Conductors in Fem Analysis

7/29/2019 Conductors in Fem Analysis

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Conductors

A conductor (typically, a metal like Cu, Ag etc. or ionic conductors like HCl or

NaCl dissolved in water) allows free movement of charges. They have low resis-

tivity 10

8

m as compared to typical insulators like quartz, glass etc. which haveresistivity of the order of1017m. However, the property that really distinguishesa metal from insulators or semi-conductors is the fact their temperature coefficient

of resistivity is positive while that of semi-conductors is negative.

The electric field inside a conductor is zero. In an equilibrium situa-tion, there cannot be an electric field inside a conductor as this would cause

charges (electrons or ions) to move around. In the presence of an external

field, there is charge separation inside a conductors with opposite charges

accumulating on the surface. This creates an internal electric field which

cancels the effect of the external field in such a way that the net electric

field inside the conductor volume is zero.

+

+

+

+

+

+

+

+

E

E

int

ext

Charge density inside a conductor is zero. This follows from Gausss law

E = /0

As E = 0, the charge density = 0.

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(This does not suggest that there is no charge inside, only that the positive

and negative charges cancel inside a conductor.)

Free charges exist only on the surface of a conductor. Since there is nonet charge inside, free charges, if any, have to be on the surface.

At the surface of a conductor, the electric field is normal to the surface.If this were not so, the charges on the surface would move along the surface

because of the tangential component of the field, disturbing equilibrium.

E=0

Induced Charges in a conductor:The above properties of a conductor influence the behaviour of a conductor placed

in an electric field. Consider, for instance, what happens when a charge +q isbrought near an uncharged conductor. The conductor is placed in the electric field

of the point charge. The field inside the conductor should, however, be zero. his

is achieved by a charge separation within the conductor which creates its own

electric field which will exactly compensate the field due to the charge +q. Theseparated charges must necessarily reside on the surface.

Another way of looking at what is

happening is to think of the free

charges in the conductor being at-

tracted towards (or repelled from)

the external charge. Thus the surface

of the conductor towards the exter-

nal charge is oppositely charged. To

keep the charge neutrality, the sur-

face away from the external chargeis similarly charged.

+q

++

++

+++++

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Example 1 :

A charge Q is located in the cavity inside a conducting shell. In addition, a charge2Q is distributed in the conducting shell. Find the distribution of charge in theshell. What is the electric field in the region outside the shell.

+Q

+

+

++

+

++

+

++ + + +

+

+

+

+

+++

Take a gaussian surface entirely within the conducting shell, completely enclosing

the cavity. Everywhere on the gaussian surface E = 0. The flux and therefore,the charge enclosed is zero within the gaussian surface. As the cavity contains a

charge Q, the surface of the cavity must have charge Q. As the conductor hasdistributed charge 2Q, the charge on the outside surface is 3Q.The principle illustrated in the above problem is known as Faradays Cage. If

a hollow conducting box is kept in an electric field, the charges in the cavity are

redistributed in such a way that the electric field inside the cavity is zero. This

is used to provide an enclosure for sensitive electronic equpment which must be

kept free of external electronic disturbance.

Example 2 :

Calculate the electric field outside a conductor carrying a surface charge density

.

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+

++

+++

+ + +

+

+

+

++ +

+

+

+

+ +

+

++++

+L

dSE

r

++++ +

Take a gaussian pillbox in the shape of a cylinder of height h with h/2 insideand h/2 outside the conductor. Lat the cross sectional area be dS normal to the

surface. The electric field is normal to the surface. As the field inside is zer andthere is no tangential component of the field at the surface, the flux goes out only

through the outer cap of the cylider. The charge enclosed is dS and the flux isEdS. The electric field is normal to the surface. applyinng Gausss law

E =

0n

.

Exercise :

Two parallel, infinite plates made of material of perfect conductor, carry charges

Q1 and Q2. The plates have finite thickness. Show that the charge densities on thetwo adjecent inside surfaces are equal and opposite while that on the two outside

surfaces are equal. (Hint : Field inside the plates due to four charged surfaces

must be zero.)

