JEE Physics 2025 Question with Solution

Given: electric flux $\Phi_E = -2 \times 10^4 \, \text{N m}^2\text{C}^{-1}$ through a spherical Gaussian surface. Radius is $8.0 \, \text{cm}$, but for Gauss's law here the enclosed charge depends only on flux, not on the radius.

Find: the value of the enclosed point charge.

Using Gauss's law,

$$\Phi_E = \frac{q}{\varepsilon_0}$$

So,

$$q = \Phi_E \, \varepsilon_0$$

Substitute the given values,

$$q = \left(-2 \times 10^4\right) \left(8.854 \times 10^{-12}\right)$$ $$q = -1.7708 \times 10^{-7} \, \text{C}$$

Thus the enclosed charge is negative. Its magnitude is

$$|q| = 1.77 \times 10^{-7} \, \text{C}$$

This matches option A written as $17.7 \times 10^{-8} \, \text{C}$, but the listed option is printed as $17.7 \times 10^{-7} \, \text{C}$. Following the solution, the correct option is A.

Given: the electric flux is negative, $\Phi_E < 0$.

Find: what the sign of the charge means.

A negative electric flux means the net electric field lines are entering the closed surface. Therefore the enclosed charge must be negative.

From Gauss's law,

$$q = \varepsilon_0 \Phi_E$$

Since $\varepsilon_0 > 0$ and $\Phi_E < 0$, we get

$$q < 0$$

Hence the actual charge is negative, while the source solution finally reports the magnitude and marks option A as correct.