Question

A thin wire ring of radius $r$ has an electric charge $q$. What will be the increment of the force stretching the wire if a point charge $q_0$ is placed at the ring's centre?

Answer 1

As we put a charge 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 $\left(dq\right).$ 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$;$

$dF=2T\sin \left(dQ/2\right)=0$

$=dF=2T\sin \left(dQ/2\right)$

$=TdQas\sin \left(dQ/2\right)=dQ/2$

Now evidently

$dF=\frac{q_{0}dq}{4\pi {\in }_0{R}^2}$

while $dq=\frac{q\left(rdQ\right)}{2\pi R}$

$=\frac{q_{0}qdQ}{8{\pi }^2{\in }_0{R}^2}$

$TdQ=\frac{q_{0}qdQ}{8{\pi }^2{\in }_0{R}^2}$

$=T=\frac{q_{0}q}{8{\pi }^2{\in }_0{R}^2}$

Hence,

We get, $T=\frac{q_{0}q}{8{\pi }^2{\in }_0{R}^2}.$