the net electric flux through each face of a die (singular of dice) has a magnitude in units of `10^(3)N.m^(2)//C` that is exactly equal to the number of sport N on the face ( 1 through 6) the flux is inward for N odd out word for N even. What is the net charge inside the die ?

Answer

To find the net charge inside the die based on the given electric flux through each face, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electric Flux**:
The electric flux (Φ) through each face of the die is given in units of \(10^3 \, \text{N m}^2/\text{C}\) and corresponds to the number of spots (N) on that face. For odd N, the flux is inward (negative), and for even N, it is outward (positive).

2. **Calculating the Total Electric Flux**:
We need to calculate the total electric flux through all six faces of the die:
- Face 1: \(Φ_1 = -1 \times 10^3 \, \text{N m}^2/\text{C}\) (inward)
- Face 2: \(Φ_2 = +2 \times 10^3 \, \text{N m}^2/\text{C}\) (outward)
- Face 3: \(Φ_3 = -3 \times 10^3 \, \text{N m}^2/\text{C}\) (inward)
- Face 4: \(Φ_4 = +4 \times 10^3 \, \text{N m}^2/\text{C}\) (outward)
- Face 5: \(Φ_5 = -5 \times 10^3 \, \text{N m}^2/\text{C}\) (inward)
- Face 6: \(Φ_6 = +6 \times 10^3 \, \text{N m}^2/\text{C}\) (outward)

Now, summing these values:
\[ \text{Total Flux} = (-1 + 2 - 3 + 4 - 5 + 6) \times 10^3 = (3) \times 10^3 \, \text{N m}^2/\text{C} \]

3. **Using Gauss's Law**:
According to Gauss's Law, the total electric flux through a closed surface is equal to the net charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀):
\[ \Phi = \frac{Q}{\epsilon_0} \]
Rearranging gives:
\[ Q = \Phi \cdot \epsilon_0 \]

4. **Calculating the Net Charge**:
We know:
- \(\Phi = 3 \times 10^3 \, \text{N m}^2/\text{C}\)
- The permittivity of free space, \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2\)

Substituting these values into the equation for Q:
\[ Q = (3 \times 10^3) \cdot (8.85 \times 10^{-12}) \]
\[ Q = 2.655 \times 10^{-8} \, \text{C} \]
Rounding gives:
\[ Q \approx 2.7 \times 10^{-8} \, \text{C} \]

### Final Answer:
The net charge inside the die is approximately \(2.7 \times 10^{-8} \, \text{C}\).