A wire of length L = 20   c m is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges + Q and - Q as shown in the figure. The magnitude of the charge on each half is Q = 10 3 ε 0 , where ε 0 is the permittivity of free the space. The net electric field at the centre O is

Step-by-step solution

L = π R ⇒ R = L π = 20 100 π m = 1 5 π m ⇒ due to a charge arc, electric field at centre is given by E = 2 K λ R sin θ 2 E 1 = E 2 = 2 k ⋋ R sin ⁡ 90 2 ⋋ = Q π R / 2 E 1 = E 2 = 2 2 K Q π R 2 Component along j ^ gets cancelled and E n e t = 2 E 1 = 4 K Q π R 2 = 4 × 1 10 3 ϵ 0 4 π ϵ 0 π R 2 = 4 × 10 3 ϵ 0 4 π 2 ϵ 0 R 2 = 10 2 R 2 = 100 1 5 π 2 = 25 × 10 3 E → n e t = 25 × 10 3 N C i ^