An electric field given by `vec(E) = 4hat(i) - (20 y^(2) + 2) hat(j)` pierces a Gussian cube of side 1 m placed at origin such that its three sides represents x,y and z axes. The net charge enclosed within the cube is `x xx 10^(-10)C`, where is _____. (take `in_(0) = 8.85 xx 10^(-12) N^(-1)m^(-2)C^(2)`)

Explanation

To solve the problem, we need to find the net charge enclosed within a Gaussian cube placed at the origin in an electric field given by \(\vec{E} = 4\hat{i} - (20y^2 + 2)\hat{j}\).

### Step-by-Step Solution:

1. **Identify the Electric Field Components**:
The electric field is given as:
\[ \vec{E} = 4\hat{i} - (20y^2 + 2)\hat{j} \]
Here, the \(x\)-component of the electric field is constant (\(E_x = 4\)), and the \(y\)-component varies with \(y\) (\(E_y = -(20y^2 + 2)\)).

2. **Determine the Area Vectors for the Cube**:
The cube has six faces, and we need to consider the area vectors for each face:
- For the face at \(y = 0\) (downward): \(\hat{A} = -\hat{j}\)
- For the face at \(y = 1\) (upward): \(\hat{A} = \hat{j}\)
- The faces at \(x = 0\) and \(x = 1\) will have area vectors in the \(\hat{i}\) direction.
- The faces at \(z = 0\) and \(z = 1\) will have area vectors in the \(\hat{k}\) direction.

3. **Calculate the Electric Flux through Each Face**:
- **For the face at \(y = 0\)**:
\[ E_y = -(20(0)^2 + 2) = -2 \quad \text{(downward)} \]
\[ \Phi_{y=0} = E \cdot A = (-2)(1 \cdot 1) = -2 \, \text{Wb} \]
- **For the face at \(y = 1\)**:
\[ E_y = -(20(1)^2 + 2) = -22 \quad \text{(upward)} \]
\[ \Phi_{y=1} = E \cdot A = (-22)(1 \cdot 1) = -22 \, \text{Wb} \]
- **For the faces at \(x = 0\) and \(x = 1\)**:
The electric field in the \(\hat{i}\) direction does not contribute to the net flux since the area vectors are perpendicular to the electric field.
\[ \Phi_{x=0} = 0, \quad \Phi_{x=1} = 0 \]
- **For the faces at \(z = 0\) and \(z = 1\)**:
Similarly, the electric field has no component in the \(\hat{k}\) direction, so:
\[ \Phi_{z=0} = 0, \quad \Phi_{z=1} = 0 \]

4. **Calculate the Total Electric Flux**:
The total electric flux through the cube is:
\[ \Phi_{total} = \Phi_{y=0} + \Phi_{y=1} + \Phi_{x=0} + \Phi_{x=1} + \Phi_{z=0} + \Phi_{z=1} = -2 - 22 + 0 + 0 + 0 + 0 = -24 \, \text{Wb} \]

5. **Use Gauss's Law to Find the Enclosed Charge**:
According to Gauss's law:
\[ \Phi = \frac{Q_{enclosed}}{\epsilon_0} \]
Rearranging gives:
\[ Q_{enclosed} = \Phi \cdot \epsilon_0 \]
Substituting the values:
\[ Q_{enclosed} = -24 \times (8.85 \times 10^{-12}) = -2.124 \times 10^{-10} \, \text{C} \]

6. **Express the Charge in the Required Form**:
The problem states that the net charge enclosed is \(x \times 10^{-10} \, \text{C}\). Thus:
\[ x = -2.124 \]

### Final Answer:
The value of \(x\) is approximately \(-2.124\).