Question

If an insulated non-conducting sphere of radius $ R $ has charge density $ \rho $ , the electric field at a point just on the surface of Sphere is:

• A. $ \frac{\rho R}{3{{\varepsilon }_{0}}} $
• B. $ \frac{\rho r}{{{\varepsilon }_{0}}} $
• C. $ \frac{\rho r}{3{{\varepsilon }_{0}}} $
• D. $ \frac{3\rho R}{{{\varepsilon }_{0}}} $

Answer 1

Answer: (A)

Since the sphere is non-conducting charge will be distributed throughout the volume of the sphere.
When $r$ is the volume charge density, then
$q=\frac{4}{3} \pi R^{2} \rho$

The electric field intensity due to a uniformly charged sphere at an external point is the same as if the entire charge on it were concentrated at the centre of the sphere
$E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$
$(r > R)$
$\therefore E =\frac{1}{4 \pi \varepsilon_{0}} \times \frac{4}{3} \frac{\pi R^{3} \rho}{r^{2}} $
$=\frac{\rho}{\varepsilon_{0}} \cdot \frac{R^{3}}{3 r^{2}} $ ( for $ r > R$ )
If the point $P$ less just on the surface of the sphere then $r=R$
then $E =\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R^{2}}$
$=\frac{\rho}{\varepsilon_{0}} \cdot \frac{R}{3}$