The electric field in a region is given by `vec(E)=E_(0)x hat(i)`. The charge contained inside a cubical volume bounded by the surface x=0, x=2m, y=0, y=2m, z=0 and z=2m is `n epsi_(0)E_(0)`. Find the va,ue of n.

Answer

To solve the problem, we need to find the value of \( n \) given the electric field and the charge contained in a cubical volume. Let's break down the steps:

### Step 1: Understand the Electric Field
The electric field is given by:
\[ \vec{E} = E_0 \hat{i} \]
This means the electric field is uniform and directed along the x-axis.

### Step 2: Determine the Volume of the Cube
The cube is bounded by the surfaces:
- \( x = 0 \) to \( x = 2 \, \text{m} \)
- \( y = 0 \) to \( y = 2 \, \text{m} \)
- \( z = 0 \) to \( z = 2 \, \text{m} \)

The volume of the cube is:
\[ V = \text{side}^3 = (2 \, \text{m})^3 = 8 \, \text{m}^3 \]

### Step 3: Calculate the Electric Flux through the Cube
According to Gauss's Law, the electric flux \( \Phi_E \) through a closed surface is given by:
\[ \Phi_E = \int \vec{E} \cdot d\vec{A} \]
Since the electric field is uniform, we can calculate the flux through each face of the cube.

1. **Flux through \( x = 0 \)**:
\[ \Phi_{E, x=0} = \vec{E} \cdot \vec{A} = E_0 \hat{i} \cdot (0) = 0 \]

2. **Flux through \( x = 2 \)**:
The area vector \( \vec{A} \) for this face points in the positive x-direction, and the area is \( 2 \times 2 = 4 \, \text{m}^2 \):
\[ \Phi_{E, x=2} = E_0 \cdot 4 = 4E_0 \]

3. **Flux through the other faces** (\( y = 0, y = 2, z = 0, z = 2 \)):
Since the electric field has no component in the y or z directions, the flux through these faces is zero:
\[ \Phi_{E, y=0} = 0, \quad \Phi_{E, y=2} = 0, \quad \Phi_{E, z=0} = 0, \quad \Phi_{E, z=2} = 0 \]

### Step 4: Total Electric Flux
The total electric flux through the cube is:
\[ \Phi_E = \Phi_{E, x=0} + \Phi_{E, x=2} + \Phi_{E, y=0} + \Phi_{E, y=2} + \Phi_{E, z=0} + \Phi_{E, z=2} = 0 + 4E_0 + 0 + 0 + 0 + 0 = 4E_0 \]

### Step 5: Relate Flux to Charge Enclosed
According to Gauss's Law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Where \( Q_{\text{enc}} \) is the charge enclosed by the cube. Therefore:
\[ 4E_0 = \frac{Q_{\text{enc}}}{\epsilon_0} \]
This implies:
\[ Q_{\text{enc}} = 4E_0 \epsilon_0 \]

### Step 6: Given Charge and Solve for \( n \)
We are given that the charge contained inside the cube is:
\[ Q_{\text{enc}} = n \epsilon_0 E_0 \]
Setting the two expressions for \( Q_{\text{enc}} \) equal gives:
\[ 4E_0 \epsilon_0 = n \epsilon_0 E_0 \]
Dividing both sides by \( \epsilon_0 E_0 \) (assuming \( E_0 \neq 0 \)):
\[ n = 4 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{4} \]