The electrostatic potential inside a charged sphere is given as `V = Ar^(2) + B`, where r is the disatance from the centre of the sphere, A and B are constant. Then, the charge density in the sphere is

Options

A. `16 A epsilon_(0)`
B. `6 A epsilon_(0)`
C. `20 A epsilon_(0)`
D. `-15 A epsilon_(0)`

Answer

b

Explanation

Let the volume charge density be `rho`.
The electric field inside the charged sphere,

`E = (k qr)/(R^(3))`
Where R = Radius of charged sphere
r = Distance of the point (P) inside the sphere
As volume charge density is `rho`, then in terms of charge density
`E = (rho r)/(3 epsilon_(0))`
`implies rho = (3 E epsilon_(0))/(f)`
Now, electric field, `E = - (dV)/(dr)`
Here, `V = Ar^(2) + B`
`:. E = - (dV)/(dr) = - 2Ar + 0 = - Ar`
`:. rho = (3 E epsilon_(0))/(r ) = - 6 A epsilon_(0))`