Question

From a uniformly charged disc of radius R having surface charge density $\sigma$, a disc of radius $\frac{R}{2}$ is Removed as shown. What is the electric potential at point P?

• A. $\frac{\sigma R}{\pi {\epsilon }_0}$
• B. $\frac{\sigma R}{2\pi {\epsilon }_0}$
• C. $\frac{2\sigma R}{\pi {\epsilon }_0}$
• D. $\frac{\sigma R}{4\pi {\epsilon }_0}$

Answer 1

The correct option is A $\frac{\sigma R}{\pi {\epsilon }_0}$

Radius $=R$

Surface charge density $=\sigma$

$V=\frac{\sigma }{{ϵ}_0}.\frac{R}{\pi }$