Electrostatics 2 Question 5
5. Three charges $Q,+q$ and $+q$ are placed at the vertices of a right angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $Q$ is (Main 2019, 11 Jan I)
(a) $-2 q$
(b) $\frac{-q}{1+\sqrt{2}}$
(c) $+q$
(d) $\frac{-\sqrt{2} q}{\sqrt{2}+1}$
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Answer:
Correct Answer: 5. (d)
Solution:
- Electrostatic energy between two charges $q_{1}$ and $q_{2}$ such that the distance between them $r$ is given as
$$ U=\frac{K q_{1} q_{2}}{r} $$
In accordance to the principle of superposition, total energy of the charge system as shown in the figure below is
$$ U=\frac{K q^{2}}{a}+\frac{K Q q}{a}+\frac{K Q q}{\sqrt{2} a} $$
It is given that, $\quad U=0$
$\therefore \quad \frac{K q}{a} q+Q+\frac{Q}{\sqrt{2}}=0$
$$ \Rightarrow \quad Q=\frac{-\sqrt{2} \times q}{(\sqrt{2}+1)} $$
