Q151 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
Four point charges $-Q$, $-q$, $2q$ and $3Q$ are placed at corners of the square respectively. The relation between $Q$ and $q$ for which the potential at the centre of the square is zero, is
- A$Q = q$
- Correct answerB$Q = -q/2$
- C$Q = -1/q$
- D$Q = 1/2q$
Solution
Let the side length of the square be $a$. The distance of the centre from each corner is $r = \dfrac{a}{\sqrt{2}}$.
The electric potential at the centre of the square is the algebraic sum of the potentials due to all four charges.
$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{-Q}{r} + \dfrac{-q}{r} + \dfrac{2q}{r} + \dfrac{3Q}{r} \right)$
Given that $V = 0$, we have:
$-Q - q + 2q + 3Q = 0$
$2Q + q = 0$
$Q = -\dfrac{q}{2}$
Answer: $Q = -q/2$
The electric potential at the centre of the square is the algebraic sum of the potentials due to all four charges.
$V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{-Q}{r} + \dfrac{-q}{r} + \dfrac{2q}{r} + \dfrac{3Q}{r} \right)$
Given that $V = 0$, we have:
$-Q - q + 2q + 3Q = 0$
$2Q + q = 0$
$Q = -\dfrac{q}{2}$
Answer: $Q = -q/2$
Hence, the correct answer is option: 'B'.