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

Gauss's Law states the total electric flux through a closed surface:

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

where:

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

Scenario:

• Charge \( Q \) is at the cube's center.
• The cube forms a closed surface.

Due to symmetry, flux is evenly distributed across the cube's six faces. Gauss's law directly yields the total flux for the entire closed surface:

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

Key Insight: The flux through each cube face can be found by:

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

The question seeks the total flux through all six surfaces, which is:

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