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Physics · Electrostatics

JEE Main 2025 — 23 January, Evening Shift — Question 49

Two charges 7μc7 \mu \mathrm{c} and −4μc-4 \mu \mathrm{c} are placed at ( -7 cm , 0,0)0,0) and ( 7 cm,0,07 \mathrm{~cm}, 0,0 ) respectively. Given, ϵ0=8.85×10−12C2 N−1 m−2\epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}, the electrostatic potential energy of the charge configuration is :

  1. Option A:

    -1.5 J

  2. Option B:

    -2.0 J

  3. Option C:

    -1.2 J

  4. Option D:

    -1.8 J

    Correct

Answer: D

Step-by-step solution

P.E. of two charges

u=14πε0q1q2r\mathrm{u}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}}

r=(x2−x1)2+(y2−y1)2+(z2−z1)2\mathrm{r}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}

=14 cm=14 \mathrm{~cm}

∴u=9×109×7×10−6×(−4)×10−614×10−2\therefore \mathrm{u}=\frac{9 \times 10^{9} \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}}

=−1.8 J=-1.8 \mathrm{~J}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Potential Energy and Electric Potential

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