question_answer

A charge is distributed with a linear density \[\lambda \] over the length L 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) \[\frac{q\lambda L}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\]        

B) \[\frac{q\lambda L}{4\pi {{\varepsilon }_{0}}R(R+L)}\]

C) \[\frac{q\lambda L}{4\pi {{\varepsilon }_{0}}RL}\]         

D) \[\frac{q\lambda L}{4\pi {{\varepsilon }_{0}}{{L}^{2}}}\]

Correct Answer: B

Solution :

[b]: As coulombs 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, \[dF=\frac{q\lambda dr}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[F=\int_{{}}^{{}}{dF}=\int\limits_{R}^{R+L}{\frac{q\lambda dr}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}=\frac{q\lambda L}{4\pi {{\varepsilon }_{0}}R(R+L)}}\]