In the figure shown an infinitely long wire of uniform linear charge density $λ$ is kept perpendicular to the plane of figure such that it extends upto infinity on both sides of the paper. Find the electrostatic force on a semicircular ring kept such that its geometrical axis coincides with the wire. The semicircular ring has a uniform linear charge density $λ$ .

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$\frac{λλ'}{{πε}_{0} R}$

$\frac{{λλ}^{'}}{4 {πε}_{0}}$

$\frac{{λλ}^{'}}{2 {πε}_{0}}$

$\frac{{λλ}^{'}}{{πε}_{0}}$

answer is D.

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Detailed Solution

The electrostatic field intensity at a point on the ring is E = $\frac{λ}{2 {πε}_{0}} \frac{1}{R}$ .
The force on the elementary charge dq is
$dF = dqE = (λ^{'} Rdθ) \cdot \frac{λ}{2 {πε}_{0}} \frac{1}{R}$
The sine component of dF will get cancelled and cosine component will get added.
Net force on the ring
$F = 2 \int_{0}^{+ π / 2} dFcos θ = 2 \int_{0}^{+ π / 2} \frac{λ \cdot λ^{'}}{2 {πε}_{0}} dθ \cdot \cos θ = \frac{{λλ}^{'}}{{πε}_{0}}$
$\frac{{λλ}^{'}}{{πε}_{0}}$