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 | \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) |
Match List-I with List-II
\[\begin{array}{|l|l|} \hline \text{List-I (Soil component)} & \text{List-II (Definition)} \\ \hline (A)~\text{Azonal soils} & (I)~\text{An individual natural aggregate of soil particles.} \\ (B)~\text{Regoliths} & (II)~\text{Organisms living in the soil or ground} \\ (C)~\text{Ped} & (III)~\text{Soils have uniformity from the top-surface to the base, and do not have well-developed soil horizons.} \\ (D)~\text{Edaphons} & (IV)~\text{Zone of loose and unconsolidated weathered rock materials.} \\ \hline \end{array}\]
Choose the correct answer from the options given below:
Match List-I with List-II
\[\begin{array}{|l|l|} \hline \text{List I Content of humus} & \text{List II Percentage of contents} \\ \hline \text{(A) Carbon} & \text{(I) 35-40\%} \\ \hline \text{(B) Oxygen} & \text{(II) ~5\%} \\ \hline \text{(C) Hydrogen} & \text{(III) 55-60\%} \\ \hline \text{(D) Nitrogen} & \text{(IV) 15\%} \\ \hline \end{array}\]
Choose the correct answer from the options given below: