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.

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

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. 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?

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

We need to only compute even m.

Take , i.e. the dipole moment. For . Recall that the first term is zero for odd value of . Further since , the dipole moment also vanishes.

Let us compute quadrupole moments for which . Since

We have , . For calculating 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 The first term only contributes for even , and,  for the second term  we have . Thus the octupole moments also vanish. The first non-zero moment is ,which is called” hexadecupole” moment.

3. A rectangular parallelepiped having dimension filled with a dielectric in which the polarization is given by , where 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.

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., . 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. 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 ε . 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 .

Let the charge density at the interface be  . If the entire region had permittivity , 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 . 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 with relative permittivity 2 and 3, there is a free charge density C/m². The electric field strength in the first medium has a magnitude V/m and its direction makes an angle of with the interface.  Find the electric field in the second medium.

At the interface between two dielectrics, the tangential component of is continuous . Since the interface has free charges, we have . The last relation gives .

Thus .  Taking F/m , we  get from the second equation,

Substituting values,

Thus , i.e. the electric field makes an angle of 45⁰ with the interface.  The field strength in the second medium is V/m.