Capacitance

President University Erwin Sitompul EEM 9/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 9

Engineering Electromagnetics

http://zitompul.wordpress.com


Chapter 6 Dielectrics and Capacitance

CapacitanceNow let us consider two conductors

embedded in a homogenous dielectric.Conductor M2 carries a total positive

charge Q, and M2 carries an equal negative charge –Q.

No other charges present the total charge of the system is zero.

• The charge is carried on the surface as a surface charge density.

• The electric field is normal to the conductor surface.

• Each conductor is an equipotential surface


CapacitanceThe electric flux is directed from M2

to M1, thus M2 is at the more positive potential.

Works must be done to carry a positive charge from M1 to M2.

Let us assign V0 as the potential difference between M2 and M1.

We may now define the capacitance of this two-conductor system as the ratio of the magnitude of the total charge on either conductor to the magnitude of the potential difference between the conductors.


0

QCV

S

dC

d

E S

E L


CapacitanceThe capacitance is independent of the potential

and total charge for their ratio is constant. If the charge density is increased by a factor,

Gauss's law indicates that the electric flux density or electric field intensity also increases by the same factor, as does the potential difference.


Sd

Cd

E S

E L

Capacitance is a function only of the physical dimensions of the system of conductors and of the permittivity of the homogenous dielectric.

Capacitance is measured in farads (F), 1 F = 1 C/V.


CapacitanceChapter 6 Dielectrics and Capacitance

We will now apply the definition of capacitance to a simple two-conductor system, where the conductors are identical, infinite parallel planes, and separated a distance d to each other.

Sz

E a

S zD a

The charge on the lower plane is positive, since D is upward.N z SD D

The charge on the upper plane is negative,N z SD D


CapacitanceThe potential difference between lower and upper planes is:


lower

0 upperV d E L

0S

ddz

S d

The total charge for an area S of either plane, both with linear dimensions much greater than their separation d, is:

SQ S

The capacitance of a portion of the infinite-plane arrangement, far from the edges, is:

0

QCV

Sd


CapacitanceChapter 6 Dielectrics and Capacitance

ExampleCalculate the capacitance of a parallel-plate capacitor having a mica dielectric, εr = 6, a plate area of 10 in2, and a separation of 0.01 in.

210 inS 2 2 210 in (2.54 10 m in)

3 26.452 10 m

0.01ind 20.01in (2.54 10 m in)

42.54 10 m

SCd

12 3

4

(6)(8.854 10 )(6.452 10 )2.54 10

1.349 nF


0SV d

CapacitanceThe total energy stored in the capacitor is:


1 22 volEW E dv

212 vol

S dv

212 0 0

S dS dzdS

212

S Sd

2

1 222SS d

d

1 202EW CV 1

02QV2

12

QC

0

QCV

SCd


1 14abQV

a b

Several Capacitance ExamplesAs first example, consider a coaxial cable or coaxial capacitor

of inner radius a, outer radius b, and length L.The capacitance is given by:


ln2L

abaVb

LQ Lab

QCV

2

ln( )L

b a

Next, consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b>a.

ab

QCV

41 1a b


If we allow the outer sphere to become infinitely large, we obtain the capacitance of an isolated spherical conductor:


Several Capacitance Examples

4C a

0.556 pFC

24rQDr

214rQEr

204Qr

1( )a r r

1( )r r

A sphere about the size of a marble, with a diameter of 1 cm, will have:

Coating this sphere with a different dielectric layer, for which ε = ε1, extending from r = a to r = r1,


Several Capacitance ExamplesWhile the potential difference is:


1

12 2

1 04 4a r

a r

Qdr QdrV Vr r

1 1 0 1

1 1 1 14Q

a r r

1 1 0 1

41 1 1 1

C

a r r

Therefore,



Several Capacitance ExamplesA capacitor can be made up of several dielectrics.Consider a parallel-plate capacitor of area S and spacing d,

d << linear dimension of S.The capacitance is ε1S/d, using a dielectric of permittivity ε1.Now, let us replace a part of this dielectric by another of

permittivity ε2, placing the boundary between the two dielectrics parallel to the plates.

• Assuming a charge Q on one plate, ρS = Q/S, while

DN1 = DN2, since D is only normal to the boundary.

• E1 = D1/ε1 = Q/(ε1S), E2 = D2/ε2 = Q/(ε2S).

• V1 = E1d1, V2 = E2d2.

0

QCV

1 2

1 2

1d dS S

1 2

QV V

1 2

11 1C C



Several Capacitance Examples

1 1 2 2S Sd

Another configuration is when the dielectric boundary were placed normal to the two conducting plates and the dielectrics occupied areas of S1 and S2.

• Assuming a charge Q on one plate, Q = ρS1S1 + ρS2S2.

• ρS1 = D1 = ε1E1, ρS2 = D2 = ε2E2.• V0 = E1d = E2d.

0

QCV

1 2C C


Homework 8D6.4D6.5

Deadline: 19.06.2012, at 08:00.


Documents

Capacitance