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. $\frac{\rho R}{3{\epsilon }_0}$
• B. $\frac{\rho r}{{\epsilon }_0}$
• C. $\frac{\rho r}{3{\epsilon }_0}$
• D. $\frac{3\rho R}{{\epsilon }_0}$

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
$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$
$(r>R)$
$\therefore E=\frac{1}{4\pi {\epsilon }_0}\times \frac{4}{3}\frac{\pi R^3\rho }{r^2}$
$=\frac{\rho }{{\epsilon }_0}\cdot \frac{R^3}{3r^2}$ ( for $r>R$ )
If the point $P$ less just on the surface of the sphere then $r=R$
then $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2}$
$=\frac{\rho }{{\epsilon }_0}\cdot \frac{R}{3}$