JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Electrostatics

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) $8 q_{2}$ ✓ Correct
(B) $6 q_{2}$
(C) $8 q_{1}$
(D) $6 q_{1}$

Answer: (A)

Solution

The charge

q_{1}

sits at the centre of the circular arc, so its distance to both

C

and

D

equals the radius

40

cm; it gives the same potential at the two points and contributes nothing to the change. The charge

q_{2}

lies on the base: its distance to

C

3 0^{2} + 4 0^{2} = 50

cm and its distance to

D

40 - 30 = 10

cm. The change in potential energy of

q_{3}

Δ U = q_{3} (V_{D} - V_{C}) = \frac{q _{3}}{4 π ϵ _{0}} q_{2} (\frac{1}{0.10} - \frac{1}{0.50}) = \frac{q _{3}}{4 π ϵ _{0}} q_{2} (10 - 2) .

Thus

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

, and the value of

K

8 q_{2}

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Electrostatics

Solution

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