The electric field In a region is given gy `vecE=(E_(0)x)/(b)hati`. Find the charge contained in the cubical volume bounded by the surfaces x=0, x=a, y=0, y=a,z=0 and z=a. take `E_(0)=5xx10^(3)NC^(-1),a=1cm and b=2cm`.

Answer

To find the charge contained in the cubical volume bounded by the surfaces \( x=0, x=a, y=0, y=a, z=0, z=a \) given the electric field \( \vec{E} = \frac{E_0 x}{b} \hat{i} \), we can use Gauss's law, which relates the electric flux through a closed surface to the charge enclosed by that surface.

### Step-by-Step Solution:

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

2. **Calculate the Electric Flux**:
The electric flux \( \Phi_E \) through a surface is given by:
\[ \Phi_E = \int \vec{E} \cdot d\vec{A} \]
For the cube, we need to calculate the flux through each of the six faces.

3. **Calculate the Flux through Each Face**:
- **Face at \( x=0 \)**: Here, \( \vec{E} = 0 \), so \( \Phi_E = 0 \).
- **Face at \( x=a \)**: Here, \( \vec{E} = \frac{E_0 a}{b} \hat{i} \). The area vector \( d\vec{A} = a^2 \hat{i} \) (outward normal).
\[ \Phi_E = \vec{E} \cdot d\vec{A} = \frac{E_0 a}{b} \cdot a^2 = \frac{E_0 a^3}{b} \]
- **Faces at \( y=0, y=a, z=0, z=a \)**: The electric field has no component in the \( y \) or \( z \) directions, so the flux through these faces is also \( 0 \).

4. **Total Electric Flux**:
The total electric flux through the cube is:
\[ \Phi_E = \frac{E_0 a^3}{b} \]

5. **Using Gauss's Law**:
According to Gauss's law:
\[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the charge enclosed and \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

6. **Calculate the Enclosed Charge**:
Rearranging Gauss's law gives:
\[ Q_{\text{enc}} = \Phi_E \cdot \epsilon_0 = \frac{E_0 a^3}{b} \cdot \epsilon_0 \]

7. **Substituting Values**:
Convert \( a \) from cm to m:
\[ a = 1 \, \text{cm} = 0.01 \, \text{m} \]
Now substitute \( E_0 = 5 \times 10^3 \, \text{N/C} \), \( a = 0.01 \, \text{m} \), and \( b = 0.02 \, \text{m} \):
\[ Q_{\text{enc}} = \frac{(5 \times 10^3) \cdot (0.01)^3}{0.02} \cdot (8.85 \times 10^{-12}) \]

8. **Calculate**:
\[ Q_{\text{enc}} = \frac{(5 \times 10^3) \cdot (1 \times 10^{-6})}{0.02} \cdot (8.85 \times 10^{-12}) = \frac{5 \times 10^{-3}}{0.02} \cdot (8.85 \times 10^{-12}) = 0.25 \cdot (8.85 \times 10^{-12}) = 2.2125 \times 10^{-12} \, \text{C} \]

### Final Answer:
The charge contained in the cubical volume is approximately:
\[ Q_{\text{enc}} \approx 2.21 \times 10^{-12} \, \text{C} \]