A ring of radius R is with 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. Its enlarged view is as shown. For equilibrium of this segment, we can write.
`F= 2DeltaT sin ((d theta)/(2))`
Here F is the repulsive force between q and elemental charge dQ
`dQ=(Q)/( 2pi R) (Rd 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)) (QRd theta)/(2piR) ~~ 2DeltaT((d theta)/(2))`
(`:. sin alpha = alpha` for small angle)
`Delta = (Qq)/( 8pi^(2) in_(0) R^(2))`