A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field `vecE` at the centre O is

Explanation

A segment of charge is selected on ring at an angle `theta` and subtending angle `d theta` at its centre . Electric field due to this segment is shown as dE and is given by :
`dE=1/(4piepsilon_0)(Q/pid theta)/R^2`
Here we have used the formula of electric field intensity due to a point charge for segment . On rearranging we get the following :
`dE=1/(4piepsilon_0)(Qd theta)/R^2`
Consider the corresponding segment on other half of ring as shown in the figure. It can be understood that components of electric field due to the charge segments along axis YY. are added but the components alon the axis XX. are cancelled by each other . Direction of net field is along the line of symmetry as shown in the figure.
`E=int_"Half-ring"1/(4pi^2epsilon_0)(Qd theta)/R^2 cos theta`
`rArr E=Q/(4pi^2 epsilon_0 R^2)int_(-pi//2)^(pi//2)cos theta d theta`
`E=int_"Half-ring" dE cos theta`
Here , the limits of integration are selected to cover the complete semicircular ring. We can also change limits from 0 to `pi//2`, but after multiplying the expression by 2.
`rArr E=2Q/(4pi^2epsilon_0R^2)int_0^(pi//2)cos theta d theta`
`rArr E=2 Q/(4pi^2 epsilon_0R^2)[sin theta]_0^(pi//2)`
`rArr E=Q/(2pi^2epsilon_0R^2)[sin pi//2 - sin 0]`
`rArr E=Q/(2pi^2 epsilon_0R^2)`
The direction of net electric field intensity is marked in the given figure.