A non-conducting sphere of radius R = 5 cm is centred at origin of coordinate system. It has a spherical cavity of radius 1 cm having centre at (0, 3) cm . The material of sphere has a uniform charge density $p=\frac{{10}^{-6}}{\pi }C{m}^{-3}$ Calculate the electric potential at point P(4 cm, 0) to the nearest two digit integer.

The electric potential at P is due to two spheres

(i) A large solid sphere of charge density (p)

(ii) A small solid sphere of charge density (- p)

The electric potential due to whole solid sphere of radius R at a distance r from centre

$V_1=\frac{1}{4\pi {\epsilon }_0}q\frac{(3R^2-r^2)}{2R^3}$

$\Rightarrow V_1=\frac{1}{4\pi {\epsilon }_0}\frac{\left(\frac{4}{3}\pi R^{3}p\right)(3R^2-r^2)}{2R^3}$

The electric potential at external point at a distance $r^1=\sqrt{x^2+y^2}=5cm$

$\Rightarrow V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}=\frac{1}{4\pi {\epsilon }_0}.\frac{\frac{4}{3}\pi r^3(-p)}{r^1}$

So, net potential  $V=V_1+V_2$

$\Rightarrow V=\frac{\pi p}{4\pi {\epsilon }_0}[\frac{2}{3}(3R^2-r^2)-\frac{4}{3}\frac{r^3}{r^1}]$

Where R = 5 cm $=5\times {10}^{-2}m.$

$r=4cm=4x{10}^{-2}m.$

$r^1=5cm=5x{10}^{-2}m.$

$p=\frac{{10}^{-6}}{\pi }$

Substituting given values, V = 35.16V