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Q.

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

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a

$- 6 a ε_{0} r$

b

$- 24 π a ε_{0}$

c

$- 6 a ε_{0}$

d

$- 24 π a ε_{0} r$

answer is C.

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Detailed Solution

detailed_solution_thumbnail

$E = - \frac{dϕ}{dt} = - 2 ar$

By Gauss's theorem,
$E (4 {πr}^{2}) = \frac{q}{ϵ_{0}}$

$q = - 8 {πϵ}_{0} {ar}^{2}$

$ρ = \frac{q}{v} = \frac{- 8 {πϵ}_{0} {ar}^{3}}{\frac{4}{3} {πr}^{3}} = - 6 a ϵ_{0}$

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The electrostatic potential inside a charged spherical ball is given by ϕ=ar2+b where r is the distance from the centre and a, b are constants. Then the charge density inside the ball is: