Question

The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$ where r is the distance from the centre a, b are constants. Then the charge density inside the ball is

• A. $ - 6 a \varepsilon_0 r$
• B. $ - 24 \pi a \varepsilon_0 $
• C. $ - 6 a \varepsilon_0 $
• D. $ - 24 \pi a \varepsilon_0 r$

Answer 1

Answer: (C)

Electric field, $E = - \frac{d \phi}{dt} = - 2ar $ ...(i)
By Gauss's theorem $E\left(4\pi r^{2}\right) = \frac{q}{\varepsilon_{0}} $ ...(ii)
From (i) and (ii),
$\Rightarrow q = - 8 \pi\varepsilon_{0 }ar^{3}$
$ \rho = \frac{dq}{dV} = \frac{dq}{dr} \times\frac{dr}{dV} $
$= \left(-24 \pi\varepsilon_{0} ar^{2}\right) \left(\frac{1}{4 \pi r^{2}}\right) = -6 \varepsilon_{0} a $