An electric charge 10⁻⁶ is placed at origin (0, 0) m of X - Y co-ordinate system. Two points P and Q are situated at (√(3), √(3)) m and (√(6), 0) m respectively. The potential difference between the points P and Q will be: (a) √(3) V (b) √(6) V (c) 0 V (d) 3 V

Answer: (c). V = kq/r depends only on distance from the charge. rP = √(3+3) = √(6) m, rQ = √(6) m rP = rQ ⇒ VP - VQ = 0 So the answer is (c) 0 V.

An electric charge 10⁻⁶ is placed at origin (0, 0) m of X - Y… | JEE Main 2024 (27 Jan, Shift 1)

Q41JEE Main 2024 · 27 JAN · SHIFT 1Easy

Electric Potential & Equipotentials

An electric charge $10^{-6}\,\mu\text{C}$ is placed at origin $(0, 0)\text{ m}$ of $X - Y$ co-ordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3})\text{ m}$ and $(\sqrt{6}, 0)\text{ m}$ respectively. The potential difference between the points $P$ and $Q$ will be:

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Answer: (c)

$V = \frac{kq}{r}$ depends only on distance from the charge.
$r_P = \sqrt{3+3} = \sqrt{6}\text{ m}$ , $r_Q = \sqrt{6}\text{ m}$
$r_P = r_Q \Rightarrow V_P - V_Q = 0$
So the answer is (c) $0\text{ V}$ .

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