Capacitance

President University Erwin Sitompul EEM 9/1

Lecture 9

Engineering Electromagnetics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 4


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 M1 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 (dS and dL) and of the permittivity of the homogenous dielectric (ε).

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



Capacitance of a Two-Wire LineThe configuration of the two-wire line consists of two parallel

conducting cylinders, each of circular cross section.We shall be able to find complete information about the electric

field intensity, the potential field, the surface charge density distribution, and the capacitance.

This arrangement is an important type of transmission line.

• Schematics of a transmission line



Capacitance of a Two-Wire LineThe capacitance, together with conductance, forms a shunt

admittance of a transmission line.The line capacitance is proportional to the length of the

transmission line.When an alternating voltage is applied to the line, the line

capacitance draws a leading sinusoidal current, called the charging current.

The charging current is negligible for lines less than 100 km long. For longer lines, the capacitance becomes increasingly important and has to be accounted for.

The value of such capacitance is significantly higher with underground cables than with overhead lines, due to the close proximity of the individual conductors.



Capacitance of a Two-Wire LineThe potential field of two

infinite line charges, with a positive line charge in the xz plane at x = a and a negative line at x = –a, is shown below.

The potential of a single line charge with zero reference at a radius of R0 is:

0ln2L R

VR

10 20

1 2

ln ln2L R RV

R R

10 2

20 1

ln2L R R

R R

The combined potential field can be written as:



Capacitance of a Two-Wire LineWe choose R10 = R20, thus placing the zero reference at equal

distances from each line.Expressing R1 and R2 in terms of x and y,

2 2

2 2

( )ln2 ( )L x a yV

x a y

141

LVK e

2 2

1 2 2

( )( )x a yKx a y

2 2

2 2

( )ln4 ( )L x a y

x a y

To recognize the equipotential surfaces, some algebraic manipulations are necessary.

Choosing an equipotential surface V = V1, we define a dimensionless parameter K1 as:

so that



Capacitance of a Two-Wire Line

2 2 21

1

12 01

Kx ax y aK

22

121

1 1

211 1

a KKx a yK K

1

1

21

a Kb

K

1

1

11

Kh aK

After some multiplications and algebra, we obtain:

The last equation shows that the V = V1 equipotential surface is independent of z and intersects the xy plane in a circle of radius b,

The center of the circle is x = h, y = 0, where:



Capacitance of a Two-Wire Line

2 2a h b

Let us now consider a zero-potential conducting plane located at x = 0, and a conducting cylinder of radius b and potential V0 with its axis located a distance h from the plane.

Solving the last two equations for a and K1 in terms of b and h,

2 2

1h h bK

b

The potential of the cylinder is V0, so that:

021

LVK e

0 0

1 1

4 2ln lnLV VK K

Therefore,


Capacitance of a Two-Wire LineChapter 6 Dielectrics and Capacitance

0

LLCV

2 2

2

ln

LCh h b b

Given h, b, and V0, we may determine a, K1, and ρL.The capacitance between the cylinder and the plane is now

available. For a length L in the z direction,

1

4lnLK

1

2ln

LK

1

2cosh ( )

Lh b

• Prove the equity by solving quadratic equation in eα, where cosh(α) = h/b.

• cosh(α) = (eα+e–α )/2


2 2a h b


Capacitance of a Two-Wire LineExample

The black circle shows the cross section of a cylinder of 5 m radius at a potential of 100 V in free space. Its axis is 13 m away from a plane at zero potential.

05, 13, 100b h V 2 213 5 12

2 2

1h h bK

b

13 125

5 1 25K

0

1

4lnLVK

1

2cosh ( )

Ch b

124 (8.854 10 )(100) 3.46 nC mln 25

12

1

2 (8.854 10 ) 34.6 pF mcosh (13 5)



We may also identify the cylinder representing the 50 V equipotential surface by finding new values for K1, b, and h.

141

LVK e 12 94 8.854 10 50 3.46 10e

5

1

1

21

a Kb

K

1

1

11

Kh aK

2 12 55 1

13.42 m

5 1125 1

18 m



2 2

2 2

( )ln2 ( )L x a y

x a y

E

2 2 2 2

2( ) 2 2( ) 22 ( ) ( )

x y x yLx a y x a yx a y x a y

a a a a

2 2 2 2

2( ) 2 2( ) 22 ( ) ( )

x y x yLx a y x a yx a y x a y

a a a aD E =

,max , , 0S x x h b yD

9

,max 2 2

3.46 10 13 5 12 13 5 122 (13 5 12) (13 5 12)S

20.165 nC m

2 22 ( ) ( )L h b a h b ah b a h b a

=



++

+

+

++

+

+

-

--

--

--

-

9

,min 2 2

3.46 10 13 5 12 13 5 122 (13 5 12) (13 5 12)S

20.073 nC m

,min , , 0 2 22 ( ) ( )L

S x x h b yh b a h b aDh b a h b a

=

,min , , 0S x x h b yD

,max , , 0S x x h b yD

,max ,min2.25S S


Capacitance of a Two-Wire LineFor the case of a conductor with b << h, then:


2 2ln ln ln 2h h b b h h b h b

2ln(2 )

LCh b

( )b h


Homework 8D6.4.D6.5.D6.6.


Deadline: Monday, 23 June 2014.

For D6.6., replace “1 cm, 20 V, 5 cm” with “StID cm, BMonth V, StID+7 cm”

Example: Rudi Bravo (002201700016) was born on 3 June 2002. Rudi will do D6.6 with ”16 cm, 6 V, 23 cm”.

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Capacitance