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 $
BETA
