question_answer

Two positive charges of magnitude 'q' are placed at the ends of a side (side 1) of a square of side '2a'. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge Q moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is : [AIEEE 11-05-2011]

A) zero                              

B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{a}\left( 1+\frac{1}{\sqrt{5}} \right)\]

C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{a}\left( 1-\frac{2}{\sqrt{5}} \right)\]

D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2qQ}{a}\left( 1-\frac{1}{\sqrt{5}} \right)\]

Correct Answer: D

Solution :

[d] Potential at point A, \[{{V}_{A}}=\frac{2Kq}{a}-\frac{2Kg}{a\sqrt{5}}\]

Potential at point B,\[{{V}_{B}}=0\]

\[\therefore \]Using work energy theorem,

\[{{W}_{AB}}{{)}_{electric}}=Q({{V}_{A}}-{{V}_{B}})\]

\[=\frac{2KqQ}{a}\left[ 1-\frac{1}{\sqrt{5}} \right]\]\[=\left( \frac{1}{4\pi {{\in }_{0}}} \right)\frac{2Qq}{a}\left[ 1-\frac{1}{\sqrt{5}} \right]\]