Physics Q38 – JEE-MAIN 2025

Home Physics Electric Charges and FieldsQuestion #116

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:

Clue 1 of 3: Potential Energy Difference

The change in potential energy of a charge $q_{3}$ moved between two points is given by $q_{3}$ times the potential difference between those points.

🥷

Ninja Strategy•Symmetry and Cancellation

Recognize that if the circular path is centered at $q_{1}$ , the potential contribution from $q_{1}$ at points C and D will be identical, leading to its cancellation in the potential difference calculation. This immediately eliminates options solely dependent on $q_{1}$ and simplifies the calculation to only consider $q_{2}$ .

🧠 Full Solution Path

Step 1: Determine initial and final positions and distances✦ Active

Let $q_{1}$ be at point A and $q_{2}$ be at point B. The distance AB is 30 cm. Point C is initially at $(0, 40)$ cm if A is at the origin $(0, 0)$ cm. The problem states that $q_{3}$ moves along a circular path of radius 40 cm from C to D. Since C is 40 cm from A, it implies that A is the center of this circular path. Therefore, the distance from $q_{1}$ to C ( $r_{1 C}$ ) is 40 cm, and the distance from $q_{1}$ to D ( $r_{1 D}$ ) is also 40 cm.

For point C $(0, 40)$ cm:

r_{1 C} = 40 cm = 0.4 m

r_{2 C} = \sqrt{(30 cm)^{2} + (40 cm)^{2}} = \sqrt{900 + 1600} = \sqrt{2500} = 50 cm = 0.5 m

For point D, assuming it's on the x-axis (e.g., $(40, 0)$ cm) as a common quarter-circle path from C, and still 40 cm from A:

r_{1 D} = 40 cm = 0.4 m

r_{2 D} = \sqrt{(40 cm - 30 cm)^{2} + (0 cm)^{2}} = \sqrt{10^{2}} = 10 cm = 0.1 m

💡 Teacher's Secret Hint

The phrase 'circular path of radius 40 cm' is crucial. If C is 40 cm from A, then A is the center of the circle, making $r_{1 C} = r_{1 D} = 40$ cm.

Step 2: Calculate the potential at points C and D○ Expand

The electrostatic potential at a point due to multiple charges is the sum of potentials due to individual charges. The potential at C ( $V_{C}$ ) and D ( $V_{D}$ ) are:

V_{C} = \frac{1}{4 π ϵ_{0}} (\frac{q_{1}}{r_{1 C}} + \frac{q_{2}}{r_{2 C}}) = \frac{1}{4 π ϵ_{0}} (\frac{q_{1}}{0.4} + \frac{q_{2}}{0.5})

V_{D} = \frac{1}{4 π ϵ_{0}} (\frac{q_{1}}{r_{1 D}} + \frac{q_{2}}{r_{2 D}}) = \frac{1}{4 π ϵ_{0}} (\frac{q_{1}}{0.4} + \frac{q_{2}}{0.1})

Step 3: Calculate the potential energy difference and find K○ Expand

The difference in potential energy $Δ U$ when charge $q_{3}$ moves from C to D is $q_{3} (V_{D} - V_{C})$ .

Δ U = q_{3} (V_{D} - V_{C}) = q_{3} \frac{1}{4 π ϵ_{0}} [(\frac{q_{1}}{0.4} + \frac{q_{2}}{0.1}) - (\frac{q_{1}}{0.4} + \frac{q_{2}}{0.5})]

Δ U = q_{3} \frac{1}{4 π ϵ_{0}} [\frac{q_{1}}{0.4} - \frac{q_{1}}{0.4} + \frac{q_{2}}{0.1} - \frac{q_{2}}{0.5}]

Δ U = q_{3} \frac{1}{4 π ϵ_{0}} [q_{2} (\frac{1}{0.1} - \frac{1}{0.5})]

Δ U = q_{3} \frac{1}{4 π ϵ_{0}} [q_{2} (10 - 2)] = q_{3} \frac{8 q_{2}}{4 π ϵ_{0}}

Comparing this with the given expression $\frac{q_{3} K}{4 π ϵ_{0}}$ , we find that $K = 8 q_{2}$ .

💡 Teacher's Secret Hint

Notice how the terms involving $q_{1}$ cancel out due to the circular path centered at $q_{1}$ .

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Physics Question 38 – JEE-MAIN 2025

Clue 1 of 3: Potential Energy Difference

🧠 Full Solution Path