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Question

# Half ring of radius a is uniformly charged with linear charge density λ. Find out electric field intensity at point O.

A
λ2πϵ0a(^i)
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B
λ4πϵ0a(+^i)
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C
λ2πϵ0a(+^i)
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D
λ4πε0a(^i)
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Solution

## The correct option is A λ2πϵ0a(−^i) Taking an infinitely small element dl on the upper side of ring which have charge dq as shown in figure. Since charge is uniformly distributed over the ring. ∴dq=λdl=λ(adθ) Now electric field at centre of ring due to dq will be dE=kdqa2=kλadθa2=kλadθ From symmetry, dEy will get cancelled on integration. ∴Ey=∫dEy=0 So, net electric field at the centre will be Enet=∫dEx=∫dEcosθ(−^i) ⇒Enet=∫π/2−π/2kλacosθdθ(−^i)=kλa(sinθ)+π/2−π/2(−^i) ⇒Enet=λ4πϵ0a.(2)(−^i)=λ2πϵ0a(−^i) Alternative: Net electric firld due to a arc subtending angle α at the centre, Enet=2kλRsinα2 Here α=180∘ Enet=2kλasin90∘ Enet=2kλa=λ4πϵ0a.(2)=λ2πϵ0a Since ring is symmetric to x− axis, Ey=0 and Ex will be along negative x− axis Hence, option (a) is correct.

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