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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/2π/2kλacosθdθ(^i)=kλa(sinθ)+π/2π/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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