Q. The potential energy of 4-particalse each of mass 1 kg placed at the four vertices of a square of side length 1 m is

A) $+ 4.0\ G$
B) $- 7.5\ G$
C) $- 5.4\ G$
D) $+ 6.3\ G$
Explanation

(c) $Here,\ \ AB = BC = CD = DA = 1m$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ BD = AC$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{1^{2} + 1^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{2}m$
Total potential energy
$\ \ U = \left\lbrack \frac{- G \times 1 \times 1}{AB} \right\rbrack + \left\lbrack \frac{- G \times 1 \times 1}{BC} \right\rbrack + \left\lbrack \frac{- G \times 1 \times 1}{CD} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \left\lbrack \frac{- G \times 1 \times 1}{DA} \right\rbrack + \left\lbrack \frac{- G \times 1 \times 1}{BD} \right\rbrack + \left\lbrack \frac{- G \times 1 \times 1}{AC} \right\rbrack$
$\ \ \ \ \ = 4 \times \left\lbrack \frac{- G \times 1 \times 1}{1} \right\rbrack + 2\left\lbrack \frac{- G \times 1 \times 1}{\sqrt{2}} \right\rbrack = - 5.4G$
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Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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

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