Electrostatics PYQ | Electric Field Due to Semicircular Ring | JEE Main

 

❓ Question

A uniformly charged semi-circular ring of radius $R$ carries total charge $Q$. A charge $q$ is placed at the center of the ring.

Find the magnitude of the electrostatic force on the charge $q$.

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✍️ Solution

For a uniformly charged arc of angle $\theta$, the electric field at the center is

$$E=\frac{2k\lambda\sin\left(\frac{\theta}{2}\right)}{R},$$

where

$$\lambda=\frac{\text{Charge}}{\text{Arc length}}.$$

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Linear Charge Density

For a semi-circular ring,

$$\text{Arc length}=\pi R.$$

Hence,

$$\lambda=\frac{Q}{\pi R}.$$

Also,

$$\theta=180^\circ=\pi.$$

Therefore,

$$\sin\frac{\theta}{2} = \sin90^\circ = 1.$$

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Electric Field at the Center

Substitute the values:

$$\begin{aligned} E &=\frac{2k\lambda}{R}\\ &=\frac{2k}{R}\left(\frac{Q}{\pi R}\right)\\ &=\frac{2kQ}{\pi R^2}. \end{aligned}$$

Using

$$k=\frac{1}{4\pi\varepsilon_0},$$
$$E = \frac{Q}{2\pi^2\varepsilon_0R^2}.$$

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Force on Charge $q$

Since

$$F=qE,$$
$$\boxed{ F=\frac{Qq}{2\pi^2\varepsilon_0R^2} }$$

The force acts along the axis of symmetry of the semicircle (downward if the ring is positively charged and lies above the center).

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✅ Final Answer

$$\boxed{ F=\frac{Qq}{2\pi^2\varepsilon_0R^2} }$$