Question

A point charge $q$ is situated at a distance $r$ on axis from one end of a thin conducting rod of length $L$ having a charge $Q$ [uniformly distributed along its length]. The magnitude of electric force between the two is _____

• A. $\frac{KQq}{r^2}$
• B. $\frac{2KQ}{r(r+L)}$
• C. $\frac{KQq}{r(r-L)}$
• D. $\frac{KQq}{r(r+L)}$

Answer 1

The correct option is D $\frac{KQq}{r(r+L)}$

$\text{Step 1: Charge on the small element}$

Let $\lambda =\frac{Q}{L}$ be the linear charge density of rod.

As shown in figure, Consider an elementary charge $dq$ having length $dx$ at a distance $x$ from the charge $q$, then
$dq=\lambda dx=\frac{Q}{L}dx$

$\text{Step 2: Force due to small element}$

By Coulomb's Law:
$dF=\frac{1}{4\pi {ϵ}_0}\frac{qdq}{x^2}=K\frac{qQdx}{L{x}^2}$ $....\left(1\right)$

$\text{Step 3: Integrate above to get the Net force}$
$\therefore$ Total force, $F=\int dF=\frac{KQq}{L}{\int }_r^{r+L}{x}^{-2}dx$

$\Rightarrow F=\frac{KQq}{L}{[-\frac{1}{x}]}_r^{r+L}$ $=\frac{-KQq}{L}[\frac{1}{r+L}-\frac{1}{r}]$

$\Rightarrow F=\frac{-KQq}{L}\left[\frac{r-r-L}{(r+L)r}\right]=\frac{KQq}{r(r+L)}$

Hence, option $\left(D\right)$ is correct.