Electric Charges and Fields
- A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is λ. The electric field at the centre of this arc is
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λ = linear charge density;
Charge on elementary portion dx = λ dx.
Electric field at O , dE = λ dx 4πε0a2
Horizontal electric field, i.e., perpendicular to AO, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of AO
= ∑ dE cos θ⇒ E = ∫ λ dx cos θ 4πε0a2 Also , dθ = dx or dx = a dθ a 
= λ [1 - (-1)] = λ 4πε0a 2πε0a Correct Option: A
λ = linear charge density;
Charge on elementary portion dx = λ dx.Electric field at O , dE = λ dx 4πε0a2
Horizontal electric field, i.e., perpendicular to AO, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of AO
= ∑ dE cos θ⇒ E = ∫ λ dx cos θ 4πε0a2 Also , dθ = dx or dx = a dθ a = λ [1 - (-1)] = λ 4πε0a 2πε0a
