1. Calculate the capacitance of two conductors carrying equal and opposite charges in terms of the coefficient of potential.

Let . We have,

Use,  to get, . The capacitance

2.Three concentric shells of radii  respectively carry charges  and . Calculate the energy of the system.

The potential on the outermost conductor is .  Potential on the central conductor is

The potential on the innermost conductor is



The matrix for the coefficient of potential is

The energy is given by, on substituting

3. Two concentric spherical shells of radii  are both grounded and a point charge q is placed at a distance r between them, being insulated from both the surfaces. Calculate the energy of the system charges.

We use Green’s reciprocal theorem by constructing an artificial problem (figure to right) where the outer sphere is grounded but the inner sphere is maintained at a constant potential  without any charge being between them.

In the second case, the potential in the region between the spheres satisfies Laplace’s equation in variable r of spherical coordinates (as it can only depend on r) which has the solution  , where A and B are constants.  Substituting boundary conditions, the potential at r is given by

Since both spheres in  given configuration is grounded, their potential is zero. Let  and  be charges induced on the inner sphere and outer sphere  respectively. We then have,

which can be solved to give

The charge induced on the outer sphere may be found by changing the boundary condition on the right to grounded inner sphere and the outer sphere being at a constant potential. It can also be inferred by realizing that the total induced charge should be .

4. Calculate the work done in doubling the separation  between the plates of a parallel plate capacitor  from d to 2d, if the potential difference between the plates remains

The work done is calculated by the difference between initial and final energies. Since the potential difference between the plates is  maintained at , the energy is given by the formula . If the initial separation is d, the capacitance is  and it becomes  when the separation is doubled.  Work done is then

For r < a, the  Gaussian surface to be a cylinder of radius r and length . The amount of charged enclosed by this surface is . The flux is only due to the curved surface because by cylindrical symmetry, the electrical field lines are perpendicular to the top and bottom caps. Thus for r < a, , which gives, . For a < r < b, the amount of charged enclosed by this surface is . The electric field is found using this easily.