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

Answer 1

Answer: (B)

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

Answer 2

Potential energy of the configuration $U = k.\frac{Qq}{a} + \frac{k.q^{2}}{a} + k.\frac{Qq}{a\sqrt{2}} = 0$ ⇒ $Q = \frac{- \sqrt{2}q}{\sqrt{2} + 1}$