Question

A ring of radius $R$ is made out of a thin metallic wire of area of cross section $A$. The ring has a uniform charge $Q$ distributed on it. A charge $q_0$ is placed at the centre of the ring. If $\text{Y}$ is the young`s modules for the material of the ring and $\Delta R$ is the change is the radius of the ring then find $\Delta R.$

• A. $\frac{q_{0}Q}{8{\pi }^2{\epsilon }_{0}RA\text{Y}}$
• B. $\frac{q_{0}Q}{8{\pi }^2{\epsilon }_0^{2}RA\text{Y}}$
• C. $\frac{q_{0}Q}{4{\pi }^2{\epsilon }_{0}RA\text{Y}}$
• D. $\frac{q_{0}Q}{4\pi {\epsilon }_{0}RA\text{Y}}$

Answer 1

The correct option is A $\frac{q_{0}Q}{8{\pi }^2{\epsilon }_{0}RA\text{Y}}$

For a ring, Linear charge density $\lambda =\frac{Q}{2\pi R}$

Tension force in the ring $T=\frac{q_0\lambda }{4\pi {\epsilon }_{0}R}=\frac{q_{0}Q}{8{\pi }^2{\epsilon }_0{R}^2}$

But, Young's modulus $\text{Y}=\frac{\text{Stress}}{\text{Strain}}=\frac{\left(\frac{T}{A}\right)}{\left(\frac{\Delta R}{R}\right)}\Rightarrow \Delta R=\frac{TR}{A\text{Y}}$

$\Delta R=\frac{q_{0}QR}{8{\pi }^2{\epsilon }_0{R}^{2}A\text{Y}}$

$\Delta R=\frac{q_{0}Q}{8{\pi }^2{\epsilon }_{0}RA\text{Y}}$

Hence, option (a) is the correct answer.