An electric field given by `vec(E )= 4hat(i)+3(y^(2)+2)hat(j)` passes through Gaussian cube of side `1m` placed at origin such that its three sides represents `x,y` and `z` axis. This net charge enclosed within the cube is

Options

A. `4 epsilon_(0)`
B. `3 epsilon_(0)`
C. `4 epsilon_(0)`
D. zero

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Electric Field
The electric field is given as:
\[ \vec{E} = 4\hat{i} + 3(y^2 + 2)\hat{j} \]
This means that the electric field has a constant component in the x-direction (4 N/C) and a variable component in the y-direction that depends on \(y\).

### Step 2: Identify the Gaussian Surface
We have a Gaussian cube of side 1 m placed at the origin. The cube's sides are aligned with the x, y, and z axes. Therefore, the cube extends from \(x = 0\) to \(x = 1\), \(y = 0\) to \(y = 1\), and \(z = 0\) to \(z = 1\).

### Step 3: Calculate the Electric Flux
The electric flux \(\Phi\) through a closed surface is given by:
\[ \Phi = \iint_S \vec{E} \cdot d\vec{A} \]
Where \(d\vec{A}\) is the area vector of the surface.

### Step 4: Analyze Each Face of the Cube
1. **Faces parallel to the x-axis** (at \(x=0\) and \(x=1\)):
- At \(x=0\): \(\vec{E} = 4\hat{i} + 3(y^2 + 2)\hat{j}\)
- At \(x=1\): \(\vec{E} = 4\hat{i} + 3(y^2 + 2)\hat{j}\)
- The flux through these faces cancels out because the electric field is constant in the x-direction.

2. **Faces parallel to the y-axis** (at \(y=0\) and \(y=1\)):
- At \(y=0\): \(\vec{E} = 4\hat{i} + 3(0^2 + 2)\hat{j} = 4\hat{i} + 6\hat{j}\)
- At \(y=1\): \(\vec{E} = 4\hat{i} + 3(1^2 + 2)\hat{j} = 4\hat{i} + 9\hat{j}\)
- The flux through the face at \(y=0\) is:
\[ \Phi_{y=0} = \vec{E} \cdot d\vec{A} = (4\hat{i} + 6\hat{j}) \cdot (0\hat{i} + 1\hat{j}) = 6 \text{ (entering)} \]
Thus, \(\Phi_{y=0} = -6\) (as it enters).
- The flux through the face at \(y=1\) is:
\[ \Phi_{y=1} = \vec{E} \cdot d\vec{A} = (4\hat{i} + 9\hat{j}) \cdot (0\hat{i} + 1\hat{j}) = 9 \text{ (exiting)} \]
Thus, \(\Phi_{y=1} = 9\).

3. **Faces parallel to the z-axis** (at \(z=0\) and \(z=1\)):
- There is no z-component of the electric field, so the flux through these faces is zero.

### Step 5: Calculate the Net Flux
Now, we can calculate the net flux through the cube:
\[ \Phi_{net} = \Phi_{y=0} + \Phi_{y=1} = -6 + 9 = 3 \]

### Step 6: Relate Flux to Charge Enclosed
According to Gauss's law:
\[ \Phi = \frac{Q_{enclosed}}{\epsilon_0} \]
Where \(\epsilon_0\) is the permittivity of free space. Therefore, we can find the enclosed charge:
\[ Q_{enclosed} = \Phi \cdot \epsilon_0 = 3\epsilon_0 \]

### Final Answer
The net charge enclosed within the cube is:
\[ Q_{enclosed} = 3\epsilon_0 \]