Chapter 23: Electrostatic Energy and Capacitance

Capacitors and Capacitance Capacitor

• Any two conductors separated by an insulator (or a vacuum) form a capacitor• In practice each conductor initially has zero net charge and electrons are transferred from one conductor to the other (charging the conductor)

• Then two conductors have charge with equal magnitude and opposite sign, although the net charge is still zero

• When a capacitor has or stores charge Q , the conductor with the higher potential has charge +Q and the other -Q if Q>0

Capacitors and Capacitance Capacitance

• One way to charge a capacitor is to connect these conductors to opposite terminals of a battery, which gives a fixed potential difference Vab between conductors ( a-side for positive charge and b-side for negative charge). Then once the charge Q and –Q are established, the battery is disconnected.• If the magnitude of the charge Q is doubled, the electric field becomes twice stronger and Vab is twice larger.

• Then the ratio Q/Vab is still constant and it is called the capacitance C.

• When a capacitor has or stores charge Q , the conductor with the higher potential has charge +Q and the other -Q if Q>0

ltcoulomb/vo 1C/V 1 farad 1F 1units abV

QC

-Q Q

Calculating Capacitance Parallel-plate capacitor in vacuum

• Charge density:A

Q

• Electric field:A

QE

00

• Potential diff.: A

QdEdVab

0

1

• Capacitance:d

A

V

QC

ab0

• The capacitance depends only on the geometry of the capacitor. • It is proportional to the area A.• It is inversely proportional to the separation d• When matter is present between the plates, its properties affect the capacitance.

-Q+Q

+Q

-Q

a

b

b

abaab EddEdEVVV

Calculating Capacitance Units

1 F = 1 C2/N m (Note [C2/N m2)

0 = 8.85 x 10-12 F/m

1 F = 10-6 F, 1 pF = 10-12 F

Example 24.1: Size of a 1-F capacitor

F 0.1 , mm 1 Cd

2812

3

0

m 101.1F/m 1085.8

m) 100.1F)(0.1(

Cd

A

Calculating Capacitance Example 24.2: Properties of a parallel capacitor

kV 10.0 V 000,10 , m 00.2 , mm 00.5

in vacuumcapacitor palte-parallelA 2 VAd

F 0.00354F 1054.3

m1000.5

)m F/m)(2.00 1085.8(

5

3

212

0

d

AC

C 35.4C1054.3

V) 10C/V)(1.00 1054.3(5

49

abCVQ

N/C 1000.2

)m 00.2)(mN/C 1085.8(

C 1054.3

6

22212

5

00

A

QE

Calculating Capacitance Example 24.3: A spherical capacitor

ra

rb-

--

--

-

-

-

+ ++

+ +

+

+

+

-Q

Qr

From Gauss’s law: 0enclQ

AdE

surfaceGaussian a as sphere aon point every at

toparallel and magnitudein constant is AdE

200

2

4)4(

r

QE

QrE

This form is the same as that for a point charge

r

QV

04

ba

ab

babaab rr

rrQ

r

Q

r

QVVV

000 444

ab

ba

ab rr

rr

V

QC

04

Calculating Capacitance Example 24.4: A cylindrical capacitor (length L)

Signal wire

Outer metal braid

r r

line charge density

Q -Q

23.10 Example from ln2

0

0 r

rV

a

b

a

bab

rrL

rr

L

V

QC

ln

2

ln2

0

0

Capacitors in Series and Parallel Capacitors in series

Capacitors in Series and Parallel Capacitors in series (cont’d)

a

b

cVVab 1VVac

2VVcb

Q

QQ

Q

1C

2C

22

11 C

QVV

C

QVV cbac

2121

11

CCQVVVVab

21

11

CCQ

V

The equivalent capacitance Ceq of the series combination is defined asthe capacitance of a single capacitor for which the charge Q is the sameas for the combination, when the potential difference V is the same.

i

ieqeqeqeq CCCCCQ

V

CV

QC

111111

21

Capacitors in Series and Parallel Capacitors in parallel

a

b

VVab 1Q 1C 2C

VCQVCQ 2211

VCCQQQ )( 2121

21 CCV

Q

The parallel combination is equivalent to a single capacitor with thesame total charge Q=Q1+Q2 and potential difference.

i ieqeq CCCCC 21

2Q

Capacitors in Series and Parallel Capacitor networks

Capacitors in Series and Parallel Capacitor networks (cont’d)

Capacitors in Series and Parallel Capacitor networks 2

C

A

B

A

B

C C

CCC

CCC

C C

C C

C C C3

1

A

B

C C

C C

C C3

4A

B

C41

15

Energy Storage and Electric-field Energy

Work done to charge a capacitor

• Consider a process to charge a capacitor up to Q with the final potential difference V.

C

QV

• Let q and v be the charge and potential difference at an intermediate stage during the charging process.

C

q

• At this stage the work dW required to transfer an additional element of charge dq is:

C

qdqdqdW

• The total work needed to increase the capacitor charge q from zero to Q is:

QW

C

Qqdq

CdWW

0

2

0 2

1

Energy Storage and Electric-field Energy

Potential energy of a charged capacitor

• Define the potential energy of an uncharged capacitor to be zero.

• Then W in the previous slide is equal to the potential energy U of the charged capacitor

QVCVC

QU

2

1

2

1

22

2

The total work W required to charge the capacitor is equal to the totalcharge Q multiplied by the average potential difference (1/2)V duringthe charging process

Energy Storage and Electric-field Energy

Electric-field energy

• We can think of the above energy stored in the field in the region between the plates.

