ElectroMechanics

Consider an circular arc element of charge on Ring , So when we place an 'Q' charge at the centre of ring Then due to Electrostatic Repulsion b/w charges an Extra Tension is created in Ring . which balance the repulsion b/w them .

$\displaystyle{{ \therefore \quad \cfrac { Kq(dq) }{ { R }^{ 2 } } =2dT\sin { (d\theta ) } \\ \quad \sin { (d\theta ) } \quad \approx \quad d\theta \\ \because \quad dq=\cfrac { Q }{ 2\pi R } R(2d\theta )\quad \quad \\ dT=\cfrac { KQq }{ 2\pi { R }^{ 2 } } \quad .\quad .\quad .\quad (1) }}$ .

Now due to this extra Tension an Stress is developed in ring which further Cause the Expansion of radius (Due to Strain ) Now Using Hooks Law (By assuming that Young's modulus is so high so that Radius is Expand till Below it's Fracture point)

$\displaystyle{Stress=Y\times Strain}$ .

$\\ \cfrac { dT }{ A } =Y\times (\cfrac { 2\pi (R+dR)-2\pi R }{ R } )\quad .\quad .\quad .(2)$ .

Using above equation we should get :

$\displaystyle{\boxed { dR\quad =\quad \cfrac { KQq }{ 2\pi RAY } } }$ .