Step 1: Understanding the Concept:
This problem is a direct application of Gauss's Law of electrostatics. Gauss's Law states that the total electric flux (\(\Phi\)) through any closed surface is equal to the total net electric charge (\(Q_{\text{in}}\)) enclosed within that surface, divided by the permittivity of free space (\(\epsilon_0\)).
Step 2: Key Formula or Approach:
According to Gauss's Law:
\[ \Phi = \frac{Q_{\text{in}}}{\epsilon_0} \]
To find the total charge inside the cube, we can rearrange the formula:
\[ Q_{\text{in}} = \Phi \times \epsilon_0 \]
Step 3: Detailed Explanation:
Given:
Net electric flux, \( \Phi = 1.05 \, \text{N m}^2 \text{C}^{-1} \)
Permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)
Now, we calculate the total charge enclosed, \(Q_{\text{in}}\):
\[ Q_{\text{in}} = (1.05 \, \text{N m}^2 \text{C}^{-1}) \times (8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2}) \]
\[ Q_{\text{in}} = 9.2925 \times 10^{-12} \, \text{C} \]
Rounding to two decimal places, we get:
\[ Q_{\text{in}} \approx 9.29 \times 10^{-12} \, \text{C} \]
Step 4: Final Answer:
The total charge inside the cube will be \(9.29 \times 10^{-12}\) C.

| LIST I | LIST II | ||
| A | Gauss's Law in Electrostatics | I | \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\) |
| B | Faraday's Law | II | \(\oint \vec{B} \cdot d \vec{A}=0\) |
| C | Gauss's Law in Magnetism | III | \(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\) |
| D | Ampere-Maxwell Law | IV | \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) |

Fill in the blank with the correct option.
The teacher believed that the student’s sudden lack of interest in class was an ..........., as he had always been enthusiastic and attentive.