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} $