Question

A charge is distributed with a linear density $\lambda$ over a rod of the length $L$ placed along radius vector drawn from the point where a point charge $q$ is located. The distance between $q$ and the nearest point on linear charge is $R$. The electrical force experienced by the linear charge due to $q$ is

• A. $\frac{q\lambda L}{4\pi {\epsilon }_0{R}^2}$
• B. $\frac{q\lambda L}{4\pi {\epsilon }_{0}R(R+L)}$
• C. $\frac{q\lambda L}{4\pi {\epsilon }_{0}RL}$
• D. $\frac{q\lambda L}{4\pi {\epsilon }_0{L}^2}$

Answer 1

The correct option is B $\frac{q\lambda L}{4\pi {\epsilon }_{0}R(R+L)}$

As coulomb's force is an action reaction pair, so force experienced by the linear charge is equal and opposite to the force experienced by point charge $q$. Here, we are computing electric force experienced by $q$ due to the charge.

Considered a small element on line charge as shown, then force experienced by $q$ due to this element is,

$F=\int dF={\int }_R^{R+L}\frac{q\lambda dr}{4\pi {\epsilon }_0{r}^2}=\frac{q\lambda L}{4\pi {\epsilon }_{0}R(R+L)}$

$dF=\frac{q\lambda dr}{4\pi {\epsilon }_0{r}^2}$

On integrating between the given limits, we will get

$F=\frac{q\lambda L}{4\pi {\epsilon }_{0}R(R+L)}$