Chapter 25 Capacitance-IIIn the last lecture:

we calculated the capacitance C of a system of two isolated conductors.

We also calculated the capacitance for some simple geometries.

In this chapter we will cover the following topics:

-Methods of connecting capacitors (in series , in parallel). -Equivalent capacitance. -Energy stored in a capacitor. -Behavior of an insulator (a.k.a. dielectric) when placed in the electric field created in the space between the plates of a capacitor. -Gauss’ law in the presence of dielectrics.

(25 - 1)

HITT

A. the work done by the field is positive and the potential energy of the electron-field system increases

B. the work done by the field is negative and the potential energy of the electron-field system increases

C. the work done by the field is positive and the potential energy of the electron-field system decreases

D. the work done by the field is negative and the potential energy of the electron-field system decreases

E. the work done by the field is positive and the potential energy of the electron-field system does not change

An electron moves from point i to point f, in the direction of a uniform electric field. During this displacement:

i j

This means that if we apply the same voltage across the capacitors in fig.a and fig.b (either right or left) by connectingto a battery, the same charge is provided by the battery. Alternatively,i

V

qf we place the same charge on plates of the capacitors in fig.a

and fig.b (either right or left), the voltage across them is identical. This can be stated in the following manner: If we place the

qV

capacitorcombination and the equivalent capacitor in separate black boxes, by doing electrical mesurements we cannot distinguish between the two.

Consider the combination of capacitorsshown in the figure to the left and to the right (upper part). We will substitutethese combinations of capacitor with a single capacitor eqC

Equivalent Capacitor

that is

"electrically equivalent" to the capacitor group it substitutes.

(25 - 9)

The fig.a we show three capacitors connected in parallel. This means that the plate of each capacitor is connectedto the terminals of a battery of voltage . We will substitute

V

Capacitors in parallel

the parallel combination of fig.a with a single equivalent capacitor shown in fig.b which is also connectedto an identical battery

1 1 1 2 2 2

3 3 3 1 2 3 1

The three capacitors have the across their plates.The charge on is: . The charge on is: .The charge on is: . The net charge

C q C V C q C VC q C V q q q q C

same potential difference V

2 3

1 2 31 2 3

1 21

The equivalent capacitance

For a parallel combination of n capacitors is given by the expression:

...n

e n

eq

q jj

C C V

C C

C C C C C

C VqC C C CV V

1 2 3eqC C C C

(25 - 10)

The fig.a we show three capacitors connected in series. This means that one capacitor is connected after the other.The combination is connected to the terminals of a battery of vol

Capacitors in series

tage . We will substitute the series combination of fig.a with a single equivalent capacitor shown in fig.b which is also connected to an identical battery.The three capacitors have the

V

same charg

1 1 1

2 2 2

3 3 3

1 2 3

on their plates.The voltage across is: / . The voltage across is: / . The voltage across is: / .The net voltage across the combina

tion

Thus we have:

C V q CC V q CC V q C

V V V V

V q

e q

1 2 3

1 2 3

1 1 1

The equivalent capacitance 1 1 1eq

C C Cq qCV

qC C C

1 2 3

1 1 1 1

eqC C C C (25 - 11)

In general a capacitor systemmay consist of smaller capacitorgroups that can be identified asconnected "in parallel" or "in series"

More complex capacitor systems

12 1 2

1 2

12 3

In the example of the figure and in fig.a are connected . They can be substituted by the equivalent capacitor as shown in fig.b. Cap

in parallel

acitors and in fig.b are cC C C

C C

C C

123

123123 12 3

onnected .They can be substituted by a single capacitor as shown in fig.c

is given by the equation:

in series

1 1 1

C

CC C C

(25 - 12)

dAC o /

A questionWhat is the equivalent capacitance between the points A and B?

A.1 μFB. 2 μFC.4 μFD.10μFE.None of these

A B

What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 μC

---

--

+++++

dq'

q'-q'

V'

q'q

V' V

Charge

VoltageO

A

B

Consider a capacitor which is has a charge .We can calculate the work required to chargethe capacitor by assuming that we transfer a charge

from the negative

C qW

dq

Energy stored an an electric field

plate to the positive plate.We assume that the capacitor charge is and thecorresponding voltage . The work required

for the charge transfer is given by:

We continue this proc

qV dW

qdW V dq dqC

0

ess till the capacitor charge is

1equal to . The total work

q

q W V dq q dqC

2 2

0

21 If we substitute we get: or 2

Work can also be calculated by determining the area of triangle OAB which is

equal to . Area

2 2

2

2

qq CVqW q CV

C

W AVqV d

q

q W

VW WC

(25 - 13)

2 2

The work spent to charge a capacitor is stored in the form of potential energy that can be retrieved when is capacitor is

discharged. Thus 2 2

WU W

q CV qVUC

Potential energy stored in a capacitor

2

We can ask the question: where is the potential energy of a chargedcapacitor stored? The answer is counter intuitive. The energy is stored in the space between the capacitor plate

Energy density

s where a uniform electric field / is generated by the capacitor charges.In other words the electric field can store energy in empty space!

E V d

2 2

2 2q CVUC

---

--

+++++

q-q

d

E

A A

2 2

We define as energy densiry (symbol ) the potential energy per unit volume.

