The electrostatic potential inside a charged spherical ball is given by `phi=ar^2+b` where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is:

Options

A. `-24pi a in_(0)`
B. `-6 a in_(0)`
C. `-24pi a in_(0)r`
D. `-6a in_(0)r`

Answer

B

Explanation

`E_(r)=-(del phi)/(del r)=-2 ar`
By gauss theorem `4pi r^(2)E_(r)=q/epsilon_(0)`
By differention `4 pid (r^(2)E_(r))=(dq)/epsilon_(0)=((4pir^(2)dr)rho)/epsilon_(0)`
`rArr r^(2)dE_(r)+2rE_(r)dr=1/epsilon_(0)rhor^(2)dr`
`rArr (delE_(r))/(delr)+2/r E_(r)=rho/epsilon_(0)= rho=-6epsilon_(0) a`