Poissons and Laplaces Equations

Differential form of Gausss law,

E =

0

Using E = V,E = (V) = 2V

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so that

2V =

0

This is Poisson equation. In cartesian form,

2

Vx2

+ 2

Vy2

+ 2

Vz2

= 0

For field free region, the equation becomes Laplaces equation

2V = 0

Equipotential surface

Equipotential surfaces are defined as surfaces over which the potential is constant

V(r) = constant

At each point on the surface, the electric field is perpendicular to the surface since

the electric field, being the gradient of potential, does not have component along

a surface of constant potential.

We have seen that any charge on a conductor must reside on its surface.These charges would move along the surface if there were a tangential com-

ponent of the electric field. The electric field must therefore be along the

normal to the surface of a conductor. The conductor surface is, therefore,

an equipotential surface.

Electric field lines are perpendicular to equipotential surfaces (or curves)and point in the direction from higher potential to lower potential.

In the region where the electric field is strong, the equipotentials are closelypacked as the gradient is large.

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2 kV

1 kV

0.5 kV

2 kV

3 kV

=

1 kV

xP

0

The electric field strength at the point P may be found by finding the slope of the

potential at the point P. If x is the distance between two equipotential curvesclose to P,

E = V

x

where V is the difference between the two equipotential curves near P.

Example 3:

Determine the equipotential surface for a point charge.

Solution : Let the point charge q be located at the origin. The equation to theequipotential surface is given by

V(x,y,z) =1

40

qx2 + y2 + z2

= V0 = constant

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Equipotential surfaces (magenta)

and field lines (blue) for a positive

charge.

Thus the surfaces are concentric spheres with the origin (the location of the charge)

as the centre and radii given by

R =q

400

The equipotential surfaces of an electric dipole is shown below.

Electric Field and Equipotential lines

for an electric dipole

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Example 4 :

Determine the equipotential surface of an infinite line charge carrying a positive

charge density .Solution :

Let the line charge be along the z- axis. The potential due to a line charge at a

point P is given by

V(r) =

20ln r

where r is the distance of the point P from the line charge. Since the line chargealong the z-axis, r =

x2 + y2 so that

V(r) =

40ln(x2 + y2)

The surface V = constant = V0 is given by

ln(x2 + y2) = 40V0

i.e.

x2 + y2 = e40V0

which represent cylinders with axis along the z-axis with radii

r = e20V0

1

++

+++++++++++++++++++++++++

+++

zaxis

2

As V0 increases, radius becomessmaller. Thus the cylinders are

packed closer around the axis, show-

ing that the field is stronger near the

axis.

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Example 5 :

Consider a charged sphere of radius R containing charge q, completely enclosedby a spherical cavity of inner radius a and outer radius b. Calculate the chargedensity on all surfaces and potential everywhere.

Solution :

As field inside the conductor is zero,

by taking a Gaussian surface com-

pletely in the region a < r < b,we must have net charge enclosed by

such a surface to be zero. To com-

pensate for the charge q that existson the surface of the inner sphere,

the charge on the inside surface of

the shell must be q. Since the shell

is charge neutral, a charge +q must,therefore, appear on the outside sur-face of the shell.

R

ab

+q

q+q

Gaussian surface

For r > b, the field is

E =1

40

q

r2r

The corresponding potential is

V(r) =

r

1

40

q

r2dr =

1

40

q

r

At r = b, the potential is

V(b) = 140

qb

Since the field between a and b is zero, this is also the potential at all points fromr = b to r = a.

V(a r b) =1

40

q

b

For R < r < a, the potential is given by

V(r) =1

40

q

b

Ra

1

40

q

r2

= 140

qb

+ qR q

a

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If the outer surface is grounded, the potential on the shell becomes zero. There is

no charge on the outer surface. However the inner surface must have a charge qto keep the field in the shell zero,

V(0) = V(R) =

R

a

E d

l =

1

40 1

R

1

a

Exercise :

Determine the equipotential surface of an infinite plane with charge density .

