Electric field due to semicircular charged ring

We all do comb in the morning.

Sometimes we observed when we bring the comb just after doing the comb, near to the small pieces of paper.

Then, the pieces of paper get attracted by the comb.

But, the pieces of paper are not got attracted by the comb which is far from the comb.

Because, when we do comb, then due to the rubbing operation, the comb gets charged.

And the charged particles always create a field around it within which it exerts a force on other particles.

That field is called the electric field. And out of the electric field, the charged particle cannot exert any force.

Now, we are interested to know the electric field due to a semicircular charged ring.

Let's discuss the electric field due to a semicircular charged ring.

Suppose we have a uniformly charged semicircular ring.

And we want to find the electric field at point P.

Now, we cut a small strip of length $Δ S$ on the semicircular ring.

And let the charge per unit length along the semicircle be $λ$ . Hence,

Hence, the net charge on the entire length will be,

As the force between the charged particle is calculated by Coulomb's law.

And in the terms of the electric field (E), the force (F) on the charged particles is calculated as,

Hence, it can also be written as,

As the small strip makes an angle $θ$ with the point $P$ .

Hence, there will be two components of force applied by small strip: $E_{x}$ and $E_{y}$ . As

Now, we take a small section on semicircle as $Δ S_{i}$ with charge $Δ q_{i}$ . Hence,

Hence, the electric field at P will be

Now, due to symmetry, the vertical component will be canceled out.

So now, we will only deal with the horizontal component. Hence, the electric field of horizontal component,

Hence, the net electric field ( $E_{x}$ ) can be calculated by summing up the horizontal components.

Now, we take the limit as $Δ S$ tends to zero.

As the arc length (s) is a product of angle ( $θ$ ) and radius (r). So,

After putting the value of $d s$ ,

Now, to get the net electric field due to semicircle, we have to integrate the equation from $\frac{- π}{2}$ to $\frac{π}{2}$ .

After simplification, it will be as,

Hence, the net electric field due to semicircular charged ring will be,

Revision

First, we cut a small strip on the semicircular ring.

And we find the electric field as,

So, after integrating from $\frac{- π}{2}$ to $\frac{π}{2}$ , we get the net electric field due to semicircular ring.

The End