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 :

<p>$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$</p> <p>Here, </p> <p><p>$q_1 = 7 \times 10^{-6} \, C$</p></p> <p><p>$q_2 = -4 \times 10^{-6} \, C$</p></p> <p><p>The separation between the charges is along the x-axis from $-7 \, \text{cm}$ to $7 \, \text{cm}$, so </p> <p>$r = 0.14 \, m$</p></p> <p><p>The Coulomb constant is approximated as </p> <p>$\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \frac{Nm^2}{C^2}$</p></p> <p>Substitute these values into the equation:</p> <p>$U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-4 \times 10^{-6})}{0.14}$</p> <p>Calculate the product of the charges:</p> <p>$(7 \times 10^{-6})(-4 \times 10^{-6}) = -28 \times 10^{-12} \, C^2$</p> <p>Now, divide by the distance:</p> <p>$\frac{-28 \times 10^{-12}}{0.14} = -2 \times 10^{-10} \, C^2/m$</p> <p>Finally, multiply by the Coulomb constant:</p> <p>$U = 9 \times 10^9 \times (-2 \times 10^{-10}) = -1.8 \, J$</p> <p>Thus, the electrostatic potential energy of the configuration is:</p> <p>$-1.8 \, J$</p> <p>This corresponds to Option D.</p>