Question

Three charges $Q,q,-q$ are placed at the vertices of right angled isosceles triangle as shown in the figure. The net electrostatic energy of the configuration is zero if $Q$ is equal to :

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

Answer 1

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

The electrostatic energy $U$ of the system is given by:

$U=\frac{1}{4\pi {ϵ}_0}(\frac{Qq}{a}+\frac{q^2}{a}+\frac{Qq}{\sqrt{2a}})$

When $U=0$, then:

$\frac{Qq}{a}+\frac{q^2}{a}+\frac{Qq}{\sqrt{2a}}=0$

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