Question

A sphere is uniformly charged with charge per unit volume as $\rho$ and radius $R$. The electrostatic potential energy stored inside the sphere is $\frac{4\pi {\rho }^2{R}^5}{n{ϵ}_0}$. Fill the value of $n$

• A. 90.0
• B. 90.00
• C. 90

Answer 1

Answer: (A), (B), (C)

As we know that energy per unit volume of a conductor is given by, $\frac{dU}{dV}=\frac{1}{2}{ϵ}_0{E}^2$ ....(1)

Now energy stored inside a solid sphere:
Let us assume that a small element of $dr$ thickness at a distance $r$ from point $O$.



Electric field at point $P$, $E=\frac{KQ}{R^3}.r$
From equation (1) energy of small element,
$dU=\frac{1}{2}{ϵ}_0{E}^{2}dV$

$dU=\frac{1}{2}{ϵ}_0{(\frac{KQ}{R^3}.r)}^2.4\pi r^2.dr$

Now total energy of solid sphere-
$U={\int }_0^{R}dU=\frac{4\pi {ϵ}_0{Q}^2{K}^2}{2R^6}{\int }_0^R{r}^4.dr$

$\Rightarrow U=\frac{4\pi {ϵ}_0}{2R^6}.\frac{Q^2}{16{\pi }^2{ϵ}_0^2}{\left[\frac{r^5}{5}\right]}_0^R=\frac{4\pi Q^2{R}^5}{10{ϵ}_{0}16{\pi }^2{R}^6}$

$\Rightarrow U=\frac{4\pi R^5}{10{ϵ}_0}.\frac{Q^2}{\frac{16{\pi }^2{R}^6}{9}\times 9}=\frac{4\pi R^5}{90{ϵ}_0}{\left(\frac{Q}{V}\right)}^2\{\rho =\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi R^3}\}$

$\Rightarrow U=\frac{4\pi {\rho }^2{R}^5}{90{ϵ}_0}=\frac{4\pi {\rho }^2{R}^5}{n{ϵ}_0}$

$\therefore n=90$