The electric field in a region is given by `E = (E_0x)/lhati`. Find the charge contained inside a cubical 1 volume bounded by the surfaces `x=0, x=a, y=0, yh=, z=0` and z=a. Take `E_0=5xx10^3N//C` l=2cm and a=1m.

Explanation

To find the charge contained inside a cubical volume bounded by the surfaces \(x=0\), \(x=a\), \(y=0\), \(y=a\), \(z=0\), and \(z=a\) in the presence of an electric field given by \(E = \frac{E_0 x}{L} \hat{i}\), we will use Gauss's law.

### Step-by-Step Solution:

1. **Identify the Electric Field**:
The electric field is given as \(E = \frac{E_0 x}{L} \hat{i}\). Here, \(E_0 = 5 \times 10^3 \, \text{N/C}\) and \(L = 2 \, \text{cm} = 0.02 \, \text{m}\).

2. **Determine the Boundaries of the Cube**:
The cube is defined by the surfaces:
- \(x = 0\)
- \(x = a\) (where \(a = 1 \, \text{m}\))
- \(y = 0\)
- \(y = a\)
- \(z = 0\)
- \(z = a\)

3. **Calculate the Electric Flux through Each Surface**:
According to Gauss's law, the electric flux \(\Phi_E\) through a closed surface is given by:
\[ \Phi_E = \int E \cdot dA \]
We will analyze the flux through each face of the cube.

- **For the surfaces at \(x = 0\) and \(x = a\)**:
- At \(x = 0\), \(E = 0\) (since \(E = \frac{E_0 \cdot 0}{L} = 0\)).
- At \(x = a\), \(E = \frac{E_0 a}{L} = \frac{5 \times 10^3 \times 1}{0.02} = 2.5 \times 10^5 \, \text{N/C}\).
- The area of the face at \(x = a\) is \(A = a^2 = 1^2 = 1 \, \text{m}^2\).
- Thus, the flux through the face at \(x = a\) is:
\[ \Phi_E = E \cdot A = \left(2.5 \times 10^5 \, \text{N/C}\right) \cdot (1 \, \text{m}^2) = 2.5 \times 10^5 \, \text{N m}^2/\text{C} \]

- **For the surfaces at \(y = 0\), \(y = a\), \(z = 0\), and \(z = a\)**:
- The electric field is parallel to these surfaces, hence the flux through these surfaces is zero.

4. **Total Electric Flux**:
The total electric flux through the cube is:
\[ \Phi_E = \Phi_{x=0} + \Phi_{x=a} + \Phi_{y=0} + \Phi_{y=a} + \Phi_{z=0} + \Phi_{z=a} = 0 + 2.5 \times 10^5 + 0 + 0 + 0 + 0 = 2.5 \times 10^5 \, \text{N m}^2/\text{C} \]

5. **Apply Gauss's Law**:
According to Gauss's law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
Rearranging gives:
\[ Q_{\text{enc}} = \Phi_E \cdot \epsilon_0 \]
Where \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\).

6. **Calculate the Enclosed Charge**:
Substituting the values:
\[ Q_{\text{enc}} = (2.5 \times 10^5) \cdot (8.85 \times 10^{-12}) \approx 2.21 \times 10^{-6} \, \text{C} = 2.21 \, \mu\text{C} \]

### Final Answer:
The charge contained inside the cubical volume is approximately \(2.21 \, \mu\text{C}\).