  Question

# A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ=kra, where k and a are constants and r is the distance from its centre. If the electric field at r=R2is18 times that at r = R, find the value of a.

Solution

## Let E be the electric field due to solid sphere at any point r. Let qin be the charge enclosed in the solid sphere of radius r=R2 From the given condition E(r=R2)=18E(r=R) ⟹kqin(R2)2=18kQ(R)2  ⟹Qqin=32-----(i) Let us consider a elementary solid sphere of radius r inside the solid sphere of radius R Consider the thickness of this small sphere be dr. Let dV be the volume of this small sphere. Volume dV can be given by  dV=4πr2dr----(ii) The charge dq on this volume can be given as dq=ρ4πr2dr ⟹dq=Kra4πr2dr ⟹dq=4πKr(a+2)dr-----(iii) To get qin we need to integrate dr from 0 to R2 ⟹∫qin0=4πK∫R20ra+2dr  qin=4πKRa+3(a+3)2a+3 IIy on integrating equ-(iii) from 0 to R we get Q=4πKRa+3a+3 On putting values of Q and qin in (i) we get 2a+3=32 ∴a=2  Suggest corrections   