JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2025
- Medium
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) -1.8 J
Step-by-step Solution
Detailed explanation
P.E. of two charges
\(\begin{aligned}
& \mathrm{u}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\
& \mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2} \\
& =14 \mathrm{~cm} \\
& \therefore \mathrm{u}=\frac{9 \times 10^9 \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}} \\
& =-1.8 \mathrm{~J}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{u}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\
& \mathrm{r}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2} \\
& =14 \mathrm{~cm} \\
& \therefore \mathrm{u}=\frac{9 \times 10^9 \times 7 \times 10^{-6} \times(-4) \times 10^{-6}}{14 \times 10^{-2}} \\
& =-1.8 \mathrm{~J}
\end{aligned}\)
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