Q132 / 150 · MHT CET 2026 (17 April Shift 2) · Physics
Three point charges +q, +2q and +Q are placed at the vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of Q in terms of q is
- A$-\dfrac{q}{3}$
- Correct answerB$\dfrac{-2q}{3}$
- C$\dfrac{q}{2}$
- D$\dfrac{3q}{2}$
Solution
Let the side length of the equilateral triangle be $a$.
The total potential energy of the system of three charges is the sum of the potential energies of all pairs of charges:
$U = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q \cdot 2q}{a} + \dfrac{2q \cdot Q}{a} + \dfrac{Q \cdot q}{a} \right)$
Given that the potential energy of the system is zero, $U = 0$.
$\dfrac{1}{4\pi\epsilon_0 a} (2q^2 + 2qQ + qQ) = 0$
$2q^2 + 3qQ = 0$
$q(2q + 3Q) = 0$
Since $q \neq 0$, we have:
$2q + 3Q = 0$
$Q = -\dfrac{2q}{3}$
Answer: $\dfrac{-2q}{3}$
The total potential energy of the system of three charges is the sum of the potential energies of all pairs of charges:
$U = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q \cdot 2q}{a} + \dfrac{2q \cdot Q}{a} + \dfrac{Q \cdot q}{a} \right)$
Given that the potential energy of the system is zero, $U = 0$.
$\dfrac{1}{4\pi\epsilon_0 a} (2q^2 + 2qQ + qQ) = 0$
$2q^2 + 3qQ = 0$
$q(2q + 3Q) = 0$
Since $q \neq 0$, we have:
$2q + 3Q = 0$
$Q = -\dfrac{2q}{3}$
Answer: $\dfrac{-2q}{3}$
Hence, the correct answer is option: 'B'.