What are dielectrics?
So far we have been discussing electrostatics in either vacuum or in a conductor. Conductors, as we know are substances which have free electrons, which would move around the material if we applied electric field. Under equilibrium conditions, the electric field inside is zero and free charges, if they exist, can only exist on the surface of the conductor. This, of course, is not true if we connect a piece of conductor across the terminals of a battery, which is not an equilibrium situation.
We have seen that in vacuum, in absence of charges Laplace’s equation is satisfied and in the preceding lectures we have discussed several techniques for solving the Laplace’s equation.
In this lecture we discuss a different class of material which go by the name of “insulators” or “dielectrics”. The primary difference between a conductor and a dielectric is that in the latter the electric charges (primarily the valence electrons) which could be free in conductors are tightly bound to the parent atoms or molecules. As a result when a dielectric is subjected to an electric field, the electrons do not wander around but respond to such a field in a different way. The response of dielectrics to an applied electric field are of two different kinds. In what are known as “polar molecules”, even in the absence of an electric field the charge centres for positive and negative charges do not coincide. This results in the molecules developing a permanent dipole moment. In “non-polar molecules”, in the absence of an electric field, the positive and the negative charge centres coincide and there is no dipole moment. However, when an electric field is applied, there is a charge separation and a dipole moment is induced. The induced dipole moment vanishes once the electric field is withdrawn.
As a result of the above, a dielectric is capable of producing an electric field both inside and outside the material, though the matter as a whole is electrically neutral. In this lecture we will consider the effect of such charge separation in a material on the electrostatic field and also discuss how it modifies the Maxwell’s equations that we have obtained so far.
Consider a material which consists of spherically symmetric atoms and molecules. The charge distribution inside an atom or a molecule consists of positively charged nuclei and a negatively charged electron cloud charge centres of which coincide. Recall that we had defined a dipole as an entity which has equal positive and negative charge separated by a distance. The dipole moment of an individual dipole was defined as a vector which has a magnitude equal to the product of the magnitude of either charge and the distance between them and which has a direction from the negative charge towards the positive charge. This means the individual atom or the molecule do not have a dipole moment and the net dipole moment of the dielectric, which is defined as the vector sum of the dipole moments of the constituents atoms and molecules is zero. This is of course true excepting when one looks at submicroscopic level.
In bulk material, even when the material has a permanent dipole moment, the molecules are randomly oriented and the bulk dipole moment vanishes. It is only when an electric field is applied the molecules align in the direction of the electric field, either partially or fully depending on the strength of the electric field and the material develops a dipole moment.
Before calculating the potential due to a piece of dielectric, let us look at how one mathematically handles the inverse of the distance between a point in a charge distribution and the point when the potential is to be calculated. Consider an arbitrary origin with respect to which a point inside a charge distribution is at a position the point P at which the potential is to be calculated is at
Let the angle between Let us assume We expand the distance between the two points in terms of inverse powers of r in a binomial series and in the coefficient of a given power of 1/r we write the angle term in terms of powers of the cosine function.
We will simplify this last expression by making some observation about its structure. Now, cos θ can be written as
We will express the dot product in terms of components of the two vectors,
Here we have introduced the notation in which x, y and z are respectively denoted by x1,x2 and x3 respectively. Putting r’=r we write the square of the distance r as,
where δi,j is the Kronecker delta function which has the value 1 if i=j and is zero otherwise.
Multipole Expansion of Coulomb Potential
We start with the general expression for the potential due to a continuous charge distribution and insert into it the expansion obtained above,
The omitted terms are of higher order in inverse distance r. Let us examine the individual terms written above.
The first term just has an integral of the charge density over the entire volume which is simply the total charge Q in the distribution. The contribution to the potential by this term is
This is just the potential due to a point charge at a distance r. This is the term which dominates when the distances are very large so that even the second and the third terms of the above expression can be neglected. It may be observed that at such large distances, the charge distribution looks like a point as viewed from P. This term vanishes if the material does not have a net charge.
The second term has an integral where the integrand is charge density multiplied with the vector distance. This term has the dimension of a dipole moment and we define the dipole moment of a charge distribution by the relation,
The second term then gives,
This is identical to the expression for the potential due to a point dipole at the origin, except that instead of the dipole moment of a point dipole, we have the dipole moment of the charge distribution. This terms dominates for the case of a charge neutral dielectric.
We will make some brief comments about the third term. It has the dimension of charge multiplied by square of distance and is known as a “quadrupole” term. The simplest quadrupole is four charges of identical magnitude but a pair having opposite sign to the other pair arranged at the four corners of a square. There are other ways of arranging them. The term is quite messy and we will worry about it later. The next higher order is known as an octupole term etc.
What makes the expression for the potential messy is that in the original form for the coulomb potential due to a continuous charge distribution, the coordinate of the point of observation r is not decoupled from the integration variable r’ and as a result one has to perform a separate integration at each point where one desires to calculate the potential.
Recall that while expanding we had used the angle Instead, if we used the coordinates of the positions we can expand in terms of “spherical harmonics” which we had occasion to deal with earlier. Assuming that the expansion is given by,
Using this the expression for the potential due to the charge distribution is written as
Note that in this expression the integration variable is completely decoupled from the position vector of the point where the potential is being calculated. This, in turn, means that the integral is to be calculated only once and the result of integration can be used for evaluating the potential at all points in the field. This is known as the multiplole expansion of the potential. We define Multipole moments by the relation
where l can take any non-negative integral value and corresponding to a given l, m takes values from - l to + l in steps of 1. Let us examine a few lower order multipoles. In terms of these moments, we have,
We had seen earlier that the some of the spherical harmonics are given by
Using these, we have
The monopole moment,
which is just the total charge excepting for a constant factor. The next moments are
Realising that we have the following expressions for Cartesian coordinates
we can write the above as
You can calculate the other two terms corresponding to l = d and show that
The next terms are components of quadrupole moment tensor.
Because of charge separation, if we look at a small volume element in a dielectric, it can have a dipole moment and as a result a macroscopic body can have a net dipole moment. We define polarization vector of a macroscopic body as the net dipole moment per unit volume,
If the body is charge neutral, from the expression for potential due to a point dipole at the origin (i.e. we can write the following expression for the case of a continuous charge distribution,
We will now simplify this expression using some vector calculus. First we observe that
using which we can write,
the negative sign is removed because the gradient is taken with respect to the primed coordinate. We know that the divergence of a scalar times a vector is given by the following relation
Using this we rewrite the expression for the potential as sum of two terms,
Using the divergence theorem we can convert the first term into a surface integral and we get
where we have defined the “bound” surface charge density and the “bound” volume charge densities by the following relations,
Thus the effect of the dielectric is to replace the original problem with an effective surface and a volume charge densities. These being bound to the atoms and molecules in the material are known as “bound” as opposed to free charges. The charges in a body could consist of both free charges and bound charges. The electric field is the field that a test charge experiences. This consists of contribution due to both types of charges. Thus the Maxwell’s equation for the electric field,
One defines a “displacement vector” the relation,
It is easily seen that satisfies
Note that is a vector which we made up which artificially tries to separate the effect of the bound charges from the problem and gives the effect due to free charge. In practice it is not possible to do so and the electric field is the one that we can really measure. Nevertheless introduction of leads to some mathematical simplification. It may be noted that does not have the same dimension as that of
The bound volume charge is given by the divergence of the polarization vector. One can visualize this by looking at the way the dipoles are arranged when an electric field is applied. It can be seen that the central region tends to get negative charges which become a source.
Similarly, the charges inside the material, other than in the central region compensate , leaving net charges on the surface which is the origin of the surface charge density.
1. The charge distribution inside a sphere of radius R is given by the expression Find an approximate expression for the potential along the z axis far away from the sphere by retaining the first non-vanishing term in a multipole expansion.
2. A ring of radius R has a uniform, linear charge density λ. Calculate the potential at large distances by retaining up to the quadrupole term in a multipole expansion of the potential.
Solutions to Tutorial assignment
1. Multipole moments are given by
Let us calculate the total charge
We next calculate the dipole term
Since the integration over vanishes for m=1, leaving us with
Because the angle integral vanishes. Let us calculate the quadrupole term:
The integral ove vanishes for m ≠ 0
The general expression for potential is
Retaining only the leading term l = 2, m = 0, along the z axis, θ = 0
2. The linear charge density can be written as a volume charge density by the following expression
It is straightforward to show that the volume integral of this expression gives the total charge Q = 2πR.
The multipole moments are given by
Forc m ≠ 0, the spherical harmonics have as a factor, which on integration, would give zero. Thus only m=0 terms survive, a fact also obvious from the azimuthal symmetry of the problem. The monopole term is equal to the total charge, except for a multiplying factor of The dipole term vanishes because Y10 ∝ cos θ which makes the integral over θ vanish because of the delta function. The quadrupole term,
Self Assessment Quiz
1. Show that for a uniformly charged sphere, all multipole moments other than q00 vanishes.
2. In the multipole expansion of Coulomb potential, there are three dipole terms corresponding to Add these three terms to explicitly show that the expression for the potential is given by
3. A charge distribution is given by
Expand the potential due to this charge density in a multipole expansion and obtain its leading term for large distances.
