Question

A thin insulating ring of radius $R$ has a uniform linear charge density $\lambda$. Now a charge $Q_0$ is placed at centre of this ring. The increment in the stretching force of the wire due to the charge $Q_0$ is:-

• A. $\frac{Q_0\lambda }{2\pi {\epsilon }_{0}R}$
• B. $\frac{Q_0\lambda }{4\pi {\epsilon }_{0}R}$
• C. $\frac{3Q_0\lambda }{8\pi {\epsilon }_{0}R}$
• D. $\frac{3Q_0\lambda }{4\pi {\epsilon }_{0}R}$

Answer 1

The correct option is B $\frac{Q_0\lambda }{4\pi {\epsilon }_{0}R}$

$\begin{array}{c}F=Q_o\left[\frac{2K\lambda }{R}\sin \left(\frac{2d\varphi }{2}\right)\right]\\ =Q_o\left[\frac{2K\lambda }{R}sind\varphi \right]\\ \therefore 2T\sin \left(d\varphi \right)=F\\ \Rightarrow 2T\sin \left(d\varphi \right)=Q_o.\frac{2K\lambda }{R}\sin d\varphi \\ \Rightarrow T=\frac{Q_o\lambda }{4\pi {\epsilon }_{o}R}\\ Hence,\\ optionBiscorrectanswer.\end{array}$