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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 2πε0a -
λ 2πε0a2 -
λ 4π2ε0a -
λ2 2πε0a
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Correct Option: A
λ = linear charge density;
Charge on elementary portion dx = λ dx.
| Electric field at O , dE = | ||
| 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 = ∫ | cos θ | |
| 4πε0a2 |
| Also , dθ = | or dx = a dθ | |
| a |
| = | [1 - (-1)] = | ||
| 4πε0a | 2πε0a |
