Question

Two thin rods of length L lie along x-axis, one between $x=\frac{a}{2}tox=\frac{a}{2}+L$ and the other between $x=-\frac{a}{2}tox=-\frac{a}{2}-L$.
Each rod has positive charge Q distributed uniformly along the length. Find the magnitude of the force which one rod exerts on the other.

• A. $\frac{Q^2}{4\pi {\epsilon }_0{L}^2}lo{g}_e\frac{L+a}{L-a}$
• B. $\frac{Q^2}{4\pi {\epsilon }_0{L}^2}lo{g}_e\frac{(L+a{)}^2}{a(L-a)}$
• C. $\frac{Q^2}{4\pi {\epsilon }_0{L}^2}lo{g}_e\frac{(L+a{)}^2}{a(2L+a)}$
• D. $\frac{Q^2}{4\pi {\epsilon }_0{L}^2}lo{g}_e\frac{(L+a{)}^2}{L(2a+L)}$

Answer 1

Answer: (C)

$\frac{Q^2}{4\pi {\epsilon }_0{L}^2}lo{g}_e\frac{(L+a{)}^2}{a(2L+a)}$

Electric field at point P due to a small element dx of the rod on left side is $\int dE=\int \frac{Qdx}{L4\pi {\epsilon }_0{x}^2}$
$=\frac{Q}{4\pi {\epsilon }_{0}L}\times [\frac{1}{x+\frac{a}{2}}-\frac{1}{x+\frac{a}{2}+L}]$
Force exerted on a small element dx is
$dF=\frac{QQ}{4\pi {\epsilon }_0{L}^2}[\frac{1}{x+\frac{a}{2}}-\frac{1}{x+\frac{a}{2}+L}]dx$ or
$F=\frac{Q^2}{4\pi {\epsilon }_0{L}^2}[lo{g}_{e}x+\frac{a}{2}{\int }_{a/2}^{L+a/2}-lo{g}_{e}x+\frac{a}{2}+L{\int }_{a/2}^{L+a/2}]$

$=\frac{Q^2}{4\pi {\epsilon }_0{L}^2}[lo{g}_e\frac{L+a}{a}-log\frac{2L+a}{a}]$

$=\frac{Q^2}{4\pi {\epsilon }_0{L}^2}lo{g}_e\left[\frac{(L+a{)}^2}{a(2L+a)}\right]$