A plastic disc is charged on one side with a uniform surface charge density $\sigma$ and then three quadrant…

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 $\hat{V}=0$ at infinity, the potential due to the remaining quadrant at point $P$ is
  1. $\frac{\sigma}{2 \epsilon_{0}}\left[\left(r^{2}+R\right)^{1 / 2}-r\right]$
  2. $\frac{\sigma}{2 \epsilon_{0}}[R-r]$
  3. $\frac{\sigma}{8 \epsilon_{0}}\left[\left(r^{2}+R^{2}\right)^{1 / 2}-r\right]$
  4. None of these

Solution

The potential at $P$ due to whole disc is $V=\frac{\sigma}{2 \in_{0}}\left[\sqrt{R^{2}+r^{2}-r}\right]$
Now potential due to quarter disc, $V=\frac{V}{4}=\frac{\sigma}{8 \in_{0}}\left[\sqrt{R^{2}+r^{2}}-r\right] .$ ,

Asked in: JEE Mains - Electrostatics - Test 3

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