Correct Answer - Option 2 :
\(\frac{{4\pi {\varepsilon _0}xy}}{{x - y}}\)
The radius of the outer shell = x
The radius of the inner shell = y
The inner surface of the outer shell has charge +Q.
The outer surface of the inner shell has induced charge −Q.
The potential difference between the two shells is given by,
\(V = \frac{Q}{{4\pi {\varepsilon _0}y}} - \frac{Q}{{4\pi {\varepsilon _0}x}}\)
Where ε0 is the permittivity of free space
\(V = \frac{Q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{y} - \frac{1}{x}} \right]\)
\( \Rightarrow \frac{Q}{V} = \frac{{4\pi {\varepsilon _0}}}{{\left[ {\frac{1}{y} - \frac{1}{x}} \right]}}\)
\( \Rightarrow C = \frac{{4\pi {\varepsilon _0}xy}}{{x - y}}\)