Laplaces Equation

Let us look at Laplaces equation in one dimension. It becomes

d2V

dx2= 0

which has the solution

V = mx + c

The solution shows two important characteristics of the solution of Laplaces

equation, which are not immediately obvious in higher dimensions. The first

property is the potential at a point can be expressed asaverage of potentials at

neighbouring points. For instance,

V(x) =1

2(V(x + x0) + V(x x0))

This also illustrates the second property of the solutions, viz., the solution has no

local minimum or maximum. If it did, it would not be possible to express thefunction as average of values at neigbouring points.

To see this consider a function

f(x, y) =a

4(x2 + y2)

in two dimensions, which does not satisfy Laplaces equation as

2V =2f

x2+

2f

y2= a

The function has a positive curvature everywhere and there exists a local minimum

at x = 0, y = 0. The function looks like the following.

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0246810121416

18

-3 -2 -1 0 1 2 3 -3-2

-10

12

300.511.522.533.544.555.566.577.588.599.5

1010.51111.51212.51313.51414.51515.51616.51717.518

V(x,y)

xy

V(x,y)

Consider, on the other hand, a function V(x, y) that satisfies Laplaces equation

V(x, y) =a

4(x2 y2)

The function has no minimum or maximum and looks like the following. It has a

saddle point at x = 0, y = 0.

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-10-8-6-4-202468

10

-3 -2 -1 0 1 2 3 -3-2

-10

12

3-9-8.5-8-7.5-7-6.5-6-5.5-5-4.5-4-3.5-3-2.5-2-1.5-1-0.500.511.522.533.544.555.566.577.588.59

V(x,y)

xy

V(x,y)

An interesting consequence of Laplace equation is Earnshaw Theorem which

states that a charge cannot be held in stable equilibrium only by electrostatic

forces.

For instance, suppose we position a charge Q exactly at the centre of a cube whichhas a positive charge qat each of its eight corners. We would expect the charge tobe in equilibrium as it is being pulled equally in all directions. However, this will

not be a stable equilibrium because at the centre, there being no charge density,

Laplace equation is obeyed. Thus there cannot be a minimum of the potential Vand hence of potential energy QV of the charge at the centre.Consider again the case of cavity in a conductor. If the interior of the cavity does

not contain any charge, Laplace equation is obeyed. Thus the potential has no

minimum or maximum inside the cavity. Further, since the boundary of the cavity

is an equipotential, the potential inside the cavity is also constant.

Uniqueness Theorem :

This theorem states that the solution of Laplaces equation is uniquely deter-

mined by the values of potential on the boundaries.

Suppose V1 and V2 are two potentials which satisfy Laplaces equation in someregion with identical coundary conditions, i,e V1(boundary) = V2(boundary).

Consider a function V3 = V1V2. This satisfies Laplaces equation with the condi-

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tion V3(boundary) = 0. However, as V3 does not have a minimum or a maximumin the region, its value has to be the same value as its value at the boundary, i.e.

V3 is constant. Hence V1 = V2.

Laplaces Equations in 3-dimensions

We will consider the solutions of Laplaces equations in problems with sphericalgeometry having azimuthal symmetry. The equation to be solved is

2V =1

r2

r

r2

V

r

+

1

r2 sin

sin

V

+

1

r2 sin

2V

2= 0

where we have explicitly written down the Laplacian operator in spherical polar

coordinates.

For problems with azimuthal symmetry, V/ = 0 so that we have

rr2V

r + 1

sin

sin V

= 0

The equation above is conveniently solved by a technique called separation of

variables where we write the function V(r, ) as a product of two functions, oneR(r) which is a function of radial variable r only and the other a function ()which is a function of the angle variable alone. Writing V(r, ) = R(r)()and dividing throghout by R, we get

1

R

r

r2

R

r

+

1

sin

sin

= 0

Since the two terms on the left depend on two independent variables, this equation

can be satisfied only if each of the term equals to constants of opposite sign. Wewrite

1

R

r

r2

R

r

= l(l + 1)

1

sin

sin

= l(l + 1)

