Question

A non-conducting infinite rod is placed along the $z$-axis, the upper half of the rod (lying along $z\ge 0$ ) is chárged positively with a uniform linear charge density $+\lambda$ while the lower half $(z<0)$ is charged negatively with a uniform linear charge density $-\lambda$. The origin is located at the junction of the positive and negative halves of the rod. A uniformly charged annular disc (surface charge density: ${\sigma }_0$ ) of inner radius $R$ and outer radius $2R$ is placed in the $x-y$ plane with its centre at the origin. The force on the rod due to the disc is $\frac{y{\sigma }_0\lambda R}{8{\epsilon }_0}$. Find the value of $y$.

Answer 1

Answer: 8

The force on the rod due to the annular disc is equal and opposite to that on the disc due to rod.
We take an annular strip of radius $r$ and width $dr$ and find the force acting on it.
$dF=\left({\sigma }_{0}2\pi rdr\right)\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{\lambda {\sigma }_0}{{\epsilon }_0}dr$
Total force, $F=\frac{{\sigma }_0\lambda }{{\epsilon }_0}\underset{R}{\overset{2R}{\int }}dr=\frac{{\sigma }_0\lambda R}{{\epsilon }_0}$