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, $\epsilon_0=8.85 imes 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the electrostatic potential energy of the charge configuration is :

Question

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, $\epsilon_0=8.85 	imes 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the electrostatic potential energy of the charge configuration is :

Aicharya · Accepted Answer

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 $7 \mu \mathrm{c}$ and $-4 \mu \mathrm{c}$ are placed at $(-7 \mathrm{~cm}, 0,0)$ and $(7…

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