The electrostatic potential inside a charged sphere is given as $V=A r^2+B$, where $r$ is the distance from the centre of the sphere, $A$ and $B$ are constants. Then, the charge density in the sphere is

• A

  $16 A \varepsilon_0$
• Correct answer
  B

  $-6 A \varepsilon_0$
• C

  $20 A \varepsilon_0$
• D

  $-15 A \varepsilon_0$

Solution

Let the volume charge density be $\rho$. The electric field inside the charged sphere,
$$ E=\frac{K q r}{R^3} $$ 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, $$ \begin{aligned} E & =\frac{\rho r}{3 \varepsilon_0} \\ \quad \quad \quad \rho & =\frac{3 E \varepsilon_0}{r} \end{aligned} $$ Now, electric field, $E=-\frac{d V}{d r}$ Here, $V=A r^2+B$ $$ \begin{array}{ll} \therefore & E=-\frac{d V}{d r}=-2 A r+0=-2 A r \\ \therefore & \rho=\frac{3 E \varepsilon_0}{r} \end{array} $$ $$ =\frac{3\left(-2 A r \cdot \varepsilon_0\right)}{r}=-6 A \varepsilon_0 $$

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