Electrostatics — Easy JEE physics MCQ
A cube of side $a$ is placed in a uniform electric field $\vec{E} = E_0\hat{i}$. The total electric flux through the cube is:
- A. $E_0 a^2$
- B. $2E_0 a^2$
- C. $6E_0 a^2$
- D. Zero
Solution
For a uniform electric field, the flux entering through one face equals the flux leaving through the opposite face.
The flux through the face perpendicular to $\hat{i}$ (left): $\Phi_1 = -E_0 a^2$
The flux through the opposite face (right): $\Phi_2 = +E_0 a^2$
The flux through all other faces (parallel to $\vec{E}$): $\Phi = 0$
Total flux: $\Phi_{total} = -E_0 a^2 + E_0 a^2 + 0 = 0$
Alternatively, since there's no charge enclosed, by Gauss's law: $\Phi = \frac{q_{enc}}{\varepsilon_0} = 0$