Question

A thin non-conducting disc of radius R is uniformly charged with surface charge density $\sigma$. If potential at any point on the circumference of disc be $V$ and total electrostatic potential energy of the disc be $U$, then

• A. $V=\frac{2\sigma R}{\pi {\epsilon }_0}$
• B. $U=\frac{4}{3}\frac{{\sigma }^2{R}^3}{{\epsilon }_0}$
• C. $V=\frac{\sigma R}{\pi {\epsilon }_0}$
• D. $U=\frac{2{\sigma }^2{R}^3}{3{\epsilon }_0}$

Answer 1

Answer: (C), (D)

$U=\frac{2{\sigma }^2{R}^3}{3{\epsilon }_0}$

$dV=\frac{\sigma xd\theta dx}{4\pi {\epsilon }_{0}x}$
$V=\frac{\sigma \underset{-\pi /2}{\overset{{}^{\pi }/{}_2}{\int }}d\theta {\int }_0^{l}dx}{4\pi {\epsilon }_0}$
$V=\frac{\sigma \underset{{}^{-\pi }/{}_2}{\overset{{}^{\pi }/{}_2}{\int }}2r\cos \theta d\theta }{4\pi {\epsilon }_0}$
$V=\frac{\sigma r}{\pi {\epsilon }_0}$
$U=\underset{0}{\overset{R}{\int }}\frac{\sigma r}{\pi {\epsilon }_0}\left(\sigma 2\pi rdr\right)$
$U=\frac{2}{3}\frac{\pi {\sigma }^2{R}^3}{\pi {\epsilon }_0}$