Question

A charge Q is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is

• A. $ \frac{Q}{6 \varepsilon_0}$
• B. $ \frac{Q} {8 \varepsilon_0}$
• C. $ \frac{Q} { \varepsilon_0}$
• D. $ \frac{Q} {2 \varepsilon_0}$

Answer 1

The Correct Option is B

To determine the electric flux through all the six faces of a cube when a charge Q is situated at one of its corners, we start by understanding the application of Gauss's Law.

Gauss's Law is given by the formula:

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}

Where \vec{E} is the electric field, d\vec{A} is the differential area vector, Q_{\text{enc}} is the charge enclosed by the surface, and \varepsilon_0 is the permittivity of free space.

Since the charge Q is placed at a corner of the cube, a portion of the charge’s influence extends into a quarter of a smaller cube enclosed by 8 such cubes forming a larger imaginary cube around Q at the corner. Thus, the charge Q is effectively distributed equally among these 8 small cubes.

Therefore, the charge enclosed by one of these smaller cubes, which corresponds to the original cube where faces are to be evaluated for flux, is \frac{Q}{8}.

Using Gauss's Law, the flux through the larger cube is:

\oint \vec{E} \cdot d\vec{A} = \frac{Q}{8\varepsilon_0}

However, flux through a smaller cube (one of the original problem cubes) is due to this fraction of the total charge. Since the original setup only allows for consideration of the smaller cube itself, each face of the original cube will see a total flux:

\frac{Q}{8\varepsilon_0}

Thus, the electric flux through all six faces of the cube when the charge Q is at a corner is correctly given by:

Option: \frac{Q}{8 \varepsilon_0}