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  Solved  Paper (Held On 11 May  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 :

                                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 theroem, \[{{W}_{AB}}{{)}_{electric}}=Q({{V}_{A}}-{{V}_{B}})\] \[=\frac{2KqQ}{a}\left[ 1-\frac{1}{\sqrt{5}} \right]\]\[=\left( \frac{1}{4\pi {{\in }_{0}}} \right)\frac{2Qq}{a}\left[ 1-\frac{1}{\sqrt{5}} \right]\]