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. `-24piaepsi_(0)r`
B. `-6piepsi_(0)r`
C. `-24piaepsi_(0)`
D. `-6aepsi_(0)`

Answer

D

Explanation

The potential `phi=ar^(2)+b`
`therefore`The electric field inside the spherical ball is
`E=-(dphi)/(dr)=-2ar[E=-(dV)/(dx)]`
The electric flux`=ointE*ds=Exx4pir^(2)`
`=-2arxx4pir^(2)=-8piar^(3)`
By Gauss's theorem,`-8piar^(3)=(q)/(epsi_(0))`
`therefore q=-8piar^(3)epsi_(0)`
`rho=(q)/(V)=(q)/((4)/(3)pir^(3))=(3xx(-8piar^(3)epsi_(0)))/(4pir^(3))`
`therefore=6epsi_(0)a`