Question

Three point charges Q, \(+2q\) and \(+q\) are placed at the vertices of a right-angled isosceles triangle of length \(\sqrt{2}a\) as shown in figure. The net electrostatic potential energy of the configuration is zero, if Q is equal to

• A. \(-\frac{1}{2\sqrt{3}}q\)
• B. \(-\frac{\sqrt{2}}{3}q\)
• C. \(-\frac{\sqrt{3}}{2}q\)
• D. \(-\sqrt{\frac{2}{3}}\,q\)

Hint:

Add the potential energy of each of the three pairs: Q with 2q, Q with q, and 2q with q. Set the sum to zero.

Answer 1

The Correct Option is B

Step 1: Pair by pair:
Distances: $Q$ to $2q$ is $\sqrt2a$. $Q$ to $q$ is $\sqrt2a$. $2q$ to $q$ is $2a$ (the hypotenuse of the right isosceles triangle with equal legs $\sqrt2a$).

Step 2: Write energy in units of $\frac{q}{4\pi\epsilon_0 a}$:
$Q$-$2q$ pair: $\frac{2Q}{\sqrt2} = \sqrt2Q$.
$Q$-$q$ pair: $\frac{Q}{\sqrt2}$.
$2q$-$q$ pair: $\frac{2q^2}{2a}\to q$ (in the same units, the term is $q$).

Step 3: Zero total:
$\sqrt2Q + \frac{Q}{\sqrt2} + q = 0$, so $Q\cdot\frac{3}{\sqrt2} = -q$ and $Q = -\frac{\sqrt2}{3}q$.

Final Answer:
Option (B). \[ \boxed{-\frac{\sqrt2}{3}q \text{ (B)}} \]