Question

A point charge $q$ is situated at a distance $d$ from one end of a thin non-conducting rod of length $L$ having a charge $Q$ (uniformly distributed along its length) as shown in fig. Then the magnitude of the electric force between them is:

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{2d(d+L)}$
• B. $\frac{1}{\pi {\epsilon }_0}\frac{2qQ}{d(d+L)}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{3d(d+L)}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{d(d+L)}$

Answer 1

Answer: (D)

$F={\int }_d^{d+L}\frac{kqdq}{x^2}={\int }_d^{d+L}\frac{kq\left(\frac{Q}{L}dx\right)}{x^2}=\frac{kqQ}{L}{\int }_d^{d+L}\frac{1}{x^2}dx$
$F=\frac{kqQ}{d(d+L)}=\frac{1}{4\pi {\left(ϵ\right)}_0}\frac{qQ}{d(d+L)}$
Or

The linear charge density of rod $\lambda =\frac{Q}{L};$
$dq=\lambda dx$
Force due to small element $dx$ to charge $q$
$dF=\frac{1(q)(\lambda )dx}{4\pi {ϵ}_0{x}^2}$
Total force in-between point charge and rod
$F_{net}=\frac{qQ}{L}{\int }_d^{L+d}\frac{x^{-2}dx}{4\pi {ϵ}_0}=\frac{qQ}{4\pi {ϵ}_{0}L}{\left[-\frac{1}{x}\right]}_d^{L+d}$
$=\frac{qQ}{4\pi {ϵ}_{0}L}\left[\frac{1}{d}-\frac{1}{L+d}\right]$
$=\frac{qQ}{4\pi {ϵ}_0}\left[\frac{1}{(d)(L+d))}\right]$