A ring of radius R is with a uniformly distributed charge Q 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 Delta T sin ((d theta)/(2))`
Here F is the repulsive force between q and elemental charge
`dQ [ because dQ = (Q)/(2pi R) (R d theta)]`
The electric outward force on element is `F= (1)/(4pi in_(0))(qdQ)/(R^(2))`
From the above three equations, we can write
`(1)/(4pi in_(0)) (q)/(R^(2)) (QR d theta)/(2pi R) ~~ 2 Delta T ((d theta)/(2)) ( because sin alpha ~~ alpha " for small angle")`
`Delta T= (Qq)/(8pi^(2) in_(0)R^(2))`