Question

A semicircular rod charged with a charge $Q$ is placed as shown in figure. Radius of the wire is $R$ and the infinitely long line of charge with linear density $\lambda$ is passing through its centre and perpendicular to the plane of wire. If the force experienced by the semicircular rod is $\frac{\lambda Q}{x{\pi }^2{\epsilon }_{0}R}$. Then the value of $x$ is

Answer 1

$\begin{array}{cc}dF& =dqE\\ dF& ={\lambda }^{\prime }Rd\theta \frac{2k\lambda }{R}\\ dF& =\frac{2k\lambda }{R}Q\frac{d\theta }{\pi }\end{array}$



Horizontal component of force on the right half will get cancelled due to horizontal component of force on left half.
$⟹F_{net}={\int }_0^{\pi }dF\sin \theta =\frac{2k\lambda Q}{\pi R}{\int }_0^{\pi }\sin \theta d\theta$

$\therefore F=\frac{\lambda Q}{{\pi }^2{\epsilon }_{0}R}$