Q. Potential difference between centre and surface of the sphere of radius R and uniform volume charge density ρ within it will be

(a) $ \displaystyle \frac{\rho R^2}{6\epsilon_0} $

(b) $ \displaystyle \frac{\rho R^2}{4\epsilon_0} $

(c) $ \displaystyle \frac{\rho R^2}{3\epsilon_0} $

(d) $ \displaystyle \frac{\rho R^2}{2\epsilon_0} $

**Click to See Answer : **

$ \displaystyle q = \frac{4}{3}\pi R^3 \rho $

Electric Potential at the centre

$\displaystyle V_o = \frac{1}{4\pi \epsilon_0}\frac{3q}{2R} $

Electric Potential at the Surface

$ \displaystyle V_P = \frac{1}{4\pi \epsilon_0}\frac{q}{R} $

Potential difference between centre and surface

$ \displaystyle = \frac{1}{4\pi \epsilon_0}\frac{3q}{2R} – \frac{1}{4\pi \epsilon_0}\frac{q}{R} $

$ \displaystyle = \frac{1}{4\pi \epsilon_0}\frac{q}{2R} $

$ \displaystyle = \frac{1}{4\pi \epsilon_0 (2R)}(\frac{4}{3}\pi R^3 \rho) $

$ \displaystyle = \frac{\rho R^2}{6\epsilon_0} $