Electric Flux PYQ | Gauss's Law Cube Question | JEE Main Physics 2025

❓ Question

A point charge $Q$ is placed at a corner of a cube of side $d$.

Find the electric flux through the shaded 2D square sheet shown in the figure.

---

✍️ Solution

Imagine 8 identical cubes joined together so that the charge $Q$ lies at the center of the resulting larger cube.

Hence, the total electric flux through the larger cube is

$$\Phi_{\text{total}}=\frac{Q}{\varepsilon_0}.$$

Since the larger cube has 6 identical faces,

$$\Phi_{\text{one face}} =\frac{Q}{6\varepsilon_0}.$$

---

Flux Through One Small Square

Each face of the larger cube is made up of

$$2\times2=4$$

identical small square sheets.

Therefore, the flux through one small square is

$$\Phi_{\text{square}} = \frac{1}{4}\cdot\frac{Q}{6\varepsilon_0} = \frac{Q}{24\varepsilon_0}.$$

Equivalently,

$$Q_{\text{effective}} = Q\left(\frac18\right)\left(\frac13\right),$$

where

• $\frac18$ is the contribution of the charge to one small cube,
• $\frac13$ is the contribution to one of the three outer faces of that cube.

Thus,

$$\Phi = \frac{Q}{8\times3\,\varepsilon_0} = \frac{Q}{24\varepsilon_0}.$$

---

✅ Final Answer

$$\boxed{\Phi=\frac{Q}{24\varepsilon_0}}$$