Consider a length L of two coaxial conductors of inner radiu

Consider a length L of two coaxial conductors of inner radius a and outer radius b (b>a) separated by a homogeneous dielectric with permittivity sigma as shown below. Assume that the conductors carry a charge +Q and -Q uniformly distributed on their respective surfaces. Use Gauss\'s Law to obtain the electric field E vector, between them. From the expression for energy in an electric field obtain C.

Solution

We have two coaxial cylinders of length L and radii a and b with permittivity being . The cylinders have charges +Q and -Q on them.

a.) Let us assume a Gaussian surface of radius R such that a < R < b and is coaxial with the given cylinders.

We will evaluate the net electric flux through this surface and then resolve to find the electric field between them.

The net flux through a cylinderical cross section of length L would be gievn as: E(2RL)

Now by Gauss\'s law we know that the net flux through a surface is given as Charge enclosed / Permittivity

Hence,E(2RL) = Q/

Therefore electric field between them is given as: E = Q/2RL

b.) For determination of capacitance, we need to determine the potential difference across the conductors and then we can find the capacitance as Q./ V

For potential difference, we can write Va - Vb = E.dr

or, Va - Vb = Q dR/2RL = (Q/2L) ln(b/a)

or, Q / (Va - Vb) = 2L / ln(b/a)

Therefore the required capacitance is 2L / ln(b/a)

Also the energy stored in a capacitor of capacitance C and potential difference V is given as 0.5 VQ

hence the required expression for energy = (Q²/4L) ln(b/a)