Question

A plastic disc is charged on one side with a uniform surface charge density $\sigma$ and then three quadrant of the disk are removed. The remaining quadrant is shown in figure, with $V=0$ at infinity, the potential due to the remaining quadrant at point $P$ is

• A. $\frac{\sigma}{2 \epsilon_0}\left[\left(r^2+R\right)^{1 / 2}-r\right]$
• B. $\frac{\sigma}{2 \epsilon_0}[R-r]$
• C. $\frac{\sigma}{8 \epsilon_0}\left[\left(r^2+R^2\right)^{1 / 2}-r\right]$
• D. None of these

Answer 1

Answer: (C)

The potential at $P$ due to whole disc is
$V=\frac{\sigma}{2 \epsilon_0}\left[\sqrt{R^2+r^2-r}\right] .$
Now potential due to quarter disc,
$V=\frac{V}{4}=\frac{\sigma}{8 \epsilon_0}\left[\sqrt{R^2+r^2}-r\right] .$