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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 23 January · Evening Shift · Physics · Electrostatics

Two charges 7 μ c and -4 μ c are placed at (-7 ~cm, 0,0) and (7...

Two charges 7μc7 \mu \mathrm{c} and −4μc-4 \mu \mathrm{c} are placed at (−7 cm,0,0)(-7 \mathrm{~cm}, 0,0) and (7 cm,0,0)(7 \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 :

Options

  1. A-1.8 J
  2. B-2.0 J
  3. C-1.5 J
  4. D-1.2 J

Correct answer

A. -1.8 J

Step-by-step solution

P.E. of two charges u=14πε0q1q2rr=(x2−x1)2+(y2−y1)2+(z2−z1)2=14 cm∴u=9×109×7×10−6×(−4)×10−614×10−2=−1.8 J\begin{aligned} & \mathrm{u}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\ & \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 \mathrm{~cm} \\ & \therefore \mathrm{u}=\frac{9 \times 10^9 \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}} \\ & =-1.8 \mathrm{~J} \end{aligned}
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