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 solve the problem, we will follow these steps:

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

### Step 2: Identify the Volume and Boundaries
We need to find the charge contained inside a cubical volume bounded by the surfaces:
- \( x = 0 \)
- \( x = a \)
- \( y = 0 \)
- \( y = h \)
- \( z = 0 \)
- \( z = a \)

Given \( a = 1 \, \text{m} \) and assuming \( h = 1 \, \text{m} \) for the y-boundary.

### Step 3: Calculate the Electric Flux
Using Gauss's Law, the electric flux \( \Phi \) through a surface is given by:
\[ \Phi = \int E \cdot dA \]

Since the electric field is only in the x-direction, we will only consider the flux through the surfaces at \( x = 0 \) and \( x = a \).

1. **Flux through \( x = 0 \)**:
\[ E(0) = \frac{E_0 \cdot 0}{L} = 0 \]
Therefore, the flux through \( x = 0 \) is:
\[ \Phi_{x=0} = E(0) \cdot A = 0 \]

2. **Flux through \( x = a \)**:
\[ E(a) = \frac{E_0 \cdot a}{L} = \frac{5 \times 10^3 \cdot 1}{0.02} = 2.5 \times 10^5 \, \text{N/C} \]
The area \( A \) of the face at \( x = a \) is:
\[ A = h \cdot a = 1 \cdot 1 = 1 \, \text{m}^2 \]
Thus, the flux through \( x = a \) is:
\[ \Phi_{x=a} = E(a) \cdot A = 2.5 \times 10^5 \cdot 1 = 2.5 \times 10^5 \, \text{N m}^2/\text{C} \]

### Step 4: Apply Gauss's Law
According to Gauss's Law:
\[ \Phi = \frac{q}{\epsilon_0} \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

Setting the flux equal to the charge divided by \( \epsilon_0 \):
\[ \frac{q}{\epsilon_0} = 2.5 \times 10^5 \]
Thus, we can solve for \( q \):
\[ q = 2.5 \times 10^5 \cdot \epsilon_0 \]
\[ q = 2.5 \times 10^5 \cdot 8.85 \times 10^{-12} \]
\[ q \approx 2.21 \times 10^{-6} \, \text{C} \]

### Final Answer
The charge contained inside the cubical volume is approximately:
\[ q \approx 2.21 \times 10^{-6} \, \text{C} \]

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