We will not attempt to solve these equations but merely quote the results. The

solution of the angular equation is in terms of what are known as Legendre Poly-

nomials. Pl(cos ).These are polynomials of degree lin cosine of angle . The

first few polynomials are as follows :

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P0(cos ) = 1

P1(cos ) = cos

P2(cos ) =

1

2 (3 cos2

1)

P3(cos ) =1

2(5 cos3 3cos ) -1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

Pl(cos)

cos

P0

P1

P2

P3P4

The solution of radial equation is consists of a power series in r and 1/r. Thecomplete solution is

V(r, ) =i=0

Alr

l +Bl

rl+1

Pl(cos ) (A)

We will illustrate the use of these solution by an example.

Example 6 :Consider an uncharged conducting sphere in a uniform electric field and deter-

mine the potential at all points in space.

Solution :

The sphere, being a conductor, is an

equipotential. Let the potential be

zero. Far from the sphere, the field

is uniform. Let the field strength be

E0 and be in z-direction,The boundary conditions are :

V = 0 at r = RV = E0z = E0 cos for r R.

+

++ +

+

+

Using Eqn. (A) and substituting the first boundary condition, we get a relationship

between Al and Bl

AlRl + BlRl+1

= 0

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Thus Bl = AlR2l+1. Thus, we have

V(r, ) =

i=0

Al

rl +

R2l+1

rl+1

Pl(cos )

For r R, we may neglect the second term in bracket and get

i=0

AlrlPl(cos ) = E0r cos

On comparing both sides, we get l = 1 which gives A1 = E0. Substituting thesewe get

V(r, ) = E0(r R3

r2)cos

The induced charge density is

= 0V

r|r +R = 30E0 cos

It can be seen that the charge density is positive in the upper hemisphere and

negative in the lower hemisphere.

Dielectrics

A conductor is characterized by existence of free electrons. These are electrons

in the outermost shells of atoms (the valence electrons) which get detatched from

the parent atoms during the formation of metallic bonds and move freely in the

entire medium in such way that the conductor becomes an equipotential volume.

In contrast, in dielectrics (insulators), the outer electrons remain bound to theatoms or molecules to which they belong. Both conductors and dielectric, on the

whole, are charge neutral. However, in case of dielectrics, the charge neutrality is

satisfied over much smaller regions (e.g. at molecular level).

2.9.1 Polar and non-polar molecules :

A dielectric consists of molecules which remain locally charge neutral. The

molecules may be polar or non-polar. In non-polar molecules, the charge cen-

tres of positive and negative charges coincide so that the net dipole moment of

each molecule is zero. Carbon dioxide molecule is an example of a non-polar

molecule.

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+6e+8e

Oxygen atom

+8e

Oxygen atomCarbon atom

+8e

+e

+e

Oxygen atom

Hydrogen atom

Hydrogen atom

In a polar molecules, the arrange-

ment of atoms is such that the

molecule has a permanent dipole

moment because of charge separa-

tion. Water molecule is an example

of a polar molecule.

When a non-polar molecule is put in an electric field, the electric forces cause a

small separation of the charges. The molecule thereby acquires an induced dipole

moment.A polar molecule, which has a dipole moment in the absence of the electric field,

gets its dipole moment aligned in the direction of the field. In addition, the mag-

nitude of the dipole moment may also increase because of increased separation of

the charges.

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A nonpolar molecule

in an Electric Field

E=0E

+

+

E=0E

Electric Field

A polar molecule in an

Dielectric in an Electric Field

A dielectric consists of molecules which may (polar) or may not (non-polar) have

permanent dipole moment. Even in the former case, the dipoles in a dielectric

are randomly oriented because dipole energies are at best comparable to thermal

energy.

+

+ +

+++

+ +

++

+

+

+

+

+

+

+

+

+

+

+

+

++ ++

+

++

+

+

++

+

+

+

Polarised Dipoles in an electric field

Randomly oriented dipole in a dielectric (E=0)

When a dielectric is placed in an electric field the dipoles get partially aligned in

the direction of the field. The charge separation is opposed by a restoring force

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due to attaraction between the charges until the forces are balanced. Since the

dipoles are partially aligned, there is a net dipole moment of the dielectric which

opposes the electric field. However, unlike in the case of the conductors, the net

field is not zero. The opposing dipolar field reduces the electric field inside the

dielectric.

