Electrostatics Question 355

Question: Three particles, each having a charge of $ 10\mu C $ are placed at the corners of an equilateral triangle of side $ 10cm $ . The electrostatic potential energy of the system is (Given $ \frac{1}{4\pi {\varepsilon _{0}}}=9\times 10^{9}N-m^{2}/C^{2} $ ) [MP PMT 1994]

Options:

A) Zero

B) Infinite

C) $ 27J $

D) $ 100J $

Show Answer

Answer:

Correct Answer: C

Solution:

For pair of charge $ U=\frac{1}{4\pi {\varepsilon _{0}}}.\frac{q _{1}q _{2}}{r} $

$ U _{System}=\frac{1}{4\pi {\varepsilon _{0}}}[ \frac{10\times {{10}^{-6}}\times 10\times {{10}^{-6}}}{10/100} . $

$ . +\frac{10\times {{10}^{-6}}\times 10\times {{10}^{-6}}}{10/100}+\frac{10\times {{10}^{-6}}\times 10\times {{10}^{-6}}}{10/100} ] $

$ =3\times 9\times 10^{9}\times \frac{100\times {{10}^{-12}}\times 100}{10}=27J $



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