Electrostatics 2 Question 13
14. Three charges $Q,+q$ and $+q$ are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if $Q$ is equal to
$(2000,2 M)$
(a) $\frac{-q}{1+\sqrt{2}}$
(b) $\frac{-2 q}{2+\sqrt{2}}$
(c) $-2 q$
(d) $+q$
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Answer:
Correct Answer: 14. (a, d)
Solution:
- Net electrostatic energy of the configuration will be
$$ U=K \frac{q \cdot q}{a}+\frac{Q \cdot q}{\sqrt{2} a}+\frac{Q \cdot q}{a} $$
Here, $K=\frac{1}{4 \pi \varepsilon _0}$
Putting $\quad U=0$ we get, $Q=\frac{-2 q}{2+\sqrt{2}}$
