Question

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 :

• 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. $\frac{1}{4\pi\varepsilon_{0}} \frac{2qQ}{a}\left(1-\frac{1}{\sqrt{5}}\right)$

Answer 1

Answer: (D)

$\frac{1}{4\pi\varepsilon_{0}} \frac{2qQ}{a}\left(1-\frac{1}{\sqrt{5}}\right)$

Answer 2

Potential at point A, $V_{A} = \frac{2kq}{a}- \frac{2ka}{a\sqrt{5}}$ Potential at point B, $V_{B} = 0$ $\therefore\quad$ Using work energy theroem $W_{AB )\,electric} = Q\left(V_{A} - V_{B}\right)$ $=\quad \frac{2KqQ}{a}\left[1-\frac{1}{\sqrt{5}}\right]\quad\quad=\quad\left( \frac{1}{4\pi\varepsilon_{0}}\right) \frac{2Qq}{a}\left[1-\frac{1}{\sqrt{5}}\right]$