Physics · Electrostatics

JEE Main 2025 — 7 April, Morning Shift — Question 61

Two charges $q_{1}$ and $q_{2}$ are separated by a distance of 30 cm . A third charge $q_{3}$ initially at ' $C$ ' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of $q_{3}$ from $C$ to $D$ is given by $\frac{q_{3} K}{4 \pi \varepsilon_{0}}$ the value of $K$ is

Option A:
$8 q_{1}$
Option B:
$8 q_{2}$
Correct
Option C:
$6 q_{2}$
Option D:
$6 q_{1}$

Answer: B

Step-by-step solution

$u(C)=\left(\frac{k q_{1}}{40}+\frac{k q_{2}}{50}\right) q_{3}$

$u(D)=\left(\frac{k q_{1}}{40}+\frac{k q_{2}}{10}\right) q_{3}$

So, $\Delta u=|u(D)-u(C)|=k q_{2}\left[\frac{1}{10}-\frac{1}{50}\right] q_{3}$

$\Rightarrow \quad \Delta u=\frac{k q_{2} 4}{50} q_{3}=\frac{4 q_{2} q_{3}}{4 \pi \varepsilon_{0} 50}$

$\Rightarrow\left(\frac{4 q_{2} q_{3} \times 2}{4 \pi \varepsilon_{0}}\right)$ SI unit

$\Leftrightarrow \frac{q_{3} k}{4 \pi \varepsilon_{0}}$ $\Rightarrow k=8 q_{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential