Question

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

• A. $-24\pi a{\epsilon }_{0}r$
• B. $-6\pi a{\epsilon }_{0}r$
• C. $-24\pi a{\epsilon }_0$
• D. $-6\pi a{\epsilon }_0$

Answer 1

The correct option is C $-24\pi a{\epsilon }_0$

Electrostatic potential inside a charge spherical ball $=\varphi ={ar}^2+b$

the distance $=r$

The change density

$\frac{d\varphi }{dr}=2ar$

$=2a\times (-12\pi {ϵ}_0)$

$=-24\pi q{ϵ}_0$