Electrostatics Question 749

Question: Charge +q and -q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is

Options:

A) $ \frac{qQ}{2\pi {\varepsilon _{0}}L} $

B) $ \frac{qQ}{3\pi {\varepsilon _{0}}L} $

C) $ -\frac{qQ}{6\pi {\varepsilon _{0}}L} $

D) $ \frac{qQ}{4\pi {\varepsilon _{0}}L} $

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Potential at $ C=V _{C}=0 $

Potential at $ D=V _{D} $

$ =k( \frac{-q}{L} )+\frac{kq}{3L} $

$ =-\frac{2}{3}\frac{kq}{L} $

Potential difference $ V _{D}-V _{C}=\frac{-2}{3}\frac{kq}{L}=\frac{1}{4\pi {\varepsilon _{0}}}( -\frac{2}{3}.\frac{q}{L} ) $

$ \Rightarrow \text{ Work done}=Q( V _{D}-V _{C} ) $

$ =-\frac{2}{3}\times \frac{1}{4\pi {\varepsilon _{0}}}\frac{qQ}{L}=\frac{-qQ}{6\pi {\varepsilon _{0}}L} $



NCERT Chapter Video Solution

Dual Pane