Question

A ring radius 0.1 m is made out of a thin metallic wire of area of cross section ${10}^{-4}{m}^2$. The ring has a uniform charge of $\pi$ coulombs. Find the change in the radius of the ring when a charge of ${10}^{-8}$C is placed at the center of the ring. Young's modulus of the metal is $2\times {10}^{11}N{m}^{-2}$.

Answer 1

$q={10}^{-8}C,\lambda =\frac{Q}{2\pi r}=\frac{\pi }{2\pi \left(0.1\right)}=5C{m}^{-1}$

Take an arc of ring subtending small angle $\theta$ at the center. Charge on this arc is $dq=\lambda r\theta$.

For net force to be zero on this arc,

$2Tsin\left(\frac{\theta }{2}\right)=F$

or $2Tsin\left(\frac{\theta }{2}\right)=\frac{kq\left(\lambda r\theta \right)}{r^2}$

or $T=\frac{kq\lambda }{r}$

$=\frac{9\times {10}^9\times {10}^{-8}\times 5}{0.1}=4500N$

$\Delta r=\frac{Tr}{AY}=\frac{4500\times 0.1}{{10}^{-6}\times 2\times {10}^{11}}$

$=225\times {10}^{-5}m=2.25mm$