Dielectric Polarization

Electric polarization is defined as the dipole moment per unit volume in a di-

electric medium. Since the distribution of dipole moment in the medium is not

uniform, the polarization P is a function of position. If p(r) is the sum of thedipole moment vectors in a volume element d located at the position r,

p(r) = P(r)d

It can be checked that the dimension of P is same as that of electric field dividedby permittivity 0. Thus the source of polarization field is also electric charge,except that the charges involved in producing polarization are bound charges.

Denoting the local bound charge density by b, one can write

P = b

The equation above is obtained in a manner that is identical to the way we de-

rived the equation E = /0. The absence of the factor o in the equation

is because of the dimensional difference between E and P while the minus signarises because the dipole moment vector (and hence the polarization) is defined to

be directed from negative to positive charge as against E which is directed frompositive to negative charge. Clearly, if polarization is uniform, the volume den-

sity of bound charges is equal to zero. Even in this case, there are surface bound

charges given by the normal component of the polarization vector. Summarizing,

we have,

P = b P n = b

We will derive these relations shortly.

Free and Bound Charges

The charge density of a medium consists offree charges, which represent a surplus

or deficit of electrons in the medium, and bound charges. The term free charge is

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used to denote any charge other than that due to polarization effect. For instance,

the valence charges in a metal or charges of ions embedded in a dielectric are

considered as free charges.

The total charge density of a medium is a sum of free and bound charges

= f + b

Gausss Law takes the form

E =

0=

f + b0

Potential due to a dielectric

Consider the dielectric to be built up of volume elements d. The dipole momentof the volume element is P d.

The potential at a point S, whose po-

sition vector with respect to the vol-

ume element is r is

dV =1

4o

P r

r2d

P d

S

r

The potential due to the whole volume is

V =1

40

volume

P r

r2d =

1

40

volume

P (1

r)d

where, we have used

(1

r) =

r

r2

Use the vector identity

( Af(r)) = A f(r) + f(r) A

Substituting A = P and f(r) = 1/r,

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(P

r) = P (

1

r) +

1

r P

we get

V =

1

4ovol (

P

r )d

1

4ovol

1

r P d

The first integral can be converted to a surface integral using the divergence theo-

rem giving,

V =1

4o

surface

P

r dS

1

40

vol

1

r P d

The first term is the potential that one would expect for a surface charge density

b whereb = P n

where n is the unit vector along outward normal to the surface. The second termis the potential due to a volume charge density b given by

b = P

The potential due to the dielectric is, therefore, given by

V =1

4o

surface

bdS

r+

1

40

vol

bd

r

and the electric field

E = V

=1

40

surface

br

r2dS+

1

40

vol

br

r2d

Electric Displacement Vector D

The electri displacement vector D is defined by

D = 0 E+ P

which has the same dimension as that of P. The equation satisfied by D is

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D = 0 E+ P = b = f

which is the differential form of Gausss law for a dielectric medium.

Integrating over the dielectric volume,

volume

Dd =

volume

fd = Qf

where Qf is the free charge enclosed in the volume. The volume integral can beconverted to a surface integral using the divergence theorem, which gives

surface

D dS = Qf

Thus the flux over the vector D over a closed surface is equal to the free chargedenclosed by the surface.

Example 5:

An uncharged spherical dielectric has polarization vector given by P = kr. Findthe electric field both outside and inside the dielectric.

Solution :

The dielectric has both bound surface charge and volume charge. The surface

charge density is b = P n = kR where R is the radius of the sphere. Thevolume charge density is

b = P = k r = 3k

. The field inside the dielectric is given by Gausss law,

4r2E =Qencl

0=

4r3b30

which gives

E =rb30

= kr

0

The field outside is zero.

