JEE Main 2025 — Electrostatics Question with Solution

$$U=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r}$$

Here,

$$q_1=7\times 10^{-6}C$$

$$q_2=-4\times 10^{-6}C$$

The separation between the charges is along the x-axis from $$-7\text{cm}$$ to $$7\text{cm}$$, so

$$r=0.14m$$

The Coulomb constant is approximated as

$$\frac{1}{4\pi {ϵ}_0}\approx 9\times 10^9\frac{N{m}^2}{C^2}$$

Substitute these values into the equation:

$$U=9\times 10^9\times \frac{(7\times 10^{-6})(-4\times 10^{-6})}{0.14}$$

Calculate the product of the charges:

$$(7\times 10^{-6})(-4\times 10^{-6})=-28\times 10^{-12}{C}^2$$

Now, divide by the distance:

$$\frac{-28\times 10^{-12}}{0.14}=-2\times 10^{-10}{C}^2/m$$

Finally, multiply by the Coulomb constant:

$$U=9\times 10^9\times (-2\times 10^{-10})=-1.8J$$

Thus, the electrostatic potential energy of the configuration is:

$$-1.8J$$

This corresponds to Option D.