A semicircular arc of radius R is charged uniformly and its linear charge density is $λ$. What will the electric field be at its centre?

$λ/(2 π ε_0 R)$

The correct answer is Option (1) → $λ/(2 π ε_0 R)$

Given: A uniformly charged semicircular arc of radius $R$ and linear charge density $λ$.

Each small element of charge $dq = λ\,R\,dθ$ produces an electric field $dE = \frac{1}{4πϵ_0}\frac{dq}{R^2}$ at the centre.

Due to symmetry, the horizontal components of $dE$ cancel, and only the vertical components add up.

Vertical component: $dE_y = dE \sinθ = \frac{1}{4πϵ_0}\frac{λR\,dθ}{R^2}\sinθ = \frac{λ}{4πϵ_0R}\sinθ\,dθ$

Integrating over the semicircle ($θ = 0$ to $π$):

$E = \frac{λ}{4πϵ_0R}\int_0^π \sinθ\,dθ$

$E = \frac{λ}{4πϵ_0R}[2]$

∴ $E = \frac{λ}{2πϵ_0R}$

Direction: Perpendicular to the diameter and directed toward the concave side of the arc.

Answer: $E = \frac{λ}{2πϵ_0R}$, directed toward the centre of curvature.