A cube has sides of length `L = 0.300 m`. It is placed with one corner at the origin ( refer to Fig.2.119).The electric field is not uniform but is given by `vec( E) = ( -5.00 NC^(-1)) x hat(i) + (3.00 NC^(-1)) z hat(k)`.
The total electric charge inside the cube is

Options

A. `- 0.054 epsilon_(0) C`
B. `0.081 epsilon_(0) C`
C. `0.135 epsilon_(0) C`
D. ` 0.054 epsilon_(0) C`

Explanation

To find the total electric charge inside the cube given the electric field, we can use Gauss's law, which states:

\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]

where \(\Phi_E\) is the electric flux through a closed surface, \(Q_{enc}\) is the total charge enclosed by the surface, and \(\epsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\).

### Step 1: Calculate the Electric Flux through Each Face of the Cube

The electric field is given by:

\[ \vec{E} = (-5.00 \, \text{N/C}) \hat{i} + (3.00 \, \text{N/C}) \hat{k} \]

The cube has 6 faces, and we need to calculate the electric flux through each face.

#### Face 1 (x = 0):
- Area vector \(\vec{A_1} = L^2 \hat{i}\)
- \(\Phi_1 = \vec{E} \cdot \vec{A_1} = (-5.00) \cdot (0.3^2) = -0.45 \, \text{N m}^2/\text{C}\)

#### Face 2 (x = L):
- Area vector \(\vec{A_2} = -L^2 \hat{i}\)
- \(\Phi_2 = \vec{E} \cdot \vec{A_2} = (-5.00) \cdot (0.3^2) = -0.45 \, \text{N m}^2/\text{C}\)

#### Face 3 (y = 0):
- Area vector \(\vec{A_3} = L^2 \hat{j}\)
- \(\Phi_3 = \vec{E} \cdot \vec{A_3} = 0\) (since \(\hat{j}\) is perpendicular to \(\vec{E}\))

#### Face 4 (y = L):
- Area vector \(\vec{A_4} = -L^2 \hat{j}\)
- \(\Phi_4 = \vec{E} \cdot \vec{A_4} = 0\)

#### Face 5 (z = 0):
- Area vector \(\vec{A_5} = L^2 \hat{k}\)
- \(\Phi_5 = \vec{E} \cdot \vec{A_5} = (3.00) \cdot (0.3^2) = 0.27 \, \text{N m}^2/\text{C}\)

#### Face 6 (z = L):
- Area vector \(\vec{A_6} = -L^2 \hat{k}\)
- \(\Phi_6 = \vec{E} \cdot \vec{A_6} = (3.00) \cdot (0.3^2) = 0.27 \, \text{N m}^2/\text{C}\)

### Step 2: Sum the Electric Flux

Now, we sum the flux through all the faces:

\[ \Phi_E = \Phi_1 + \Phi_2 + \Phi_3 + \Phi_4 + \Phi_5 + \Phi_6 \]

\[ \Phi_E = (-0.45) + (-0.45) + 0 + 0 + 0.27 + 0.27 = -0.36 \, \text{N m}^2/\text{C} \]

### Step 3: Use Gauss's Law to Find the Charge

Using Gauss's law:

\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]

Rearranging gives:

\[ Q_{enc} = \Phi_E \cdot \epsilon_0 \]

Substituting the values:

\[ Q_{enc} = -0.36 \cdot (8.85 \times 10^{-12}) = -3.19 \times 10^{-12} \, \text{C} \]

### Final Answer

The total electric charge inside the cube is approximately:

\[ Q_{enc} \approx -3.19 \, \text{pC} \]