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

Show Answer

Answer:

Correct Answer: 5. (d)

Solution:

  1. 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, $\quad U=0$

$\therefore \quad \frac{K q}{a} q+Q+\frac{Q}{\sqrt{2}}=0$

$$ \Rightarrow \quad Q=\frac{-\sqrt{2} \times q}{(\sqrt{2}+1)} $$



NCERT Chapter Video Solution

Dual Pane