• Define the energy density u to be the energy per unit volume

20

2

2

121

EAd

CVu

d

AC 0

field volume

This relation works for any electric field

Energy Storage and Electric-field Energy

Example 24.9: Two ways to calculate energy stored• Consider the spherical capacitor in Example 24.3.

• The energy stored in this capacitor is:ab

ba

rr

rrC

04

ba

ab

rr

rrQ

C

QU

0

22

82

• The electric field between two conducting sphere: 204 r

QE

• The electric field inside the inner sphere is zero.

• The electric field outside the inner surface of the outer sphere is zero.

40

2

22

20

02

0 3242

1

2

1

r

Q

r

QEu

b

a

b

a

r

rba

abr

r rr

rrQ

r

drQdrr

r

QudVU

0

2

20

22

40

2

2

884

32

Energy Storage and Electric-field Energy

Example : Stored energy

R

QdV

r

QU

Eur

QE

R )4(24

1

2

1

2

1

4

1

0

2

4

22

00

202

0

Dielectrics

Dielectric materials

• Experimentally it is found that when a non-conducting material (dielectrics) between the conducting plates of a capacitor, the capacitance increases for the same stored charge Q.

• Define the dielectric constant K in the textbook) as:

0C

C

• When the charge is constant, VVCCCVVCQ // 0000

00 E

EV

V

Material Material

vacuum 1air(1 atm) 1.00059

Teflon 2.1

Polyethelene 2.25

Mica 3-6

Mylar 3.1

Plexiglas 3.40Water 80.4

Dielectrics

Induced charge and polarization• Consider a two oppositely charged parallel plates with vacuum between the plates.• Now insert a dielectric material of dielectric constant .

constant is Qwhen /0 EE • Source of change in the electric field is redistribution of positive and negative charge within the dielectric material (net charge 0). This redistribution is called a polarization and it produces induced charge and field that partially cancels the original electric field.

0

000

EEEE ind

0ty permittivi thedefine and 1

1

ind

E 22000 2

1

2

1EEu

d

A

d

ACC

Dielectrics

Molecular model of induced charge

Dielectrics

Molecular model of induced charge (cont’d)

Dielectrics

Why salt dissolves

Normally NaCl is in a rigid crystal structure, maintained by the electrostatic attraction between the Na+ and Cl- ions.

Water has a very high dielectric constant (78). This reduces the field between the atoms, hence their attraction to each other. The crystal lattice comes apart and dissolves.

Dielectrics

Gauss’s law in dielectrics

cond

ucto

r

diel

ectr

ic

+

+

+

+

+

-

-

ind

Gauss’s law:0

)(

A

EA ind

indind or

11

00

or

A

EAA

EA

0

freeenclQAdE

enclosed freecharge

ExercisesProblem 1

(a) If the distance d is halved, how much does the capacitance changes?(b) If the area is doubled, how much does the capacitance changes?(c) For a given stored charge Q, to double the amount of energy stored how much should the distance d be changed?Now a metal slab of thickness a (< d) and of the same area A is insertedbetween the two plates in parallel to the plates as shown in the figure(the slab does not touch the plates). (d) What is the capacitance of this arrangement?(hint:serial connection)(e) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present.

adAn air capacitor is made by using two flat plateseach with area A separated by a distance d.

Problem 1 Solution

(a) doubled. is C so ,0 d

AC

(b) doubled. is C so ,0 d

AC

(c) doubled. be should d and 2

so ,2

and 0

22

0 A

dQU

C

QU

d

AC

(d)

ad

ACCC

ad

AC

ad

eqeq

0eq

0

/2/1 :is C ecapacitanc equivalent the

Therefore .2

ecapacitanc thehascapacitor two theseofEach

plates. ebetween th 2/)( of gap a has which ofeach series,in connected

capacitors twoof system a be toconsidered becan t arrangemen This

(e)000 therefore, C

ad

dC

d

AC eq

Problem 2

(a) Express the capacitance C0 in terms of the potential difference V0

between the plates and the charge Q if air is between the plates.(b) Express the dielectric constant in terms of the capacitance C0 (air gap) and the capacitance C with material of the dielectric constant ). (c) Using the results of (a) and (b), express the ratio of the potential difference V/V0 if Q is the same, where V is the potential difference between the plates and a dielectric material dielectric constant is fills the space between them.(d) A voltmeter reads 45.0 V when placed across the capacitor. When dielectric material is inserted completely filling the space, the voltmeter reads 11.5 V. Find the dielectric constant of this material. (e) What is the voltmeter read if the dielectric is now pulled partway out so that it fills only one-third of the space between the plates? (Use the formula for the parallel connection of two capacitors.)

In this problem you try to measure dielectric constant of a material. Firsta parallel-plate capacitor with only air between the plates is charged byconnecting it to a battery. The capacitor is then disconnected from thebattery without any of the charge leaving the plates.

Problem 2

(a)00 /VQC

(b) 0/CC(c) /1/// 000 VVCCVV(d) From (c) 91.35.11/0.45/0 VV

(e) In the new configuration the equivalent capacitor is where C1/3 is the contribution from the part that has the dielectric material and C0,2/3 is the part that has air gap. because the capacitance is proportional to the area.

3/2,03/1 CCCeq

03/2,03/1 (2/3) and )3/1( CCCC

)]3/2()3/1[()3/2()3/1( 003/2,03/1 CCCCCCeq

V 8.2291.5

3)V 0.45()]3/2()3/1/[(

)]3/2()3/1[(//

0

00

VV

CCVV eq

Using the results from (c)