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Question

A charge +Q is enclosed by a cube of side L. Which of the following statements are true ?
A. The total electric flux through all surfaces = $\frac{\text{Q}}{\epsilon_0}$
B. The electric flux through one surface = $\frac{\text{QL}^2}{\epsilon_0}$
C. The electric flux through one surface = $\frac{\text{Q}}{6\epsilon_0}$
D. The electric fields come out through the surfaces
Choose the correct answer from the options given below :

The correct answer is
A, C and D only

Gauss's Law for Electric Flux

Gauss's Law states that the total electric flux ($\Phi_E$) through any closed surface is equal to the net charge enclosed ($Q_{enc}$) divided by the permittivity of free space ($\epsilon_0$). Mathematically, this is expressed as:

$\Phi_{total} = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$

Analyzing the Statements

  • Statement A: The total electric flux through all surfaces = $\frac{\text{Q}}{\epsilon_0}$

    The cube is a closed surface enclosing the charge +Q. According to Gauss's Law, the total electric flux passing through this closed surface is indeed $\frac{Q}{\epsilon_0}$. Therefore, statement A is true.

  • Statement B: The electric flux through one surface = $\frac{\text{QL}^2}{\epsilon_0}$

    The flux depends on the enclosed charge and the properties of the closed surface, not directly on the area squared ($L^2$) in this manner for a single surface unless related to a specific electric field setup. For a symmetrical enclosure like a cube with a central charge, the flux is distributed equally among its faces. This formula does not align with Gauss's Law principles for flux distribution per surface. Therefore, statement B is false.

  • Statement C: The electric flux through one surface = $\frac{\text{Q}}{6\epsilon_0}$

    Since the total flux through the cube (6 surfaces) is $\frac{Q}{\epsilon_0}$ (from Statement A), and the cube is symmetrical, the electric flux distributes equally among its 6 identical faces. Thus, the flux through one surface is $\frac{1}{6}$ of the total flux: $\frac{1}{6} \times \frac{Q}{\epsilon_0} = \frac{Q}{6\epsilon_0}$. Therefore, statement C is true.

  • Statement D: The electric fields come out through the surfaces

    The enclosed charge is +Q, which is positive. Electric field lines originate from positive charges and point outwards. Since the charge is inside the cube, the electric field lines must pass through the surfaces of the cube to emerge from the enclosed volume. Therefore, statement D is true.

Conclusion

Based on the analysis, statements A, C, and D are true.

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