Question

Three charges $Q,+q$ and $+q$ are placed at the vertices of a right angle triangle (isosceles triangle) as shown. If the net electrostatic potential energy of the configuration is zero, value of $Q$ is:

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

Answer 1

Answer: (D)

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

We know that the electrostatic potential energy of a two charge system separated by a distance r is given by,

$E=\frac{k{q}_1{q}_2}{r}$

Hence, for the configuration in figure to have zero energy,

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

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

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