A long charged cylinder of linear …

Two co-axial cylindrical shells A and B of radii a and b are possessed by a cylindrical capacitor. Assume l be length of the cylindrical shell. Due to the introduction of +q charge on the inner cylindrical shell A, equal but opposite charge -q is induced on the inner surface of the outer cylindrical shell B. The induced charge +q on its outer surface will flow to earth if the shell B is earthed.

The capacitance of the cylindrical capacitor is given as follows if V is potential difference between the cylindrical shell A and B.
{tex}C = \frac{q}{V}{/tex}
By applying the Gaussian a theorem, we first need to find electric field E in the space between two shells to find out potential difference between the cylindrical shells A and B . Let a cylinder of radius r (such that b > r > a) and length l as the Gauss surface. Charge enclosed by the Gaussian surface is {tex}\lambda{/tex}1 and if {tex}\lambda{/tex} is charge per unit length on the shell A.
The electric flux will cross through only curved surface of the cylinder (Gauss surface). As the area of curved surface of cylinder is {tex}2\pi rl{/tex}, we have by Gaussian theorem.
{tex}\oint E \cdot ds = \frac{q}{{{\varepsilon _0}}}{/tex}
{tex}E \cdot 2\pi rl = \frac{{\lambda l}}{{{\varepsilon _0}}}{/tex}
{tex}E = \frac{\lambda }{{2\pi {\varepsilon _0}r}}{/tex}
The field lines are radial and normal to the axis of charged cylinder.