Question

Consider a circular disc of radius R having a uniform surface charge density \sigma. two discs of radius R/2 are removed as shown in figure Find potential at point O.​

Answer 1

$V = \frac{\sigma R (\pi-2)}{2\pi \epsilon}$

Step-by-step explanation:

Approach

For solving this type of questions let the removed part is putted on disk again and on removed part's negatively charged plates are putted on its top.

potential at the edge of disc is given by

$V^1= \frac{\sigma R }{\pi \epsilon}$

And potential at the centre of disc is given by

$V^2 = \frac{\sigma R }{2 \epsilon}$

Potential at O is given by

$V^1= \frac{\sigma R }{2\epsilon}- 2 \times \frac{\sigma \frac{r}{2} }{\pi \epsilon}$

$= \frac{\sigma R }{2\epsilon} - \frac{\sigma R }{\pi \epsilon} = \frac{\sigma R }{\epsilon}( \frac{1}{2} - \frac{1}{\pi} )$

$V = \frac{\sigma R (\pi-2)}{2\pi \epsilon}$

Note:- refer to attachment for diagram and better explanation of the method

Additional information

The method of taking whole disk and than adding negative charge to the area which is given removed in question can be applied on all questions of annular disk as well depending upon the situation

Answer 2

This is the statagy

For solving this type of questions let the removed part is putted on disk again and on removed part's negatively charged plates are putted on its top.