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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 7 April · Morning Shift · Physics · Electrostatics

Two charges q_1 and q_2 are separated by a distance of 30 cm. A...

Two charges q1\mathrm{q}_1 and q2\mathrm{q}_2 are separated by a distance of 30 cm. A third charge q3q_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 q3q_3 from CC to DD is given by q3 K4πϵ0\frac{\mathrm{q}_3 \mathrm{~K}}{4 \pi \epsilon_0}, the value of K is :
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

Options

  1. A8q28 \mathrm{q}_2
  2. B6q26 \mathrm{q}_2
  3. C8q18 q_1
  4. D6q16 \mathrm{q}_1

Correct answer

A. 8q28 \mathrm{q}_2

Step-by-step solution

Potential at C
VC=kq10.4+kq20.5\mathrm{V}_{\mathrm{C}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.5}
Potential at D
VD=kq10.4+kq20.1\mathrm{V}_{\mathrm{D}}=\frac{\mathrm{kq}_1}{0.4}+\frac{\mathrm{kq}_2}{0.1}
ΔU=(VD−Vc)(q3)=(kq20.1−kq20.5)(q3)\Delta \mathrm{U}=\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{c}}\right)\left(\mathrm{q}_3\right)=\left(\frac{\mathrm{kq}_2}{0.1}-\frac{\mathrm{kq}_2}{0.5}\right)\left(\mathrm{q}_3\right)
ΔU=8kq2q3=8q2q34πε0\Delta \mathrm{U}=8 \mathrm{kq}_2 \mathrm{q}_3=\frac{8 \mathrm{q}_2 \mathrm{q}_3}{4 \pi \varepsilon_0}
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