Q53 / 75 · JEE Main 2025 (23 Jan Shift 2) · Physics
Two charges $7 \mu \mathrm{c}$ and $-4 \mu \mathrm{c}$ are placed at $(-7 \mathrm{~cm}, 0,0)$ and $(7 \mathrm{~cm}, 0,0)$ respectively. Given, $\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the electrostatic potential energy of the charge configuration is :
- Correct answerA-1.8 J
- B-2.0 J
- C-1.5 J
- D-1.2 J
Solution
P.E. of two charges
$\begin{aligned}
& \mathrm{u}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\
& \mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2} \\
& =14 \mathrm{~cm} \\
& \therefore \mathrm{u}=\frac{9 \times 10^9 \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}} \\
& =-1.8 \mathrm{~J}
\end{aligned}$
Hence, the correct answer is option: 'A'.