Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Electromagnetic Theory

Electrical Engineering (EE) : Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The document Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Electromagnetic Theory.
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We continue with our discussion of dielectric medium.

Example : Dielectric Sphere in a uniform Electric Field

The problem is very similar to the one we did for the case of a conducting sphere in an electric field. At far distances from the sphere, the field is uniform and is equal to  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev which corresponds to a potential of φ = -E0r cos θ.

As before we can expand the potential inside and outside in terms of Legendre polynomials. Since the distant potential only has Legendre polynomial of order 1, we need to include only to include up to this order. Outside the sphere, the field is given by the potential

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Inside the sphere, the origin is included. Thus, inside the sphere, the potential cannot have a singularity at the origin and is given by

φin (r, θ) = A2r cos θ.
Comparing with the asymptotic limit at long distances A1 = -E0.

The potential itself is continuous at r=R,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since there are no free charges on the surface, the normal component of the displacement vector is continuous across the surface. The normal component being the radial direction, we have,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solving, we get,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

so that

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

and the electric field inside the dielectric is

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

It is seen that the due to the presence of dielectric the electric field inside is reduced by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We observe that if a dipole is kept at the origin of a sphere of radius R, since the radius of the sphere can be considered very large compared with the dimensions of the dipole, the field is given by  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev Thus the effect of dielectric is the same as that of replacing the dielectric with a dipole of moment

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The polarization of the sphere is then obtained by dividing this by the volume of the sphere, i.e., by  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the field due to the dielectric can be written as

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Note that the polarization in the medium is uniform and is directed along the direction of the external field. The field lines approach the sphere from the left (negative z direction) and leave from the right, making the left face of the sphere negatively charged.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Reduction in the electric field due to the presence of a dielectric

Consider a point charge in an isotropic dielectric with permittivity ∈. Taking the position of the charge at the origin, the field must be radial. The displacement field can be found from the Gauss’s law,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

so that

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the dielectric is uniform,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus in the presence of a dielectric the electric field is reduced by a factor  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev The dielectric constant is a measure of the factor by which the electric field gets reduced in a linear dielectric medium. The polarization vector is given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Consider a Gaussian sphere of radius r. We can calculate the volume charge density inside the volume by the negative divergence of the polarization vector,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

However, there is a bound surface charge density. Since the dielectric is inside the sphere, the surface direction is inward Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus there exists a net negative bound charge. The total charge is obtained by multiplying the above with the surface area, viz., 4πr2 which gives  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev as the total charge bound to the surface. The effective charge seen by a test charge is the sum of the original charge as reduced by this amount, i.e.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

i.e., the test charge experiences the electric field due to an effective charge screened by a factor equal to the dielectric constant of the medium.

Example : A parallel plate capacitor filled with a dielectric :

From the above discussion it is clear the electric field inside the dielectric will be reduced due to polarization effect of the dielectric. Consider a Gaussian pillbox in the shape of a rectangular parallelepiped of a very small thickness but of area a (A is the area of the plates)partly inside the medium through one of the plates and partly outside.

Since the plates have real charges, using Gauss’s law for the displacement vector, we have, Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev Q here is the total free charge on the plate. Thus, Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev the direction of  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev from the negative plate towards the positive plate. The electric field in the dielectric is  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev The potential is Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev which gives the capacitance to be

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the capacitance is increased by a factor of k, the dielectric constant. This, of course, is true only for a linear dielectric.

Microscopic Theory

Before leaving the subject of electrostatics, we will briefly touch upon the molecular aspect of polarizability as opposed to a purely macroscopic treatment done so far. 

We look at the dielectric as a collection of molecules. If the molecules are well separated, as would be the case in a gas, each molecule would experience an average macroscopic field Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev site. This could consist of any external field and an average field due to all other molecules etc. The polarization induced is the written in terms of electric susceptibility  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev For the case of a gas, the electric field is primarily due to the external field.

In a dense medium, however, this is not a good description because a particular molecule will be subject to a local field, which in addition to the external field, is due to interactions with the dipoles in its immediate neigbourhood as well as due to polarization of the other molecules in the medium. This local field induces a dipole moment Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev the molecule in question and we define the “atomic polarizability” on the molecule in question and we define the “atomic polarizability” by the relation α by the relation

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

If there are n atoms per unit volume, the polarization is given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

However, since an atom cannot exert a force on itself, in computing the local field we must subtract the contribution to the polarization by the atom under consideration. For this purpose, let us divide the dielectric into two parts : a spherical volume of radius Rassigned to the atom in question which contains other atoms in the immediate neighborhood of the atom and the rest of the dielectric outside this volume. The “rest of the dielectric” will be treated in a macroscopic way which gives rise to an average field Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev taken along the z direction. To this field we must add the field which arise because of the spherical volume mentioned above.

