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Capacitance and Dielectrics

Capacitance

General Definition:

VqC /==== Special case for parallel

plates:

d

A C 0

εεεε ====

Potential Energy

• I must do work to charge up a capacitor.

• This energy is stored in the form of electric

potential energy.

• We showed that this is C

Q U

2

2

====

2

02

1 Eu εεεε====

• Then we saw that this energy is stored in the

electric field, with a volume energy density

Potential difference and

Electric field Since potential difference is work per unit charge,

∫∫∫∫====∆∆∆∆ b

a EdxV

For the parallel-plate capacitor E is uniform, so

EdV ====

Also for parallel-plate case Gauss’s Law gives

d

A

V

Q CsoVd

A

Q E 0

0

0/ εεεε

εεεε εεεεσσσσ ====================

Spherical example A spherical capacitor has inner radius a = 3mm,

outer radius b = 6mm. The charge on the inner

sphere is q = 2 C. What is the potential difference?

From Gauss’s Law or the Shell

Theorem, the field inside is 2r

kq E ====

From definition of

potential difference

V 33

33

99 103

6

1

3

1 1018

106

1

103

1 102109 ××××====

−−−−××××====

××××

−−−− ××××

××××××××××××==== −−−−−−−− −−−−

∫∫∫∫

−−−−======== b

a ba kqdr

r

kq V

11 2

What is the capacitance?

FVCVQC 4

107.6)3000/()2(/ −−−−××××============

Q.25-1

(1) 3 µF

(2) 6 µF

(3) 12 µF

(4) 24 µF

A capacitor has capacitance C = 6 µF

and charge Q = 2 nC. If the charge is

increased to 4 nC what will be the

new capacitance?

Q. 25 - 1

A capacitor has capacitance C = 6 µF and a

charge Q = 2 nC. If the charge is increased to 4

nC what will be the new capacitance?

(1) 3 µF (2) 6 µF (3) 12 µF (4) 24 µF

Solution:

Capacitance depends on the structure of

the capacitor, not on its charge.

Q.25-2

1) 10,000 V/m

2) 900 V/m

3) 90 nV/m

A parallel-plate capacitor has plates of area

0.1 m2 and carries a charge of 9 nC.

What is the electric field between the plates?

Q. 25 - 2

Solution: Field at surface

of conductor:

mVmVmV /109)3(/900)2(/101)1( 84 −−−−××××××××

?

109

1.0

9

2

====

××××====

==== −−−−

E

CQ

mA

mVE

AQ

E

/101 109

109

109 101

109 /

/

4

12

8

8

1

9

0

××××==== ×××× ××××

====

××××==== ×××× ××××

========

====

−−−−

−−−−

−−−− −−−−

−−−−

σσσσ

εεεεσσσσ

Capacitors in Parallel

If several capacitors are

connected in parallel as

shown, we can consider

the arrangement as a

single capacitor.

L++= 21 qqqtot By charge conservation:

L============ 21 VVVtot

The wires are good

conductors, so:

tottottot

tot

CVq V

qqq

V

q

V

q

V

q CC

======== ++++++++++++

====

++++++++++++====++++++++

/321

3

3

2

2

1

1 21

L

LL

Therefore:

Capacitors in Series

What if several capacitors are

connected in series as shown?

Now charge conservation

gives:

L============ 21 qqqtot

And now it is the voltages that add:

L++++++++==== 21 VVVtot

So now the voltage sum,

when written in terms of the

capacitances, gives the result

for the series case:

L

L

L

++++++++====

++++++++====

++++++++====

21

2211

21

/1/1/1

///

CCC

CqCqCq

VVV

tot

tottot

tot

Results for Capacitor Combinations

• Capacitors in parallel:

L++++++++==== 21 CCCtot

• Capacitors in series:

L++++++++==== 21 /1/1/1 CCCtot

Example: Problem 25-8

What is effective capacitance of the device?

C12 FC

FC

FC

µµµµ

µµµµ

µµµµ

0.4

0.5

10

3

2

1

====

====

====

FC

CCC

µµµµ3.33/10

10/35/110/1/1/1/1

12

2112

========

====++++====++++====

FCCCtot µµµµ3.743.3312 ====++++====++++====

Dielectric Materials

• Applied electric field polarizes material

• This produces an induced surface charge

• This reduces field within material

• This reduces potential difference

• This increases capacitance

Effect of placing a dielectric between the

plates of a charged capacitor:

E=0

electronnucleus

Polarization of atoms and molecules

Non-polar molecules:

induced dipoles

Polar molecules:

preferential orientation

Dielectric materials

Dielectric materials polarize under external electric field E0

Polarization creates the opposing electric field: –E’

Total field E=E0-E ’=E0/κ < E0

κ is called dielectric constant

It shows how the field in a dielectric is weaker than in vacuum

Dielectric Constant

• So the effect of a dielectric is to

decrease the field and increase the

capacitance.

• The factor by which E is decreased

and C is increased is defined to be the

dielectric constant κ.

0CC κκκκ==== κκκκ/0EE ====

Example: Problem 25-48

Given a parallel-plate capacitor, dielectric-

filled, with area A = 100 cm2, charge Q = 890

nC, and electric field E = 1.4 kV/mm.

(a) Find dielectric constant of the material.

(b) Find the induced charge.

(a)

(((( )))) 00 // εεεεAqE ====

If there were no dielectric the field

between the plates would be given

by Gauss’s Law as

mVE /1001.1 1085.8

10/109.8 7 12

27

0 ××××====×××× ××××

==== −−−− −−−−−−−−

2.7104.1/1001.1/ 67

0 ====××××××××======== EESoκκκκ

κ/0EE =But by definition:

0

/

εεεε Aq

E

Gauss

tot====

⇒⇒⇒⇒

nCnCnCqqqso

qqqBut

tot

tot

766124890'

'

====−−−−====−−−−====

−−−−====

nCC

EAqSo tot

124104.1210104.11085.8 82612

0

====××××====××××××××××××××××====

==== −−−−−−−−−−−−

εεεε

(b)