question_answer

A solid spherical conductor of radius R has a spherical cavity of radius \[a\left( a<R \right)\] at its centre. A charge \[+Q\] is kept at the centre. The charge at the inner surface, outer surface and at a position \[r\left( a<r<R \right)\] are respectively

A)  \[+Q\],\[-Q\], \[0\]    

B)  \[Q\], \[+Q\], \[0\]

C)  \[0\],\[-Q\], \[0\]

D)  \[+Q\], \[0\], \[0\]  

Correct Answer: B

Solution :

The solid conductor with a cavity is shown in figure. The inner surface of cavity will be \[-Q\] due to induction. At outer surface, the charge will be \[+Q\] and at a point between P at a position \[r(a<r<R)\]will be zero.