JEE Main 2025 Apr 7 Shift 1, Physics Q20: Electric Potential

JEE Main2025Apr 7, Shift 1Physics

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 $\frac{q _{3} K}{4 π ε _{0}}$ , the value of $K$ is :

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

Solution

The circular arc has radius 40 cm centred at A, so $q_{3}$ is at distance 40 cm from $q_{1}$ throughout the motion; the contribution from $q_{1}$ is unchanged. Only the contribution from $q_{2}$ changes.

Distance from $q_{2}$ (at B) to $q_{3}$ :

At C: $B C = A C^{2} + A B^{2} = 4 0^{2} + 3 0^{2} = 50$ cm $= 0.50$ m.
At D: $B D = A D - A B = 40 - 30 = 10$ cm $= 0.10$ m.

Potentials at C and D from all sources: (only the $q_{2}$ -terms differ)

$V_{C} = \frac{k q _{1}}{0.40} + \frac{k q _{2}}{0.50}$

$V_{D} = \frac{k q _{1}}{0.40} + \frac{k q _{2}}{0.10}$

Change in potential energy of $q_{3}$ :

$Δ U = q_{3} (V_{D} - V_{C}) = q_{3} k q_{2} (\frac{1}{0.10} - \frac{1}{0.50}) = q_{3} \cdot k \cdot q_{2} \cdot (10 - 2) = 8 k q_{2} q_{3}$

Writing $k = \frac{1}{4 π ε _{0}}$ :

$Δ U = \frac{8 q _{2} q _{3}}{4 π ε _{0}}$

Matching with $\frac{q _{3} K}{4 π ε _{0}}$ : $K = 8 q_{2}$ .

$∴$ Option (1).

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JEE Main 2025 Apr 7 Shift 1, Physics Q20: Electric Potential

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