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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

Options

  1. A25 × 10 3 i ^   N   C - 1
  2. B50 × 10 3 i ^   N   C - 1
  3. C25 × 10 3 j ^   N   C - 1
  4. D50 × 10 3 j ^   N   C - 1

Correct answer

A. 25 × 10 3 i ^   N   C - 1

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 undefined 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 &#

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