Solutions to Self Assessment Quiz
1. This is shown by orthogonality of spherical harmonics,
2. Since we are only interested in the dipole term, the general expression for the potential for this term is
We can calculate the three dipole moments as follows:
q1,-1 is similarly calculated and gives,
We now calculate q1,0
Substituting these into the expression for the potential we get
3. We can write The multipole moments are then given by
The integral over r gives (l + 2)!. By orthogonality of the spherical harmonics, only non zero moments are for l = 0,2. We have,
Thus, the potential is given by
In the previous lecture we obtained a multipole expansion of the potential due to a continuous charge distribution. We discussed in some detail, the dipole part of this expansion and will be doing so in further detail later in this lecture. Let us make some comments on the quadrupoleterm.
The most elementary quadrupole consists of two dipoles oriented anti-parallel. A linear dipole is one where the four charges are lined up with the two dipoles being end to end but directed oppositely, so that there is a charge 2q at the centre and –q each at the ends.
Recall that the expression for the quadrupole contribution to the potential is given by
where the quadrupole moment tensor is given by
Note that since I and j take three values each, the quadrupole moment is a tensor of rank two with 9 components. A few properties of quadrupoleare :
1. The tensor is traceless. (Recall that the trace of a matrix is sum of its diagonal components. This can be easily proved by taking i=j and summing over i.
2. At large distances quadrupolefield decreases as inverse fourth power of the distance from the source. (This is easily proved by inspection as the potential goes as inverse cube of distance).
We now return back to the dipole term which is the most important term for the case of a neutral dielectric. Observe that the dipole moment defined by depends, in general, on our choice of origin, for, if we let we would have
Thus only for the case of the dielectric not having a net charge will it be independent of the choice of origin.
We had seen that the potential due to the dipole contribution to the charge distribution is the one that we would obtain if we have a volume charge density and a surface charge density,
where the charge densities are related to the polarization vector by
It may be emphasized that the bound charges, whether in volume or on the surface, are not fictitious charges; they are very real and their only difference with the free charges which we have in conductors is that they are bound to their parent atoms or molecules.
The “displacement vector” is an auxiliary field defined by
Using the relationship of the polarization vector with the bound charge densities, we get,
Thus the field is given by mentally isolating the bound charges from free charges. There is really no physical way of doing it but defining such a displacement vector leads to substantial simplification.
We can easily derive the integral form of the equation satisfied by ,
where Qf is the total free charges in the sample.
Boundary Conditions on the surface of a dielectric
We recall that in deriving the boundary condition On the surface of a conductor, we had taken a Gaussian Pillbox of infinitesimal height, half inside the conductor And half outside. The field inside the conductor being zero, the contribution to the flux came only from the top face of the pillbox, since the thickness of the pillbox being negligible the curved surface did not contribute. In the present case, there is a difference because the field inside is not zero. Further, one has to remember that at the interface, the normal is outward for the outside face and inward for the inside face.
Let us assume that there no free charges on the surface.
The volume integral of the electric field over the pillbox is given by
Using the divergence theorem on both sides,
The contribution to the surface integral on both sides, as mentioned before, are from the surfaces only with the directions of the normal being opposite,
Where we have taken the direction of normal in this equation as going away from the surface (i.e. outward). Thus we have,
This is precisely the relationship that we had earlier for the boundary condition for a conductor except that the surface charge density was due to the free charges. Adding the free charge density, the discontinuity in the normal component of the electric field is given by
We can obtain the discontinuity in the normal component of the displacement vector from
the reason for the minus sign is that outside the material σb = 0.
Thus, if there are no free surface charges we have the normal component of to be continuous,
D1n = D2n
The tangential components can be shown to be continuous by use of Stoke’s theorem. We take a rectangular contour across the surface.
E1t = E2t
In the above, the outside medium is denoted as 1 and the inside by 2.
We define a linear dielectric as one where the polarization at a point in the dielectric is proportional to the electric field at the point.
The quantity Xe is known as the electric susceptibility. Using the relation between the displacement vector and the electric field vector, we have,
The constant k = (1 + Xe) is known as the dielectric constant of the medium and ∈ = ∈0k is known as the permittivity of the medium.
Example 1 : Image Problem with Dielectric
Consider a point charge in a semi-infinite dielectric medium with permittivity ∈1 occupying the space z > 0, the remaining half space z > 0 has a permittivity ∈2. Depending on the problem either of the media could be vacuum with permittivity ∈0.
