Q. A solid spherical conductor of radius $R$ has a spherical cavity of radius $a(a < R)$ at its centre. A charge + $\mathcal{Q}$ is kept at the centre. The charge at the inner surface, outer surface and at a position $r(a < r < R)$ are respectively
A) $+ \mathcal{Q}, - \mathcal{Q},0$
B) $- \mathcal{Q}, + \mathcal{Q},0$
C) $0, - \mathcal{Q},0$
D) $+ \mathcal{Q},0,0$
Explanation
(b) The solid conductor with a cavity is shown in figure.
[image:1_image396.emf]
The inner surface of cavity will be $\mathcal{–Q}$due to induction. At outer surface, the charge will be +$\mathcal{Q}$ and at a point between$P$ at a position $r(a < r < R)$ will be zero
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.