? (Suggestion: Since we only want Q1, work with the C23 equivalent, and so treat the system as a

two capacitor system consisting of C1 and C23.)

(c) [3] Suppose instead of performing the transformation in (b), we remove C1 from the rest of the circuit (so it has the same

charge as it did in (a)), and then pull one of the plates apart so they are now 12.0 mm apart while the dielectric still fills only

4.0 mm of the 12.0 mm space. What is the work required to do this? (Hint: Don’t try to compute the work directly. Compute the

change in something else.)