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Question

A thin wire ring of radius r has a n electric charge q. The increment in the tension in the wire if a point charge q0 is placed centre of the ring is:

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Solution


As we put a charge q0 at the centre of the ring, the wire get stretched due to mutual repulsion between the positive charge q0 and each element of the wire having positive charge (dq). Such a small element is shown in the diagram.
Tensions acting at the ends of the elements due to elasticity of the ring are shown. Its two components are also shown. Carefully note the geometry.
dF is the repulsive force between the elemental charge dq and q0
Hence, the radial forces acting on the element can be summed up as
dF2T sin(dθ2)=0
dF=2T sin(dθ2)
=T dθ as sin(dθ2)dθ2
Now evidently
dF=q0 dq4πϵ0R2
while dq=q(R dθ)2πR
=q0qdθ8π2ϵ0R2T dθ=q0q dθ8π2ϵ0R2T=q0q8π2ϵ0R2

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