Capacitance Review

and

Capacitors with Dielectrics

2

Forming a Capacitor

Any two conducting electrodes can form a capacitor, regardless of their shape.

C

QC

V≡

∆

The capacitance depends only on the geometry of the electrodes, not on their present charge or potential difference.

(In fact, one of the electrodes can be moved to infinity, so the capacitance of a single electrode is a meaningful concept.)

3

Capacitors and Capacitance

1C w1 w2 bat2

Initial: 0, .V V V V∆ = ∆ = ∆ = ∆

bat C w1 w2Final: , 0.V V Ed V V∆ = ∆ = ∆ = ∆ =

0

QE

Aε=

00 C

AQ AE V

d

εε = = ∆

0

C

AQC

V d

ε≡ =∆

Units: 1 farad 1 F 1 C/V= ≡

CQ C V= ∆

4

S

P

N

n̂

The plates have area and are separated

by a distance . The upper plate has a

charge and the lower plate a charge -

A

d

q q+

Capacitance of a parallel plate capacitor

We apply Gauss' law using the Gaussian surface S shown in the figure.

The electric flux cos0 .

From Gauss' law we have:

The potential difference between the positive ano o o

EA EA

q q qEA E

A

V

ε ε ε

Φ = =

Φ = → = → =

d the negative plate is

given by: cos0

The capacitance /

o

o

o

qdV Eds E ds Ed

A

Aq qC

V qd A d

εε

ε

+ +

− −

= = = =

= = =

∫ ∫

oAC

d

ε=

5

Example: A Spherical Capacitor

A metal sphere of radius R1 is inside and concentric with a hollow metal sphere of inner radius R2.

What is capacitance of this spherical capacitor?

6

Example: A Spherical Capacitor

A metal sphere of radius R1 is inside and concentric with a hollow metal sphere of inner radius R2.

What is capacitance of this spherical capacitor?

( ) ( )f

i

s

f i s

s

V V s V s E ds∆ = − = −∫

2

1

20 0 1 2

1 1 1

4 4

R

C

R

Q QV ds

r R Rπε πε

∆ = = −

∫

1

1 20 0

1 2 2 1

1 14 4

C

R RQC

V R R R Rπε πε

−

= = − = ∆ −

7

Combining Capacitors Parallel: Same ∆V, but different Qs.

Series: Same Q, but different ∆Vs.

1 2 3parallel

C C

1 2 3

Q Q QQC

V V

C C C

+ + += =∆ ∆

= + + +

L

L

( ) ( ) ( )

seriesC 1 2 3

1 2 3

1 2 31 2 3

1

/ / /

1|| || ||

1/ 1/ 1/

Q QC

V V V V

V Q V Q V Q

C C CC C C

= =∆ ∆ + ∆ + ∆ +

=∆ + ∆ + ∆ +

= ≡+ + +

L

L

LL

8

Energy Stored in a Capacitor

1U dq V qdq

C∆ = ∆ =

21

C 20

1Q

QU qdq

C C= =∫

21 1 2

C C2 2

QU C V

C= = ∆

9

Dielectric Materials

E off E on

There is a class of polarizable dielectric materials that have an important application in the construction of capacitors. In an electric field their tiny dipoles line up, reducing the E field and potential difference and increasing the capacitance:

0

C

AQC

V d

κε≡ =∆

10

Electric Fields and Dielectrics

In an external field Eo, neutral molecules can polarize. The induced electric field E′produced by the dipoles will be in the opposite direction from the external field Eo. Therefore, in the interior of the slab the resulting field is E = Eo-E′.

The polarization of the material has the net effect of producing a sheet of positive charge on the right surface and a sheet of negative charge on the left surface, with E′ being the field made by these sheets of charge.

11

Capacitors and DielectricsIf a capacitor connected to a battery, so that it has a charge

q, and then a dielectric material of dielectric constant κ is placed in the gap, the potential is unchanged but the charge becomes κq.

12

Capacitors and DielectricsIf a capacitor is given a charge q, and then a dielectric

material of dielectric constant κ is placed in the gap, the charge q is unchanged, but the potential drops toV/κ.

13

q

-q

q'

q

q

-q'

-q

-q

V

V

V

V'

In 1837 Michael Faraday investigated what happens to the

capacitance 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 known

as the dielectric co

airairC CCκκ

=

nstant of the material.

Faraday's experiment can be carried out in two ways:

With the voltage across the plates remaining constant

In 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 battery

This is shown in fig.b

q2.

airC Cκ=

14

airC Cκ=

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 air

q κq q C κ κC

V V V

q qκ′

= = = =

′ =

This is bacause the plates are isolated

After the dielectric is inserted between the capacitor plates

the plate voltage changes from to

The new capa

VV

Vκ

′ =

Fig.b : Capacitor charge q remains constant

citance

/ air

q q qC C

V V Vκ κ

κ= = = =

′

15

A capacitor has a capacitance of 5 x 10-6 F. If silicon (κ = 12) is inserted and fills the region between the plates, find the new capacitance of the capacitor.

16

17

18

19

20

21

22

23

24

conductor dielectric

qo

o o

In a region completely filled with an insulator of

dielectric constant , all electrostatic equations

containing the constant are to be modified by

replacing with

Electric field of

κε

ε κε

Example 1 :

2

a point charge inside

a dielectric:

The electric field outside an isolated conductor

immersed in a dielectric becomes

1

4

:

o

o

qE

r

E

πκε

σκε

=

=

Example 2 :

25

Relationship Between Induced Charge and Dielectric Constant