A ring of radius `R` is having charge `Q` uniformly distributed. A point charge `Q_(0)` is placed at the centre of the ring. Find tension developed in the ring. If radius of cross - section of the ring is `a` and Young's modulus of wire is `Y`, find increase in the radius of ring.

Explanation

To find tension//compression in the ring, find the force on small length of angle ` d theta` and remove `d theta` from the expression (refer to Circular Motion , Example 33 , page 351 Vol. 1).
Charge on small length of angle `d theta`
`Q' = (Q)/( 2 pi R) dl = (Q)/( 2 pi R) R d theta = (Q d theta)/(2 pi)`
Electric force on `Q'` ,
`F = (1)/(4 pi in_(0)) (Q' Q_(0))/(R^(2)) = (1)/( 4 pi in_(0)) . (Q Q_(0))/(2 pi R^(2)) d theta`
Tension in the string
`T = (Q Q_(0))/( 8 pi^(2) in_(0) R^(2))`
Let increase in the radius of ring is `Delta R`.
Longitudinal strain
`in = ( 2pi (R + Delta R) - 2 pi R)/( 2 pi R) = (Delta R)/(R)`
Longitudinal stress
`sigma = (T)/(A) = (T)/(pi d^(2))`
where `a`: radius of cross - section of ring.
`Y = (sigma)/(in) = (T//pi a^(2))/(Delta R//R)`
`Delta R = (T R)/(pi a^(2) Y) = (Q Q_(0))/(8 pi^(2) in_(0) R^(2)) . (R )/(pi a^(2) Y)`
`= (Q Q_(0))/(8 pi^(3) in_(0) R Y a^(2))`