A ring of radius R is with a uniformly distributed charge on it. A charge q is now placed at the centre of the ring. Find the increment in tension in the ring.

Explanation

Consider an element of the ring. Its enlarged view is as shown. For equilibrium of this segment, we can write.
`F = 2 DeltaT sin ((d theta)/(2))`
Here F is the repulsive force between q and elemental charge `dQ [ :. dQ = Q/(2 pi R) (Rd theta)]`
The electric outward force on element is `F = (1)/(4 pi epsilon_0) (q dQ)/(R^2)`
From the above three equations, we can write,
`1/(4 pi epsilon_0) q/(R^2) (QRd theta)/(2 pi R) ~~ 2 Delta T ((d theta)/(2)) ( :. sin alpha  ~~ alpha` for small angle)
`Delta T = (Qq)/(8 pi^2 epsilon_0 R^2)`