The sphere can be treated as a cavity in an otherwise uniform medium. The inside surface of this cavity has a charge density σb corresponding to the uniform polarization.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the normal is directed inwards, the bound surface charge is  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev where θ is the polar angle with respect to the z axis. The electric field due to this charge distribution is easily calculated,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Now we must look into the interaction of the dipoles within the cavity with the molecule (dipole) in question). We had seen earlier that the field due to a dipole at a position Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev respect to the dipole is given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The net field at the position of the atom is obtained by averaging the field due to all the dipoles. Since the molecules are oriented randomly, let us find the average of the field along the z direction,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the dielectric is isotropic, we have

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

We are left with

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the total field is simply the sum of the first two contributions, i.e.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Since the induced dipole moment is proportional to the local field

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

So that

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

so that

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

This is called ClausiusMossotti Relation and it relates the atomic polarizability with the dielectric constant of the medium.

Energy of a charge distribution in a dielectric

In an earlier lecture we had obtained an expression for the energy of a charge distribution in free space. This was done by computing the work done in assembling the charges which are initially located at infinity. The expression for the energy was shown to be given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

In calculating the energy of a system of charges in a dielectric, we need to also take into account the work done in establishing the state of polarization of the dielectric.

Let us assume that at certain instant of time we have a charge density p already established and we add true charges to the system so that the charge density increases byDielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev be the potential in the presence of the charge density p. Since the charges are true charges, we have,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The additional work that needs to be done to increase the charge density is Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev Note that the factor of ½ that was there in the expression for the energy is not there because it had come to avoid double counting whereas in the present case we are just calculating the work done. Let us simplify the expression for the additional work done by integrating the above expression by parts,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The first term is converted to a surface integral using the divergence theorem. If we take the integration to be over all space, as has been pointed out on several occasion, the fields and potential vanish and this integral does not contribute to the energy. Using  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The total work done in establishing the charge distribution is then obtained by integrating this expression from zero field situation to the final field configuration. For linear dielectrics, we have,  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev , which can be used to get the energy expression as

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Tutorial Assignment

1. A solid dielectric cylinder of dielectric constant k is placed in a uniform electric field Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev perpendicular to its length. Obtain an expression for the potential inside and outside the cylinder.

2. A uniform electric field Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev is set up in a medium of dielectric constant k. A spherical cavity of radius R is inside this medium. Find an expression for the electric field inside the cavity.

Solutions to Tutorial assignment

1. As the extent of the cylinder is infinite in the z direction, one can use polar coordinates to describe the potential. Since the far field pattern is φ(r) = - E0r cos θ, we take the following expressions to represent the potential (Note the difference between the solution in the polar case and that in the spherical case with azimuthal symmetry)

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

On this we impose the boundary condition at the normal component of D and the tangential component of E field are continuous.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Use these to express A and B in terms of E0. The result is

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

2. Assume the dielectric to fill the entire space with the cavity centred at r=0. The problem has aqzimuthal symmetry. The potential is given by the expression

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

On this we impose the boundary condition at the normal component of D and the tangential component of E field are continuous.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

These give (as the relations are valid for all values of θ)

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solving, we get,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the potential is Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev The corresponding electric field is  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Self Assessment Quiz
1. A parallel plate capacitor is filled with a linear dielectric whose dielectric constant varies linearly from the grounded bottom plate to the upper plate located at a distance d, by the relation k = k0 +cz. If the upper plate is at a potential of φ0, obtain expressions for (a) the potential as a function of z, (b) the electric field in the dielectric and (b) the polarization charge densities on the plates.

2. The space between the plates of a parallel plate capacitor is filled with two slabs of dielectric material. The dielectric filling the space between the grounded bottom plate at z=0 up to z=d/2 is of dielectric constant k1 while that filling the space from z=d/2 up to the upper plate at z=d which is maintained at a constant potential φis of dielectric constant k2. Find an expression for the potential between the plates and the polarization charge density on the two plates.

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Solutions to Self Assessment Quiz

1. We start with an expression for the D field in the dielectric. As the plates contain free charge densities, the D field has a strength Q/A directed from the positive plate to the negative plate. The electric field is then given by Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev One can calculate the potential at any point by evaluating the line integral of the electric field from the bottom plate to a point at a distance z,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRevDielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev
Substituting  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The expression for the electric field can be obtained by taking the negative gradient of the potential or by expressing the previously obtained expression in terms of given constants, 

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRevDielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Thus the displacement vector is given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Which is of course constant. The polarization vector is given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The polarization charge densities on the plates is given by the normal component of the polarization vector. On the upper plate, z=d

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

On the lower plate z=0,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

2. Let the charge on the plates be +Q and the area of the plates be A. The displacement vector is D=Q/A. The electric field from z=0 to z= d/2  Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev and that in the region from z=d/2 to z=d is Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev The potential in the region 1 is

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The potential at z=d/2 is Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev. The potential in the region 2 is

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The potential at z=d is then given by

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

where Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

The charge density is given by the normal component of the polarization vector,

Dielectrics (Part - 3) Electrical Engineering (EE) Notes | EduRev

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