Q.
Three charges $\mathcal{Q}_{0}, - q$ and $–q$ are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if $\mathcal{Q}_{0}$is equal to
A) $\frac{q}{4}$
B) $\frac{2q}{\sqrt{32}}$
C) $\sqrt{2q}$
D) +q
Explanation
(b) Here total electrostatic potential energy is zero
$ie,\ \ \frac{- \mathcal{Q}_{0}q}{l} - \frac{q\mathcal{Q}_{0}}{l} + \frac{q^{2}}{\sqrt{2l}} = 0$
On solving,
$\mathcal{Q}_{0} = \frac{q}{2\sqrt{2}} = \frac{2q}{\sqrt{32}}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.