Question

The electrostatics potential inside a charged spherical is given by $\varphi =a{r}^2+b$ where $r$ is the difference between centre $a,b$ are constants. Then the charge density of the ball is

• A. $-6a{\epsilon }_{o}r$
• B. $-24\pi a{\epsilon }_o$
• C. $-6a{\epsilon }_o$
• D. $-24\pi a{\epsilon }_{o}r$

Answer 1

The correct option is C $-6a{\epsilon }_o$

$E=-\frac{d\varphi }{dr}$; $E=-2ar$

and $E=\frac{1}{4\pi {ϵ}_o}\frac{q}{r^2}=-2ar$

$q=-4\pi {ϵ}_o\cdot 2a{r}^3$

Charge density, $\rho =\frac{q}{Volume}=\frac{-4\pi {ϵ}_{o}2a{r}^3}{\frac{4}{3}\pi r^3}$

$\rho =-6a{ϵ}_o$