Eight charges, 1muC, -7muC , -4muC, 10muC, 2muC, -5muC, -3muC, and 6muC`, are situated at the eight cornersof a cube of side 20 cm. A spherical surface of radius 80 cm encloses this cube. The center of the sphere coincides with the center of the cube. Then, the total outgoing flux from the spherical surface (in units of Vm) is

Options

A. `36 pi xx 10^3`
B. `684 pi xx 10^3`
C. zero
D. none of these

Explanation

To solve the problem, we need to calculate the total outgoing electric flux from a spherical surface that encloses a cube with charges located at its corners. We will follow these steps:

### Step 1: Identify the Charges
The charges located at the corners of the cube are:
- \( Q_1 = 1 \mu C \)
- \( Q_2 = -7 \mu C \)
- \( Q_3 = -4 \mu C \)
- \( Q_4 = 10 \mu C \)
- \( Q_5 = 2 \mu C \)
- \( Q_6 = -5 \mu C \)
- \( Q_7 = -3 \mu C \)
- \( Q_8 = 6 \mu C \)

### Step 2: Calculate the Total Charge Enclosed
We need to sum all the charges to find the total charge enclosed within the spherical surface.

\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 + Q_6 + Q_7 + Q_8 \]

Substituting the values:

\[ Q_{\text{total}} = 1 + (-7) + (-4) + 10 + 2 + (-5) + (-3) + 6 \]

Calculating step-by-step:

1. \( 1 - 7 = -6 \)
2. \( -6 - 4 = -10 \)
3. \( -10 + 10 = 0 \)
4. \( 0 + 2 = 2 \)
5. \( 2 - 5 = -3 \)
6. \( -3 - 3 = -6 \)
7. \( -6 + 6 = 0 \)

So, the total charge enclosed \( Q_{\text{total}} = 0 \mu C \).

### Step 3: Apply Gauss's Law
According to Gauss's Law, the electric flux \( \Phi \) through a closed surface is given by:

\[ \Phi = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]

Where \( \epsilon_0 \) is the permittivity of free space. Since \( Q_{\text{total}} = 0 \):

\[ \Phi = \frac{0}{\epsilon_0} = 0 \]

### Conclusion
The total outgoing electric flux from the spherical surface is \( 0 \, \text{Vm} \).

### Final Answer
The total outgoing flux from the spherical surface is \( 0 \, \text{Vm} \).

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