Electrostatics Question 419

Question: Three charges $ Q,+q $ and $ +q $ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if Q is equal to [IIT-JEE (Screening) 2000]

Options:

A) $ \frac{-q}{1+\sqrt{2}} $

B) $ \frac{-2q}{2+\sqrt{2}} $

C) $ -2q $

D) $ +q $

Show Answer

Answer:

Correct Answer: B

Solution:

Net electrostatic energy $ U=\frac{kQq}{a}+\frac{kq^{2}}{a}+\frac{kQq}{a\sqrt{2}}=0 $

$ \Rightarrow \frac{kq}{a}( Q+q+\frac{Q}{\sqrt{2}} )=0 $

therefore $ Q=-\frac{2q}{2+\sqrt{2}} $



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