A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is :

(Take 1/4π∈₀ = 9 × 10⁹SI Units)

For a thin spherical shell that carries a total charge \(Q\) uniformly distributed over its surface, the electric potential at any point on the surface is

$V\_\{\backslash text\{surface\}\}\; \backslash ;=\backslash ;\; \backslash frac\{1\}\{4\backslash pi\backslash varepsilon\_\{0\}\}\backslash ,\backslash frac\{Q\}\{R\}$

where \(R\) is the radius of the shell and \(1/4\pi\varepsilon_{0}=9\times10^{9}\ \text{SI}\).

Because the shell is a conductor, the electric field inside the cavity of the shell is zero. Consequently the electric potential inside the cavity is the same everywhere and equals the surface potential:

$V\_\{\backslash text\{inside\}\}\; =\; V\_\{\backslash text\{surface\}\}$

In the figure the two points \(C\) and \(P\) are both located on the shell (or one on the surface and the other inside the cavity). Since the potential is constant throughout the shell and its interior, the potential difference between them is

$\backslash Delta\; V\; =\; V\_\{C\}\; -\; V\_\{P\}\; =\; 0$

Thus the required potential difference is zero volts.