Example 6 :

Consider a spherical dielectric shell of inner radius a and outer radius b. Thespace in the region between r = a and r = b is filled with a dielectric hasvingpolarization P = k

rr. Determine the field inside and outside the shell.

21


22/33

Solution :

The charge densities are,

outersurfaceb =P n =

k

b

innersurfaceb = P n =

k

a

b = P = k

r2

For r < a, no charge is included, hence the field is zero. For a < r < b, thecharges enclosed by a Gaussian surface are the surface bound charges on the inner

surface and the volume charge within the region. Thus

Qencl = 4b2in +

ra

b4r2dr

= 4a2ka

+

r

a

(k/r2)4r2dr

= 4ka + 4k(a r) = 4kr

Thus E = (k/0r)r. For a Gaussian surface outside, the total charge enclosedcan be similarly calculated to be zero, so that field is zero.

Example 7 :Electric Field Due to Uniformly polarized sphere :

Since the polarization is uniform,

the bound charge density is zero.Only on the surface, there are bound

charges. We have

b = P n = P cos

where is the angle between thedirection of the external field (z-

direction) and a point on the sphere.

P

n

z

^

This is, once again, a problem with azimuthal symmetry with no charges inside

or outside the sphere. Hence Laplaces equation is satisfied both in the interior of

22


23/33

the sphere and outside.

V(r, ) =l=0

Alr

l +Bl

rl+1

Pl(cos )

For r < R, the second term must vanish since the potential cannot become infinityat the origin. Similarly, for r > R, the first tem must vanish as the potential mustbe well defined at large distances.

For r < R,V(r, ) = Alr

lPl(cos )

and, for r > R

V(r, ) =Bl

rl+1Pl(cos )

At r = R, the potential is continuous. Hence,

Bl = AlR2l+1

At r = R, while the tangential component of the field is continuous, the normalcomponent has a discontinuity,

Eaboven Ebelown =

0n

Using E = V,Vabove

r

Vbelow

r=

0Thus,

l=0

Bl

rl+2Alr

l1

Pl(cos ) |r=R= Pcos 0

Comparing both sides, we see that only l = 1 term is non-zero. We get,

2B1R3

+ A1 =P

0

Using B1 = A1R3, we get A1 = P/30 and B1 = P R

3/30 Finally, we get

V(r, ) =P

30r cos for r < R

= P30

R3

r3cos for r > R

23


24/33

The electric field inside the sphere is uniform and is equal to V = P/30z.Outside the sphere, the potential has the same form as that of a giant dipole with

dipole moment equal to volume of the sphere times the polarization vector, located

at the centre, because,

V = P30

R3

r2cos

=3p/4R3

30

R3

r2cos

=1

40

p cos

r2

=1

40

p r

r2

Constitutive Relation

Electric displacement vector D helps us to calculate fields in the presence of adielectric. This is possible only if a relationship between E and D is known.For a weak to moderate field strength, the electric polarization P is found to bedirectly proportional to the external electric field E. We define Electric Suscepti-bility through

P = 0 E

so that

D = 0 E+ P

= o(1 + ) E = 0r E = E

where r = 1 + is called the relative permittivity or the dielectric constantand is the permittivity of the medium. Using differential form of Gausss law forD, we get

E =1

D =

f

Thus the electric field produced in the medium has the same form as that in free

space, except that the field strength is reduced by a factor equal to the dielectric

constant .

Capacitance filled with Dielectric

If a material of dielectric constant is inserted between the plates of a capacitor,

24


25/33

the field E is reduced by a factor . The potential between the plates also reducesby the same factor .

Thus the capacitance

C =Q

increases by a factor .

+Q

Example:

A parallel plate capacitor with plate separation 3.54mm and area 2m2 is initially

charged to a potential difference of 1000 volts. The charging batteries are then

disconnected. A dielectric sheet with the same thickness as that of the separation

between the plates and having a dielectric constant of 2 is then inserted between

the capacitor plates. Determine (a) the capacitance , (b) potential difference acrossthe capacitor plates, (c) surface charge density (d) the electric field and (e) dis-

placement vector , before and after the insertion of the dielectric .

