A charge Q is situated at the centre of a cube. The electric flux through one of the faces of the cube is

Options

A. `Q//epsi_(0)`
B. `Q//2epsi_(0)`
C. `Q//4 epsi_(0)`
D. `Q//6 epsi_(0)`

Answer

To find the electric flux through one face of a cube when a charge \( Q \) is situated at the center, 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 total electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \varepsilon_0 \):
\[ \Phi = \frac{Q}{\varepsilon_0} \]

### Step 2: Identify the Closed Surface
In this case, the closed surface is the cube, and the charge \( Q \) is located at its center. Since the cube is symmetric, the electric flux will be uniformly distributed across all six faces of the cube.

### Step 3: Calculate Total Electric Flux
Using Gauss's Law, the total electric flux through the entire cube is:
\[ \Phi_{\text{total}} = \frac{Q}{\varepsilon_0} \]

### Step 4: Distribute the Flux Across the Faces
Since the cube has six faces and the flux is uniformly distributed, the electric flux through one face of the cube can be calculated by dividing the total flux by the number of faces:
\[ \Phi_{\text{one face}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q}{6\varepsilon_0} \]

### Step 5: Conclusion
Thus, the electric flux through one face of the cube is:
\[ \Phi_{\text{one face}} = \frac{Q}{6\varepsilon_0} \]

### Final Answer
The electric flux through one of the faces of the cube is \( \frac{Q}{6\varepsilon_0} \).

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