Q. Tow positive charges of magnitude$\ q$ are placed at the ends of a side1 of a square of side 2$a$. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge $\mathcal{Q}$ moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is
A) $\frac{1}{4\pi\varepsilon_{0}}\frac{2q\mathcal{Q}}{a}\left( 1 - \frac{1}{\sqrt{5}} \right)$
B) Zero
C) $\frac{1}{4\pi\varepsilon_{0}}\frac{2q\mathcal{Q}}{a}\left( 1 + \frac{1}{\sqrt{5}} \right)$
D) $\frac{1}{4\pi\varepsilon_{0}}\frac{2q\mathcal{Q}}{a}\left( 1 - \frac{2}{\sqrt{5}} \right)$
Explanation
(a) $U_{i} = \frac{2kq\mathcal{Q}}{a} + \frac{2k( - q)\mathcal{Q}}{\sqrt{5a}} = \frac{1}{4\pi\varepsilon_{0}\ }.\frac{2q\mathcal{Q}}{a}(1 - \frac{1}{\sqrt{5}})$
$U_{f} = 0$
[image:1_image278.jpeg]
By conservation of energy
Gain in KE=loss in PE
$K = \frac{1}{4\pi\varepsilon_{0}}.\frac{2q\mathcal{Q}}{a}(1 - \frac{1}{\sqrt{5}})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.