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 will use Gauss's law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀). The electric field given is:

\[ \vec{E} = 4 \hat{i} - (20y^2 + 2) \hat{j} \]

### Step 1: Determine the flux through each face of the cube

The cube is placed at the origin with sides along the x, y, and z axes, and has a side length of 1 m. We will calculate the electric flux through each face of the cube.

**Face 1 (ABCD):** (z = 0)
- The area vector \( \vec{A} = dA \hat{k} \)
- The electric field at z = 0 is \( \vec{E} = 4 \hat{i} - (20y^2 + 2) \hat{j} \)
- The flux through this face is:
\[ \Phi_{ABCD} = \int \vec{E} \cdot \vec{A} = \int_0^1 \int_0^1 (4 \hat{i} - (20y^2 + 2) \hat{j}) \cdot (dA \hat{k}) = 0 \]
Since the electric field is perpendicular to the area vector, the flux is zero.

**Face 2 (EFGH):** (z = 1)
- The area vector \( \vec{A} = dA (-\hat{k}) \)
- The electric field at z = 1 is still \( \vec{E} = 4 \hat{i} - (20y^2 + 2) \hat{j} \)
- The flux through this face is:
\[ \Phi_{EFGH} = \int \vec{E} \cdot \vec{A} = 0 \]
Again, the electric field is perpendicular to the area vector, so the flux is zero.

**Face 3 (ADHE):** (y = 0)
- The area vector \( \vec{A} = dA (-\hat{i}) \)
- The electric field at y = 0 is \( \vec{E} = 4 \hat{i} - 2 \hat{j} \)
- The flux through this face is:
\[ \Phi_{ADHE} = \int_0^1 \int_0^1 (4 \hat{i} - 2 \hat{j}) \cdot (-dA \hat{i}) = -\int_0^1 4 dA = -4 \]

**Face 4 (BCGF):** (y = 1)
- The area vector \( \vec{A} = dA \hat{j} \)
- The electric field at y = 1 is \( \vec{E} = 4 \hat{i} - 22 \hat{j} \)
- The flux through this face is:
\[ \Phi_{BCGF} = \int_0^1 \int_0^1 (4 \hat{i} - 22 \hat{j}) \cdot (dA \hat{j}) = -22 \cdot 1 = -22 \]

**Face 5 (ABFE):** (x = 0)
- The area vector \( \vec{A} = dA (-\hat{i}) \)
- The electric field at x = 0 is \( \vec{E} = 4 \hat{i} - 2 \hat{j} \)
- The flux through this face is:
\[ \Phi_{ABFE} = \int_0^1 \int_0^1 (4 \hat{i} - 2 \hat{j}) \cdot (-dA \hat{i}) = -\int_0^1 4 dA = -4 \]

**Face 6 (DCGH):** (x = 1)
- The area vector \( \vec{A} = dA \hat{i} \)
- The electric field at x = 1 is \( \vec{E} = 4 \hat{i} - 2 \hat{j} \)
- The flux through this face is:
\[ \Phi_{DCGH} = \int_0^1 \int_0^1 (4 \hat{i} - 2 \hat{j}) \cdot (dA \hat{i}) = 4 \cdot 1 = 4 \]

### Step 2: Calculate the total flux through the cube

Now, we sum the fluxes through all faces:
\[ \Phi_{total} = 0 + 0 - 4 - 22 - 4 + 4 = -26 \]

### Step 3: Use Gauss's Law to find the enclosed charge

According to Gauss's law:
\[ \Phi_{total} = \frac{Q_{enclosed}}{\epsilon_0} \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{N}^{-1} \text{m}^{-2} \text{C}^2 \).

Thus, we have:
\[ -26 = \frac{Q_{enclosed}}{8.85 \times 10^{-12}} \]
\[ Q_{enclosed} = -26 \times 8.85 \times 10^{-12} = -2.31 \times 10^{-10} \, \text{C} \]

### Step 4: Express the charge in the required format

The problem states that the charge is in the form \( x \times 10^{-10} \, \text{C} \). Therefore:
\[ x = -2.31 \]

### Final Answer

The value of \( x \) is \( -2.31 \).

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