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Question

# A ring has charge Q and radius R. If a charge q is placed at its centre, then the increase in tension in the ring is :

A
Qq4πϵ0R2
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B
0
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C
Qq4π2ϵ0R2
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D
Qq8π2ϵ0R2
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Solution

## The correct option is D Qq8π2ϵ0R2Let the linear charge density of the ring be given by λThat is, λ=Q2πRwhere Q is the charge on the ring and R is its radius.Now, the force acting between a small section dx of the ring and the charge q in the center will be,F =kqλdxR2If θ be the angle subtended at the centre by dx, the force acting will be, 2T sin θ2=Tθ ( θ is very small ) Thus we get dxR=θThus force should be equal to kqλθRT=kqλRSubstituting, the value of λ we get, T =kQq2πR2=Qq8π2ϵ0R2

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