P.E. of two charges $\begin{aligned}
& \mathrm{u}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \
& \mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2} \
& =14 \mathrm{cm} \
& \therefore \mathrm{u}=\frac{9 \times 10^9 \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}} \
& =-1.8 \mathrm{J}
\end{aligned}$
Two charges 7 μ and -4 μ are placed at (-7 ~cm, 0,0) and (7 ~cm, 0,0) respectively. Given, ε_0=8.85 × 10^-12 C^2 ~N^-1 ~m^-2, the electrostatic…
JEE Main 2025 — Physics Electromagnetism
2025mcqmedium
Two charges 7μc and −4μc are placed at (−7 cm,0,0) and (7 cm,0,0) respectively. Given, ϵ0=8.85×10−12C2 N−1 m−2, the electrostatic potential energy of the charge configuration is :
Official previous-year question
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
-1.8 J
- B
-2.0 J
- C
-1.5 J
- D
-1.2 J
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