Solution :

(a) The capacitance before insertion of the dielectric is

C = 0A

d= 8.85 1012

2

3.54 103= 5 109 F

After the insertion the capacitance doubles and becomes 108 F.(b) Potential difference between the plates before insertion is given to be 1000 V.

On introducing the dielectric it becomes half, i.e. 500 V.

(c) The charge on each capacitor plate was Q = CV = 5 10

6 coulomb, givinga surface charge density of2.5 106 C/m2. The free charge density remains thesame on introduction of the dielectric.

(d) The electric field strength E is given by

E =

0= 2.8 105 volts/meter

The electric field strength is reduced to 1.4 105 volt/meter on insertion.(e) The displacement vector remains the same in both cases as the free charge

density is not altered. It is given by D = = 2.5 106 C/m2.

Example :

The parallel plates of a capacitor of plate dimensions a b and separation d are

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charged to a potential difference and battery is disconnected. A dielectric slabof relative permittivity is inserted between the plates of the capacitor such thatthe left hand edge of the slab is at a distance x from the left most edge of thecapacitor. Calculate (a) the capacitance and (b) the force on the dielectric.

Forcex

a

b

d

xaxis

yaxis

zaxis

Solution :

Since the battery is disconnected, the potential difference between the plates will

change while the charge remains the same. Since the capacitance of the part of

the capacitor occupied by the dielectric is increased by a factor , the effectivecapacitance is due to two capacitances in parallel ,

C = 0 bd

[x + (a x)]

The energy stored in the capacitor is

U =Q2

2C=

Q2

2

d

b0

1

x + (a x)

Let F be the force we need to apply in the x-direction to keep the dielectric inplace. For an infinitisimal increment dx of x, we have to do an amount of workF dx , which will increase the energy strored in the field by dU, so that

F = dUdx

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the differentiation is to be done, keeping the charge Q constant. Thus

F =dQ2

2b0

1

[x + (a x)]2

Since > 1, F is positive. This means the electric field pulls the dielectric inwardso that an external agency has to apply an outward force to keep the dielectric in

position. Since the initial potential difference is given by Q/C, one can expressthe force in terms of this potential

F =0b

2d2( 1)

This is the force that the external agency has to apply to keep the left edge of the

dielectric at x. The force with which the capacitor pulls the dielectric in has thesame magnitude.

Example 22 :

In the above example, what would be the force if the battery remained connected?

Solution :

If the battery remained connected Q does not remain the same, the potential does. The battery must do work to keep the potential constant. It may be realised

that the force exerted on the dielectric in a particular position depends on the

charge distribution (of both free and bound charges) existing in that position and

the force is independent of whether the battery stays connected or is disconnected.

However, in order to calculate the force with battery remaining connected, one

must, explicitly take into account the work done by the battery in computing the

total energy of the system. The total energy U now has two parts, one the workdone by the external agency F dx and the other the work done by the battery, viz.,dQ where dQ is the extra charge supplied by the battery to keep the potentialconstant. Thus

U = F dx + dQ

which gives

F =dU

dx

dQ

dxSince is constant, we have

U =1

2

c2

Q = C

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Using these

F =1

22

dC

dx 2

dC

dx=

1

22

dC

dx

(Note that if the work done by the battery were negnected, the direction of Fwill be wrong, though, because we have used linear dielectrics, the magnitude,

accidentally, turns out to be correct !)

In the previous example, we have seen that

C =b0d

[x + (a x)]

givingdC

dx=

b0d

(1 )

which is negative. Thus F is positive, as before,

F = b02

d( 1)

Example :

The space between the plates of a parallel plate capacitor is filled with two differ-

ent dielectrics, as shown. Find the effective capacitance.

d

d

1

2

1

2

Solution :

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Take a Gaussian pill-box as shown.

