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}} $
BETA