The interface is the z=0 plane and has no free charges on it. Based on our experience of solving similar problem using images, we will attempt to write down an expression for the potential at the point P. Since there are no free charges on the surface Dn = Dz is continuous across the surface. Thus
In addition, of course, the tangential component of continuous. Since the surface is in the x-y plane, we have
In solving this problem, we fall back on the uniqueness theorem and attempt to solve the problem by intuitive method.
Assume that in the medium 1,i.e., in the region z>0, the solution is the same as one would obtain by filling up the entire space by a dielectric of permittivity ∈1 and put an image charge q’ at z = -d. The potential at a point P (the entire space is now a uniform medium) is given by
Where we have used polar coordinates for the x-y plane, p2 = x2 + y2.
For the potential in medium 2, we assume that the solution would be the same as one would obtain by filling up the entire space by a dielectric of permittivit ∈2 and put an image charge at the location of the charge q, i.e. at z = d. We denote the sum of the original charge plus its image by q". Thus the potential at the point P is due to a single charge q" and is given by
Using the continuity of the normal component of the displacement vector,
(Remember that the direction of the outward normal at the interface was taken as uniquely defined). We have
Thus we get,
The tangential component of the electric field is obtained by differentiating the expressions for the potential with respect to x or y,
So that we have,
From (1) and (2), we obtain,
In both media Laplace’s equation and the boundary conditions are satisfied and the solution is unique.
1. Find the first few multipole moments of a linear quadrupole aligned along the z direction.
2. Consider a quadrupole in the xy plane as shown. Determine the components of the quadrupole moment tensor in Cartesian coordinates.
3. Two dielectrics having permittivity ∈1 and ∈2 have an interface which has no free charges. The electric field in medium 1 makes an angle θ1 with the interface while the field in medium 2 make an angle θ2. Find the relationship between the two angles.
Solutions to Tutorial assignment
1. In spherical polar coordinates, the charges +q are at (a,0,0) and (a,π,0). The charge densities of the distribution of charges is
The multipole moment of this charge distribution is (for l > 0,)
The multipole moment vanishes for all odd l. For even l,
Further, plm (1) = 0 except for m=0. Thus
(The above formula is valid for even values of l > 0, for l = 0, q00 = 0 because the total charge of the quadrupole is zero)
2. The charge density can be written in terms of delta functions
All other components are zero because the quadrupole is in the xy plane.
3. At the interface between two dielectrics, the tangential component of continuous
E1 cos θ1 = E2 cos θ2
In the absence of free surface charge at the interface the normal component of is continuous,
Thus, ∈1 tan θ1 = ∈2 tan θ2.
Self Assessment Quiz
1. Find the first few multipole moments of a linear quadrupole aligned along the x direction and also when it is aligned along y direction.
2. Three linear quadrupoles are arranged along the x, y and z axes with the central charge of each being at the origin and the other two charges at a distance a each from the origin. Show that the quadrupole moment of this superposition of the quadrupoles is zero. What is the lowest non-vanishing multipole of this configuration?
3. A rectangular parallelepiped having dimension is filled with a dielectric in which the polarization is given by is the position vector of a point in the dielectric with respect to the centre of the parallelepiped. Find the bound charge densities in the dielectric and show that the total bound charge is zero.
4. A line charge with a linear charge density λ is in the vacuum, at a distance d from the surface of a semi-infinite dielectric of permittivity ∈. Obtain the image charge that must be put at a distance d inside the dielectric to simulate the charge density at the boundary when the entire space has permittivity ∈0.
5. At the interface between two dielectrics with relative permittivity 2 and 3, there is a free charge density The electric field strength in the first medium has a magnitude and its direction makes an angle of tan with the interface. Find the electric field in the second medium.
Solutions to Self Assessment Quiz
1. The method is the same as that adopted for Problem 1 of the tutorial assignment. When the charges are along the x-axis, the coordinates of the charges +q are at The multipole moments are given by, for l > 0
R.h.s. is zero for odd values of m. For the quadrupole located along the y axis,
Once again, the multipole moment is zero for odd values of m.
2. To find the resulting multipole moments, we add the contribution to the multipole moment from the three cases (Problem 1 of Tutorial and Problem 1 of the Quiz) ,For l > 0
We need to only compute even m.
Take l = 1, i.e. the dipole moment. For l = 1, m = 0. Recall that the first term is zero for odd value of l. Further since p10(0) = 0, the dipole moment also vanishes.
Let us compute quadrupole moments for which We have For calculating q22 note that the first term is zero because of Kronecker delta function while the second term valishes. Thus all components of the quadrupole moment vanishes .
