Home/Question #5361
JEE Main 2025Medium

Q.A thin semicircular ring of radius R has linear charge density λ = λ₀ cosθ (θ from -π/2 to π/2). The electric field at the centre is

a
λ0πϵ0R\dfrac{\lambda_0}{ \pi \epsilon_0 R}
b
2λ0πϵ0R\dfrac{2 \lambda_0}{ \pi \epsilon_0 R}
c
zero
d
λ02πϵ0R\dfrac{\lambda_0}{2 \pi \epsilon_0 R}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

Unlock Detailed Solution

Register for Free

Already a member? Login here