Electrostatics Question 440
Question: Three charges $ Q,(+q) $ and $ (+q) $ are placed at the vertices of an equilateral triangle of side l . If the net electrostatic energy of the system is zero, then Q is equal to [MP PET 2001]
Options:
A) $ ( -\frac{q}{2} ) $
B) $ (-q) $
C) $ (+q) $
D) Zero
Show Answer
Answer:
Correct Answer: A
Solution:
Potential energy of the system $ U=k\frac{Qq}{l}+\frac{kq^{2}}{l}+\frac{kqQ}{l}=0 $
therefore $ \frac{kq}{l}(Q+q+Q)=0 $
therefore $ Q=-\frac{q}{2} $
