Question

A charge \( Q \) is placed at the center of a cube of sides \( a \). The total flux of electric field through the six surfaces of the cube is:

• A. \(\frac{6Qa^2}{\epsilon_0}\)
• B. \(\frac{Q^2a^2}{\epsilon_0}\)
• C. \(\frac{Q}{\epsilon_0}\)
• D. \(\frac{Qa^2}{\epsilon_0}\)

Hint:

Gauss’s law states that the total electric flux through a closed surface is proportional to the charge enclosed within the surface. For a cube, the flux is evenly distributed across all faces.

Answer 1

The Correct Option is C

The total electric flux through any closed surface is given by Gauss's Law:

\[ \Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enclosed}}}{\epsilon_0}, \]

where:

• \( \Phi \) is the total electric flux,
• \( Q_{\text{enclosed}} \) is the total charge enclosed within the surface,
• \( \epsilon_0 \) is the permittivity of free space.

In this case:

• The charge \( Q \) is placed at the center of the cube.
• The cube is a closed surface.

By symmetry, the flux is uniformly distributed over the six faces of the cube. However, Gauss's law directly gives the total flux through the entire closed surface:

\[ \Phi = \frac{Q}{\epsilon_0}. \]

Key Observation: The flux through each face of the cube can be calculated as:

\[ \Phi_{\text{face}} = \frac{\Phi}{6} = \frac{Q}{6\epsilon_0}, \]

but the question asks for the total flux through the six surfaces, which is simply:

\[ \Phi = \frac{Q}{\epsilon_0}. \]