Question

A non-conducting disc of radius $R$ is uniformly charged with surface charge density $\sigma $ . A disc of radius $\frac{R}{2}$ is cut from the disc, as shown in the figure. The electric potential at centre $C$ of large disc will be

• A. $\frac{\pi \sigma R}{2 \epsilon _{0}}$
• B. $\frac{\sigma R}{2 \pi \epsilon _{0}}$
• C. $\frac{\sigma R \left(\pi - 1\right)}{2 \pi \left(\epsilon \right)_{0}}$
• D. $\frac{\sigma R \left(\pi - 1\right)}{2 \left(\epsilon \right)_{0}}$

Answer 1

Answer: (C)

The electric potential at the centre of the complete disc, $V_{1}=\frac{\sigma R}{2 \epsilon _{0}}$
$r=\frac{2 R}{2} \cos \left(\frac{\theta}{2}\right)$
$d r=-\frac{2 R}{2} \sin \left(\frac{\theta}{2}\right)\left(\frac{d \theta}{2}\right)$
$\int d V_{2}=\int_{\theta=0}^{\pi} \frac{\sigma \times(r \theta)( d r)}{4 \pi \varepsilon_{0} r}=\frac{\sigma R}{2 \pi \varepsilon_{0}}=V_{2}$
$\therefore $ Net potential at centre $C$ is,
$V=\left(V_{1}-V_{2}\right)=\frac{\sigma R(\pi-1)}{2 \pi \varepsilon_{0}}$