Question

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

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

Hint:

- Always sum pairwise energies - Use geometry carefully (hypotenuse = $\sqrt{2}l$)

Answer 1

The Correct Option is C

Concept:
Electrostatic potential energy of system: \[ U = \sum \frac{k q_i q_j}{r_{ij}} \]

Step 1: Geometry of triangle
Given isosceles right triangle: \[ \text{legs} = l,\quad \text{hypotenuse} = \sqrt{2}l \]

Step 2: Write pairwise energies

• Between \( Q \) and first \( -2q \) (distance \( l \)): \[ U_1 = \frac{k(Q)(-2q)}{l} = -\frac{2kQq}{l} \]
• Between \( Q \) and second \( -2q \) (distance \( l \)): \[ U_2 = -\frac{2kQq}{l} \]
• Between two \( -2q \) charges (distance \( \sqrt{2}l \)): \[ U_3 = \frac{k(-2q)(-2q)}{\sqrt{2}l} = \frac{4kq^2}{\sqrt{2}l} = \frac{2\sqrt{2}kq^2}{l} \]

Step 3: Total energy = 0
\[ U_1 + U_2 + U_3 = 0 \] \[ -\frac{2kQq}{l} - \frac{2kQq}{l} + \frac{2\sqrt{2}kq^2}{l} = 0 \]

Step 4: Simplify
\[ -\frac{4kQq}{l} + \frac{2\sqrt{2}kq^2}{l} = 0 \] Divide by \( \frac{kq}{l} \): \[ -4Q + 2\sqrt{2}q = 0 \]

Step 5: Solve for \( Q \)
\[ 4Q = 2\sqrt{2}q \Rightarrow Q = \frac{\sqrt{2}}{2}q = \frac{q}{\sqrt{2}} \]