A point charge `4muC` is at the centre of a cubic Gaussian surface 10 cm on edge. Net electric flux through the surface is

Options

A. `2.5xx10^(5)Nm^(2)C^(-1)`
B. `4.5xx10^(5)Nm^(2)C^(-1)`
C. `4.5xx10^(6)Nm^(2)C^(-1)`
D. `2.5xx10^(6)Nm^(2)C^(-1)`

Explanation

To find the net electric flux through a cubic Gaussian surface with a point charge at its center, we can use Gauss's Law. Here is the step-by-step solution:

### Step 1: Identify the Given Values
- Point charge, \( Q = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- The edge length of the cube is given, but it is not needed for calculating the electric flux.

### Step 2: Recall Gauss's Law
Gauss's Law states that the electric flux \( \Phi_E \) through a closed surface is equal to the charge enclosed \( Q_{enc} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]

### Step 3: Substitute the Values into Gauss's Law
We know:
- \( Q_{enc} = 4 \times 10^{-6} \, C \)
- The permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)

Now substituting these values into Gauss's Law:
\[ \Phi_E = \frac{4 \times 10^{-6}}{8.85 \times 10^{-12}} \]

### Step 4: Calculate the Electric Flux
Now we perform the calculation:
\[ \Phi_E = \frac{4 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 4.52 \times 10^5 \, N \cdot m^2/C \]

### Step 5: Final Result
Thus, the net electric flux through the surface is:
\[ \Phi_E \approx 4.52 \times 10^5 \, N \cdot m^2/C \]