Question

Three charges $Q$, $+ q$ and $+ q$ are placed at the vertices of a right-angled isosceles triangle as shown. 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. $+q$

Answer 1

Answer: (B)

Net electrostatic energy = 0
$\therefore \frac{Qq}{a}+\frac{q^{2}}{a}+\frac{Qq}{a\sqrt{2}}=0$
or $Q\sqrt{2}+q\sqrt{2}+Q=0$
or $Q\left(\sqrt{2}+1\right)=-q\sqrt{2}$
or $Q=\frac{-2q}{\left(2+\sqrt{2}\right)}$.