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 $q_0$ is placed centre of the ring is:

Answer 1

As we put a charge $q_0$ at the centre of the ring, the wire get stretched due to mutual repulsion between the positive charge $q_0$ 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 $q_0$
Hence, the radial forces acting on the element can be summed up as
$dF-2Tsin\left(\frac{d\theta }{2}\right)=0$
$dF=2Tsin\left(\frac{d\theta }{2}\right)$
$=Td\theta$ as $sin\left(\frac{d\theta }{2}\right)\Rightarrow \frac{d\theta }{2}$
Now evidently
$dF=\frac{q_{0}dq}{4\pi {ϵ}_0{R}^2}$
while $dq=\frac{q\left(Rd\theta \right)}{2\pi R}$
$=\frac{q_{0}qd\theta }{8{\pi }^2{ϵ}_0{R}^2}Td\theta =\frac{q_{0}qd\theta }{8{\pi }^2{ϵ}_0{R}^2}\Rightarrow T=\frac{q_{0}q}{8{\pi }^2{ϵ}_0{R}^2}$