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: Understanding the Concept:
The electrostatic potential energy of a set of point charges is the sum, over every pair, of \(\dfrac{1}{4\pi\epsilon_0}\dfrac{q_iq_j}{r_{ij}}\).

Step 2: Key Formula or Approach:
In the figure, the charge \(Q\) is at the right-angle corner. The charge \(2q\) is on one leg at distance \(\sqrt2a\), and the charge \(q\) is on the other leg, also at distance \(\sqrt2a\). The two legs are equal, so the distance between \(2q\) and \(q\) is the hypotenuse, \(\sqrt{2}\times\sqrt2a = 2a\).

Step 3: Detailed Explanation:
\[ U = k\left[\frac{Q(2q)}{\sqrt2a} + \frac{Qq}{\sqrt2a} + \frac{(2q)(q)}{2a}\right] \]
\[ U = \frac{k}{a}\left[\frac{3Qq}{\sqrt2} + q^2\right] \]
Set \(U = 0\):
\[ \frac{3Qq}{\sqrt2} = -q^2 \Rightarrow Q = -\frac{\sqrt2}{3}\,q \]
The sign is negative, so \(Q\) attracts the other two charges and balances the repulsion between \(2q\) and \(q\). Option (A) \(-\tfrac{1}{2\sqrt3}q\) and (C) \(-\tfrac{\sqrt3}{2}q\) do not satisfy the equation, and (D) \(-\sqrt{\tfrac23}q\) is larger in magnitude by a factor \(\sqrt3\).

Final Answer:
\(Q = -\dfrac{\sqrt2}{3}q\), option (B). \[ \boxed{-\frac{\sqrt2}{3}q \text{ (B)}} \]