Three point charges Q, +2q and +q are placed at the vertices of a right-angled isosceles triangle of length √(2)a as shown in figure. The net electrostatic potential energy of the configuration is zero, if Q is equal to (a) -(1)/(2√(3))q (b) -√(2)/3q (c) -√(3)/2q (d) -√(2/3) q

Answer: (b). Q is at the right angle: legs √2 a, hypotenuse 2a. U = k[(2qQ)/(√2 a) + (qQ)/(√2 a) + (2q²)/(2a)] = 0 (3Q)/(√2) + q = 0 ⇒ Q = -(√2)/(3)q So the answer is (b).

Three point charges Q, +2q and +q are placed at the vertices of a… | MHT CET 2026 (18 Apr, Shift 2)

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Electrostatic Potential Energy

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

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Answer: (b)

Q is at the right angle: legs $\sqrt2 a$ , hypotenuse $2a$ .
$U = k\left[\dfrac{2qQ}{\sqrt2 a} + \dfrac{qQ}{\sqrt2 a} + \dfrac{2q^2}{2a}\right] = 0$
$\dfrac{3Q}{\sqrt2} + q = 0 \Rightarrow Q = -\dfrac{\sqrt2}{3}q$
So the answer is (b).

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Electrostatic Potential Energy question from MHT CET 2026 (18 Apr, Shift 2)

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