The next moment is for l = 3. The first term only contributes for even l, and, for the second term we have p3m (1) = 0. Thus the octupole moments also vanish. The first non-zero moment is l = 4, which is called” hexadecupole” moment.
3. Let us position the dielectric so that the end faces are parallel to the three Cartesian axes.
The volume density of bound charges is Since the density is constant, the total volume charge is obtained by multiplying this with the volume, i.e., qb = -3k abc. Now we need to calculate the bound charges on the surface. Note that on each face is given by the distance of that face from the centre. For instance, the top face (or the bottom face) which is at a distance b from the origin has The total bound charge on this face is obtained by multiplying the constant density with the area of the face ac giving a contribution of Each of the six faces contribute the same, giving a total of which is exactly the equal and opposite of the total volume charge
4. Let the charge density at the interface be . If the entire region had permittivity ∈0, the electric field at the interface can be calculated by taking a Gaussian cylinder perpendicular to the sheet and its ends equidistant from the sheet. The normal component of the electric field (directed outward from the respective face)due to the induced charge is given by Let the normal component of the electric field at the point P on the sheet be due to the line charge be En.
At a point P,
Since the normal components are oppositely directed, the resultant normal component is
The normal component of the net electric field on the other side of the interface is given by
Since the normal component of D field is continuous, we have (recall that the outward normal direction is uniquely defined)
Solving, we get,
This is the induced charge. The field due to this at P is
Comparing this with the expression for the normal component of the field, we see that this field is equivalently produced by locating an image line with line charge density
at a distance d from the interface, inside the dielectric.
5. At the interface between two dielectrics, the tangential component of continuous E1 cos θ1 = E2 cos θ2. Since the interface has free charges, we have D2 sin θ - D1 sin θ = σf. The last relation gives ∈2 E2 sin θ2 - ∈1E1 sin θ1 = σf.
Thus we get from the second equation,
Thus tan θ2 = 1 i.e. the electric field makes an angle of 450 with the interface.
The field strength in the second medium is
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 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
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,
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,
Solving, we get,
and the electric field inside the dielectric is
It is seen that the due to the presence of dielectric the electric field inside is reduced by
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 Thus the effect of dielectric is the same as that of replacing the dielectric with a dipole of moment
The polarization of the sphere is then obtained by dividing this by the volume of the sphere, i.e., by
Thus the field due to the dielectric can be written as
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.
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,
Since the dielectric is uniform,
Thus in the presence of a dielectric the electric field is reduced by a factor 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
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,
However, there is a bound surface charge density. Since the dielectric is inside the sphere, the surface direction is inward
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 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.
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, Q here is the total free charge on the plate. Thus, the direction of from the negative plate towards the positive plate. The electric field in the dielectric is The potential is which gives the capacitance to be
Thus the capacitance is increased by a factor of k, the dielectric constant. This, of course, is true only for a linear dielectric.
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 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 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 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
If there are n atoms per unit volume, the polarization is given by
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 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.
Since the normal is directed inwards, the bound surface charge is where θ is the polar angle with respect to the z axis. The electric field due to this charge distribution is easily calculated,
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 respect to the dipole is given by
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,
Since the dielectric is isotropic, we have
We are left with
Thus the total field is simply the sum of the first two contributions, i.e.
Since the induced dipole moment is proportional to the local field
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
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 by be the potential in the presence of the charge density p. Since the charges are true charges, we have,
The additional work that needs to be done to increase the charge density is 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,
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
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, , which can be used to get the energy expression as
1. A solid dielectric cylinder of dielectric constant k is placed in a uniform electric field perpendicular to its length. Obtain an expression for the potential inside and outside the cylinder.
2. A uniform electric field 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)
On this we impose the boundary condition at the normal component of D and the tangential component of E field are continuous.
Use these to express A and B in terms of E0. The result is
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
On this we impose the boundary condition at the normal component of D and the tangential component of E field are continuous.
These give (as the relations are valid for all values of θ)
Solving, we get,
Thus the potential is The corresponding electric field is
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 φ0 is of dielectric constant k2. Find an expression for the potential between the plates and the polarization charge density on the two plates.
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 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,
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,
Thus the displacement vector is given by
Which is of course constant. The polarization vector is given by
The polarization charge densities on the plates is given by the normal component of the polarization vector. On the upper plate, z=d
On the lower plate z=0,
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 and that in the region from z=d/2 to z=d is The potential in the region 1 is
The potential at z=d/2 is . The potential in the region 2 is
The potential at z=d is then given by
The charge density is given by the normal component of the polarization vector,