Question

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

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

Answer 1

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

Electrostatic energy of the configuration
$U=K.\frac{Qq}{a}+\frac{K.q^2}{a}+K.\frac{Qq}{a\sqrt{2}}$

Now, if $U=0$
$Qq+q^2+\frac{Qq}{\sqrt{2}}=0$
$Q+q+\frac{Q}{\sqrt{2}}=0$
$Q=\frac{-\sqrt{2}q}{\sqrt{2}+1}$