Q. Work required to set up the four charge configuration (as shown in the figure) is

A) $- {0.21q}^{2}/\varepsilon_{0}a$
B) $- 1.29q^{2}/\varepsilon_{0}a$
C) $- 1.41q^{2}/\varepsilon_{0}a$
D) $+ 2.82q^{2}/\varepsilon_{0}a$
Explanation

(a) Work is required to set up the four charge configuration
$q_{1} = + q,\ q_{2} = - q,\ q_{3} = + q\ \ and\ q_{4} = - q$
[image:2_image148.emf]
$W = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{( + q)( - q)}{AB} + \frac{( - q)( + q)}{BC} + \frac{( + q)( - q)}{CD} + \frac{( - q)( + q)}{DA} + \frac{( + q)( + q)}{AC} + \frac{( - q)( - q)}{BD} \right\rbrack$
$W = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack - \frac{q^{2}}{a} - \frac{q^{2}}{a} - \frac{q^{2}}{a} - \frac{q^{2}}{a} + \frac{q^{2}}{a\sqrt{2}} + \frac{q^{2}}{a\sqrt{2}} \right\rbrack$
$W = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{a}\left\lbrack - 4 + \sqrt{2} \right\rbrack = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{a}\lbrack - 4 + 1.414\rbrack$
$W = - 0.21 \times \frac{q^{2}}{\varepsilon_{0}a}\ \ \ \ \ \ \ \ (approx.)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.