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. $\frac{\pi \sigma R}{2{ϵ}_0}$
• B. $\frac{\sigma R}{2\pi {ϵ}_0}$
• C. $\frac{\sigma R\left(\pi -1\right)}{2\pi {\left(ϵ\right)}_0}$
• D. $\frac{\sigma R\left(\pi -1\right)}{2{\left(ϵ\right)}_0}$

Answer 1

Answer: (C)

The electric potential at the centre of the complete disc, $V_1=\frac{\sigma R}{2{ϵ}_0}$
$r=\frac{2R}{2}\cos \left(\frac{\theta }{2}\right)$
$dr=-\frac{2R}{2}\sin \left(\frac{\theta }{2}\right)\left(\frac{d\theta }{2}\right)$
$\int d{V}_2={\int }_{\theta =0}^{\pi }\frac{\sigma \times (r\theta )(dr)}{4\pi {\epsilon }_{0}r}=\frac{\sigma R}{2\pi {\epsilon }_0}=V_2$
$\therefore$ Net potential at centre $C$ is,
$V=\left(V_1-V_2\right)=\frac{\sigma R(\pi -1)}{2\pi {\epsilon }_0}$