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

Options

A. `Q/(6epsi_(0))`
B. `Q/(8epsi_(0))`
C. `Q/(epsi_(0))`
D. `Q/(2epsi_(0))`

Answer

To solve the problem of finding the electric flux passing through all six faces of a cube with a charge \( Q \) situated at one of its corners, we can use Gauss's law. Here’s a step-by-step solution:

### Step 1: Understand Gauss's Law
Gauss's law states that the electric flux \( \Phi \) through a closed surface is equal to the charge \( Q_{\text{enc}} \) enclosed by that surface divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

### Step 2: Determine the Enclosed Charge
Since the charge \( Q \) is located at one corner of the cube, we need to determine how much of this charge is effectively enclosed by the cube. A cube has 8 corners, and since the charge is at one corner, only \( \frac{1}{8} \) of the charge \( Q \) is considered to be inside the cube.

\[ Q_{\text{enc}} = \frac{Q}{8} \]

### Step 3: Calculate the Electric Flux
Now that we have the enclosed charge, we can substitute this value into Gauss's law to find the total electric flux through the cube:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{\frac{Q}{8}}{\epsilon_0} = \frac{Q}{8\epsilon_0} \]

### Step 4: Conclusion
Thus, the total electric flux passing through all six faces of the cube is:
\[ \Phi = \frac{Q}{8\epsilon_0} \]