Question

A non-conducting infinite rod is placed along the z-axis, the upper half of the rod (lying along $z\ge 0$ ) is charged positively with a uniform linear charge density $(z<0)$ 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 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 $y$.
Diagram

Answer 1

According to Newton’s third law
Force on rod due to disc $=-$Force on disc due to rod
We will consider a small annular portion of the disc of radius $r$ and with $dr$ and calculate force on this portion and then integrate it.
Charge of this portion $={\sigma }_0\left(2\pi rdr\right)$
Force on this portion $=$ charge $\times$ Electric field due to rod
$={\sigma }_0\left(2\pi rdr\right)\left(\frac{\lambda }{2\pi {\epsilon }_{0}r}\right)=\frac{\lambda {\sigma }_0}{{\epsilon }_0}dr$
Total force $F=\frac{{\sigma }_0}{{\epsilon }_0}{\int }_R^{2R}dr=\frac{\sigma \lambda R}{{\epsilon }_0}$
Also, $F=\frac{y{\sigma }_0\lambda R}{8{\epsilon }_0}$
$⟹y=8$
Hence, the value of $y$ is $8$.