A uniform charge density of `500 nC//m^(3)` is distributed throughout a spherical volume of radius 6.00cm .Consider a cubical Gaussian surface with its center at the centre of the sphere. What is the electric flux through this cubical surface if its edge length is (a) 4.00cm and (b) 14.0 cm ?

Answer

To solve the problem, we will calculate the electric flux through a cubical Gaussian surface with two different edge lengths (4.00 cm and 14.0 cm) when a uniform charge density is distributed throughout a spherical volume.

### Given Data:
- Charge density, ρ = 500 nC/m³ = \(500 \times 10^{-9} \, C/m³\)
- Radius of the sphere, R = 6.00 cm = \(0.06 \, m\)
- Edge length of the cube for part (a), a = 4.00 cm = \(0.04 \, m\)
- Edge length of the cube for part (b), b = 14.00 cm = \(0.14 \, m\)

### Step-by-Step Solution:

#### Part (a): Edge Length = 4.00 cm

1. **Calculate the Volume of the Cube:**
\[ V_{\text{cube}} = a^3 = (0.04 \, m)^3 = 0.000064 \, m^3 = 64 \, cm^3 \]

2. **Calculate the Charge Enclosed by the Cube:**
\[ Q_{\text{enclosed}} = V_{\text{cube}} \times \rho = 0.000064 \, m^3 \times 500 \times 10^{-9} \, C/m^3 \]
\[ Q_{\text{enclosed}} = 0.000064 \times 500 \times 10^{-9} = 3.2 \times 10^{-11} \, C \]

3. **Use Gauss's Law to Calculate Electric Flux:**
\[ \Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
where \(\epsilon_0 = 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)\).
\[ \Phi_E = \frac{3.2 \times 10^{-11}}{8.85 \times 10^{-12}} \approx 3.61 \, N \cdot m^2/C \]

#### Part (b): Edge Length = 14.0 cm

1. **Since the cube's edge length exceeds the sphere's radius, the charge enclosed will be the total charge in the sphere.**

2. **Calculate the Volume of the Sphere:**
\[ V_{\text{sphere}} = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (0.06 \, m)^3 \]
\[ V_{\text{sphere}} \approx \frac{4}{3} \pi (0.000216) \approx 0.000904 \, m^3 \]

3. **Calculate the Total Charge in the Sphere:**
\[ Q_{\text{total}} = V_{\text{sphere}} \times \rho = 0.000904 \, m^3 \times 500 \times 10^{-9} \, C/m^3 \]
\[ Q_{\text{total}} \approx 4.52 \times 10^{-20} \, C \]

4. **Use Gauss's Law to Calculate Electric Flux:**
\[ \Phi_E = \frac{Q_{\text{total}}}{\epsilon_0} \]
\[ \Phi_E = \frac{4.52 \times 10^{-20}}{8.85 \times 10^{-12}} \approx 0.51 \, N \cdot m^2/C \]

### Final Answers:
- (a) Electric flux through the cubical surface with edge length 4.00 cm: **3.61 N·m²/C**
- (b) Electric flux through the cubical surface with edge length 14.0 cm: **0.51 N·m²/C**