QJEEM25Phb986: JEE Main 2025 Physics - Class 12 JEEM25Phb986 Solution: Two charges and are placed at and respectively. Given, , the…

$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$

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

$r = 0.14 \, m$

The Coulomb constant is approximated as

$\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \frac{Nm^2}{C^2}$

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:

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

Thus, the electrostatic potential energy of the configuration is:

$-1.8 \, J$

This corresponds to Option D.