The volume between the plates is: where is the plate area

Thus the energy density 2 2 2

o

Uu uV

V V Ad A

U CV V VuAd Ad Ad d

22

This result, derived for the parallel plate capacitor holds in general

2oE

2

2oEu

(25 - 14)

q

-q

q'

q

q

-q'

-q

-q

V

V

V

V'

In 1837 Michael Faraday investigated what happens to thecapacitance of a capacitor when the gap between the plates is completely filled with an insulator (a.k.a. dielectri

C

Capacitor with a dielectric

c)Faraday discovered that the new capacitance is given by :

Here is the capacitance before the insertion of the dielectric between the plates. The factor is knownas the dielectric co

airairC CC

nstant of the material. Faraday's experiment can be carried out in two ways:

With the voltage across the plates remaining constantIn this case a battery remains connected to the plates . This is

V1.

shown in fig.a With the charge of the plates remaining constant.

In this case the plates are isolated from the batteryThis is shown in fig.b

q2.

airC C

(25 - 15)

q

-q

q'

q

q

-q'

-q

-q

V

V

V

V'

airC C

This is bacause the battery remains connected to the plates After the dielectric is inserted between the capacitor plates the plate charge changes from tq

Fig.a : Capacitor voltage V remains constant

o

The new capacitance airq κq q C κ κCV V V

q q

This is bacause the plates are isolated After the dielectric is inserted between the capacitor plates

the plate voltage changes from to

The ne

w capa

V VV

Fig.b : Capacitor charge q remains constant

citance / air

q q qC CV V V

(25 - 16)

conductor dielectric

qo

o o

In a region completely filled with an insulator ofdielectric constant , all electrostatic equations containing the constant are to be modified by replacing with

Electric field of

Example 1 :

2

a point charge inside

a dielectric:

The electric field outside an isolated conductor immersed in a dielectric becomes

14

:

o

o

qEr

E

Example 2 :

(25 - 17)

Dielectrics are classified into "polar" and "nonpolar"Polar dielectrics consist of molecules that have a non-zero electric dipole moment even at zero electric fielddue to

Dielectrics : An atomic view

2

the asymmetric distribution of charge within the molecule (e.g. H O ). At zero electric field(see fig.a) the electric dipole moments are randomly

oriented. When an external electric field is appoE

lied(see fig.b) the electric dipole moments tend to align

preferrentially along the direction of because thisconfiguration corresponds to a minimum of the potential energy and thus is a position

oE

of stable equilibrium. Thermal random motion opposes the alignment andthus ordering is incomplete. Even so, the partialalignment produced by the exterlal electric fieldgenerates an internal electric field that apposes

. Thus the net electric field is than

o oE E Eweaker

cosU pE

(25 - 18)

A nonpolal dielectric on the other hand consists of molecules that in the absence ofan electric field have zero electric dipole moment (see fig.a) . If we place the dielectric

between the plates of a capacitor the external electric field induces an electric dipole

moment that becomes aligned with (see fig.b). The aligned molecules do not create any net charge inside the dielectric. A

o

o

E

p E

net charge appears at the left and right surfaces of the dielectric opposite to the capacitor plates. These charges come from negative and positive ends of the electric dipoles. These surfaceinduced charges have sign to that of of the opposing plate charges. Thus the induced charges create an electric field

which opposes the applied field (see fig.c). As a result is that theoE E

opposite

net electric field between the capacitor plates is weaker.

E

oEE

(25 - 19)

S

S

n̂

n̂

In chapter 23 we formulated Gauss' law assuming that the

charges existed in vacuum. or

In this section we will write Gauss' law in a from which is suitab

o oE dA q q

Gauss' law and dielectrics

le for cases in which dielectrics are present. Consider first the parallel plate capacitor shown in fig.a.

We will the Gaussian surface S. The flux

Now we fill the space betwe

oo

oo

qE A

qEA

en the plates

with an insulator of dielectric constant (see fig.b) . We will apply Gauss' law for the same surface S. Inside S in addition to the plate charge

we also have the induced charge on the surface of the dielectric.

(eqs.1) Fro

o

o

qq qq EA

q qEA

m Faraday's experiments we have: (eqs.2)

If we compare eqs.1 with eqs.2 we have:

o

o

o

E qEA

qq A qq E d

(25 - 20)

Gauss' law in the presence of dielectrics

o E dA q

Even though the equation above was derived for the parallel plate capacitoris true in general.

The flux integral now involves The charge that is used is the plate charge, also

Eq

Note 1 : Note 2 :

known as "free charge"Using the equation above we can ignore the induced charge

The dielectric constant is kept inside the integral to desrcibe themost general case in which is not con

q

Note 3 :stant over the Gaussian surface

(25 - 21)

HITT

A parallel-plate capacitor has a plate area of 0.2m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 106 V/m between the plates, the magnitude of the charge on each plate should be:A. 8.9 x 10-7 C B. 1.8 x 10-6 C C. 3.5 x 10-6 C D. 7.1 x 10-6 C E. 1.4 x 10-5 C

A question

• Each of the four capacitors shown is 500 μF. The voltmeter reads 1000V. The magnitude of the charge, in coulombs, on each capacitor plate is:

A. 0.2 B. 0.5 C. 20 D. 50 E. none of these

Question

• A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about:

• A. 0 B. 5N C. 9N D. 1 x104 N E. 9 x 105 N

Question• A parallel-plate capacitor

has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about:

A. 0B. 5NC. 0. 9ND. 1 x104 NE. 9 x 105 N

N

AqqEF

o

71.43.01085.82

105

2

12

26

2

The electric field = σ/2εo why?