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, \[dF=\frac{q\lambda dr}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[F=\int_{{}}^{{}}{dF}=\int\limits_{R}^{R+L}{\frac{q\lambda dr}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}=\frac{q\lambda L}{4\pi {{\varepsilon }_{0}}R(R+L)}}\]You need to login to perform this action.
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