A cube of side 10 cm encloses a charge of `0.1 muC` at its center. Calculate the number of lines of force through each face of the cube.

Options

A. `1.113 xx 10^(11)`
B. `1.13 xx 10^(4)`
C. `1.13 xx 10^(9)`
D. 1883

Explanation

To solve the problem of calculating the number of lines of force through each face of a cube enclosing a charge, we can follow these steps:

### Step 1: Understand the Concept of Electric Flux
Electric flux (Φ) through a closed surface is given by Gauss's Law, which states:
\[ \Phi = \frac{Q}{\varepsilon_0} \]
where \( Q \) is the total charge enclosed and \( \varepsilon_0 \) is the permittivity of free space.

### Step 2: Identify the Given Values
- The charge \( Q = 0.1 \, \mu C = 0.1 \times 10^{-6} \, C = 1 \times 10^{-7} \, C \)
- The permittivity of free space \( \varepsilon_0 = 8.854 \times 10^{-12} \, C^2/(N \cdot m^2) \)

### Step 3: Calculate the Total Electric Flux through the Cube
Using Gauss's Law:
\[ \Phi = \frac{Q}{\varepsilon_0} = \frac{1 \times 10^{-7}}{8.854 \times 10^{-12}} \]

Calculating this gives:
\[ \Phi \approx 1.13 \times 10^{4} \, N \cdot m^2/C \]

### Step 4: Calculate the Electric Flux through One Face of the Cube
Since the cube has 6 faces, the electric flux through each face is:
\[ \Phi_{\text{face}} = \frac{\Phi}{6} = \frac{1.13 \times 10^{4}}{6} \approx 1.88 \times 10^{3} \, N \cdot m^2/C \]

### Step 5: Calculate the Number of Lines of Force
The number of lines of force is proportional to the electric flux. Therefore, the number of lines of force through each face of the cube is approximately:
\[ \text{Number of lines of force through each face} \approx 1883 \]

### Final Answer
The number of lines of force through each face of the cube is approximately **1883**.

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