At the centre of a cubical box, +Q charge is placed. The value of total flux that is coming out of a wall is :

Options

A. `(Q)/(epsi_(0))`
B. `(Q)/(3 epsi_(0))`
C. `(Q)/(4epsi_(0))`
D. `(Q)/( 6 espi_(0))`

Explanation

To solve the problem of finding the total electric flux coming out of one wall of a cubical box with a charge +Q placed at its center, we can follow these steps:

### Step 1: Understand Gauss's Law
Gauss's Law states that the total electric flux (Φ) through a closed surface is equal to the charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀). Mathematically, it is expressed as:
\[ Φ = \frac{Q_{\text{enc}}}{ε₀} \]

### Step 2: Identify the Total Charge Enclosed
In this case, the charge +Q is placed at the center of the cubical box. Therefore, the total charge enclosed by the surface of the cube is:
\[ Q_{\text{enc}} = +Q \]

### Step 3: Calculate the Total Electric Flux Through the Cube
Using Gauss's Law, we can calculate the total electric flux through the entire surface of the cube:
\[ Φ_{\text{total}} = \frac{Q}{ε₀} \]

### Step 4: Determine the Flux Through One Wall
The cube has 6 faces (walls). Since the charge is symmetrically placed at the center, the electric flux will be evenly distributed across all 6 faces of the cube. Therefore, the flux through one wall can be calculated by dividing the total flux by the number of walls:
\[ Φ_{\text{one wall}} = \frac{Φ_{\text{total}}}{6} = \frac{Q}{6ε₀} \]

### Final Answer
The total electric flux coming out of one wall of the cubical box is:
\[ Φ_{\text{one wall}} = \frac{Q}{6ε₀} \]

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