Question

Three charges $Q,+q\text{and}+q$ 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:

• A. $+q$
• B. $\frac{-\sqrt{2}q}{\sqrt{2}+1}$
• C. $\frac{-q}{1+\sqrt{2}}$
• D. $-2q$

Answer 1

The correct option is B $\frac{-\sqrt{2}q}{\sqrt{2}+1}$


Net electrostatic energy for the system,

$U=K[\frac{q^2}{a}+\frac{Qq}{a}+\frac{Qq}{a\sqrt{2}}]=0$

$\Rightarrow \frac{Kq}{a}[q+Q+\frac{Q}{\sqrt{2}}]=0$

$\Rightarrow q=-Q[1+\frac{1}{\sqrt{2}}]$

$\Rightarrow Q=\frac{-q\sqrt{2}}{\sqrt{2}+1}$

Hence, option $\left(B\right)$ is correct.