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 | \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) |
