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

Answer

B

Explanation

`E=(kqr)/( R^3) rArr E=(pr)/(3 epsilon_0)`
`rho(3E_0)/( r)`
`E=-(dv)/(dr)`
`V=A r'+B`
`rArr E=-2 A r+0=-2Ar`
`rho=(3Eepsilon_0)/( r)`
`rho =(3(-2Ar.epsilon_0))/( r)`
`rho =- 6 A epsilon _0` .