Q86 / 90 · JEE Main 2011 · Physics

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

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

Solution

Potential inside $(\phi)=a r^2+b$ $$ \therefore \mathrm{E}_{\mathrm{r}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{r}}=-2 \mathrm{ar} $$ Electric field inside uniformly charged solid volume varies with ' $r$ '. So charge density is constant $$ \begin{aligned} & \phi_{\text {net }}=(-2 \mathrm{ar}) 4 \pi \mathrm{r}^2=-8 \pi a r^3 \\ & -8 \pi \mathrm{ar}^3=\frac{\sigma \times \frac{4}{3} \pi \mathrm{r}^3}{\varepsilon_0} \\ & \therefore \sigma=-6 \mathrm{a} \varepsilon_0 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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