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

Answer

C

Explanation

Potential inside a charged spherical ball
`phi = ar^(2) + b`
`:.` Electric field, `E = (d phi)/(dr) = 2 ar` ….(i)
By Gauss's theorem, `4pi r^(2) E = (q)/(in_(0))`
Differentiating it, we get
`4 pi d(r^(2) E) = (dq)/(in_(0)) = ((4pi r^(2) dr) rho)/(in_(0))`
where `rho` is the charge density inside the ball.
`:. r^(2) dE + 2r dr xx E = (1)/(in_(0)) rho r^(2) dr`
or `(dE)/(dr) + (2)/(r )E = (rho)/(in_(0))` or `(d)/(dr) (-2ar) +(2)/(r ) (-2ar) = (rho)/(in_(0))`
or `-2a - 4a = (rho)/(in_(0))` [from (i)]
or `rho = -6 in_(0) a`