Question

A Uniformly charged solid non-conducting sphere of uniform volume charge density `rho` and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure :

(i) Point A (ii) Point B
(iii) Point C (iv) Centre of the sphere

Answer 1

Method-I :
(i) For point A : We can consider the solid part of sphere to be made of large number of spherical shells which have uniformly distributed charge on its surface. Now, since point A lies inside all spherical shells so electric field intensity due to all shells will be zero.
`vec(E_(A))=0`
(ii) For point B : All the spherical shells for which point B lies inside will make electric field zero at point B. So electric field will be due to charge present from radius r to OB.
So, `vec(E_(B))=(K 4/3 pi (OB^(3) - r^(3)) rho)/(OB^(2))vec(OB)=rho/(3 epsi_(0)) ([OB^(3)-r^(3)])/(OB^(3)) vec(OB)`
(ii) For point C, similarly we can say that for all the shell points C lies outside the shell
So, `vec(E_(C))=(K[4/3 pi (R^(3)-r^(3))])/([OC]^(3)] vec(OC) = rho/(3 epsi_(0)) (R^(3)-r^(3))/([OC]^(3)) vec(OC)`
Method-II : We can consider that the spherical cavity is filled with charge density `rho` and also `-rho`, thereby making net charge density zero after combining. We can consider two concentric solid spheres : One ofradius R and charge density `rho` and other of radius r and charge density `-rho`. applying susperposition principle :

(i) `vec(E_(A))=vec(E_(rho))+vec(E_(-rho))=(rho (vec(OA)))/(3 epsi_(0))+ ([-rho (vec(OA))])/(3 epsi_(0))=0`
(ii) `vec(E_(B))=vec(E_(rho))+vec(E_(-rho))=(rho(vec(OB)))/(3 epsi_(0))+(K[4/3 pi r^(3)(-rho)])/((OB)^(3)) vec(OB)`
`=[rho/(3 epsi_(0))-(r^(3) rho)/(3epsi_(0) (OB)^(3))]vec(OB)=rho/(3 epsi_(0))[1-r^(3)/(OB^(3))] vec(OB)`
(iii) `vec(E_(C))=vec(E_(rho))+vec(E_(-rho))=(K(4/3 pi R^(3) rho))/(OC^(3)) vec(OC)+(K(4/3 pi r^(3) (-rho)))/(OC^(3))vec(OC)=rho/(3e_(0) (OC)^(3))[R^(3)-r^(3)] vec(OC)`
(iv) `vec(E_(O))=vec(E_(rho))+vec(E_(-rho))=0+0=0`