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, we need to calculate the electric flux through each face of a cube that encloses a charge. Let's break this down step by step.

### Step 1: Understand the problem
We have a cube with a side length of 10 cm and a charge of \(0.1 \, \mu C\) (microcoulombs) at its center. We need to find the number of electric field lines (or electric flux) passing through each face of the cube.

### Step 2: Convert the charge to coulombs
The charge \(Q\) is given as \(0.1 \, \mu C\). We need to convert this into coulombs:
\[ Q = 0.1 \, \mu C = 0.1 \times 10^{-6} \, C = 1 \times 10^{-7} \, C \]

### Step 3: Use Gauss's Law
According to Gauss's Law, the total electric flux \(\Phi\) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \(Q_{\text{enc}}\) is the enclosed charge and \(\epsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, C^2/(N \cdot m^2)\).

### Step 4: Calculate the total electric flux through the cube
Substituting the values into Gauss's Law:
\[ \Phi = \frac{1 \times 10^{-7} \, C}{8.85 \times 10^{-12} \, C^2/(N \cdot m^2)} \]
Calculating this gives:
\[ \Phi \approx 1.13 \times 10^4 \, N \cdot m^2/C \]

### Step 5: Calculate the flux through each face of the cube
Since the cube has 6 faces and the charge is symmetrically placed at the center, the flux will be evenly distributed through each face. Therefore, the flux through one face \(\Phi_{\text{face}}\) is:
\[ \Phi_{\text{face}} = \frac{\Phi}{6} = \frac{1.13 \times 10^4}{6} \approx 1888.33 \, N \cdot m^2/C \]

### Final Answer
The number of lines of force (or electric flux) through each face of the cube is approximately \(1888.33 \, N \cdot m^2/C\).

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