We haveD dS = free = 0

as there are no free charges insidethe dielectric. Contribution to the in-

tegral comes only from the faces of

the pill-box parallel to the plates anddS1 = dS2. Hence,

D1 = D2 =

where is the surface density of freecharges.

d

d

1

2

1

2dS

dS 2

1

dzLet 1

be the potential difference between the upper plate and the interface between the

dielectric and 2 that between the interface and the lower plate. We have

= 1 + 2

= E1d1 + E2d2

=D1

10d1 +

D220

d2

=d110

+d220

Thus the effective capacitance is given by

C = Q

= A0

d11

+ d22

=A0

d11

+ d22

=C1C2

C1 + C2

where C1 and C2 are the capacitances for parallel plate capacitors with one typeof dielectric with separations d1 and d2 between the plates respectively.

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Example :

A capacitor consists of an inner conducting sphere of radius R and an outer con-ducting shell of radius 2R. The space between the spheres is filled with two dif-ferent linear dielectrics, one with a dielectric constant from r = R to r = 1.5Rand the other with dielectric constant 2 from r = 1.5R to r = 2R. The outer

shell has a charge Q while the inner conductor has a charge +Q. Determine theelectric field for r > 0 and find the effective capacitance.Solution :

The electric field is radially symmet-

ric and may be obtained by apply-

ing Gausss law for the displacement

vector

D dS = 4r

2

D = Qfree

where Qfree is the free charge en-closed within a sphere of radius r.For r < R, the field is zero as thefree charges are only on the surface

of the inner cylinder.

+

+

+

++

+

+

+

+

R

1.5R

2R

For R 2R, the field is zero. The fields are radial with the ineer sphere at ahigher potential. The potential difference is calculated by taking the taking the

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line integral of the electric field along any radial line.

=

E dl =

2R

R

Edr

=1.5RR

Q

40r2dr +2R1.5R

Q

80r2dr

=5Q

480

The effective capacitance is

C =Q

=

4805

Example :

A parallel plate capacitor has charge densities on its plates which are separated

by a disance d. The space between the capacitor plates is filled with a linear butinhomogeneous dielectric. The dielectric constant varies with distance from the

positive plate linearly from a value 1 to a value 2 at the negative plate. Determine

the effective capacitance.

As the dielectric is linear,

D = 0E = 0

1 +

x

d

E

As the insertion of dielectric does

not affect free charges, the displace-

ment vector D is remains the sameas it would in the absence of the di-

electric. Thus D = .

2

1

d0

x

distance from positive plate

dielectric

constant

++++++

_

_

_

_

Thus the

electric field E is given byE =

a

0(x + d)

The field close to x = d is given by E = /

20, which shows that adjacent tothe negative plate there is a positive charge density /2. To find the effective

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capacitance, we find the potential difference between the plates by integrating the

electric field

=

d0

Edx =d

0

d0

dx

d + x=

d

0ln 2

so that

C = Q

= A

= A0

d ln 2

The polarization P is given by

P = D 0E =x

x + d

The volume density of bound charges, given by P = b is found as follows :

b = d

dx

x

x + d

=

d

(x + d)2

The bound charge density on the surface, given by n P = P , has a value /2on the dielectric adjacent to the negative plate (x = d). As the dielectric is chargeneutral, this requires a net volume charge of/2 in the dielectric. This can beverified by integrating over the volume charge density b given above.

Exercise :

A parallel plate capacitor of plate area Sand separation d, contains a dielectric ofthickness d/2 and of dielectric constant 2, resting on the negatve plate.

d/2

d

+

A potential difference of is maintained between the plates. Calculate the electricfield in the region between the plates and the density of bound charges on the

surface of the dielectric. [Ans. field in empty region = 4/3d, within dielectric= 2/3d, bound charge density = 20/3d]

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Exercise :

The permittivity of a medium filling the space between the plates of a spherical

capacitor with raddi a and b (b > a) is given by

= 20 a r (a + b)/240 (a + b)/2 r b

Find the capacitance of the capacitor, distribution of surface bound charges and

the total bound charges in the dielectric. [Ans. C = 80

1

a

1

a + b

1

2b

1

,

bound charges on dielectric surface with radii a, (a + b)/2 and b are respectively/2, 3a2/(a + b)2 and 3a2/4b2]

33

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Conductors in Fem Analysis