Figure shows a Gaussian surface in the shape of a cube with edge length 1.40 m. What are:
- The net flux F through the surface
- The net charge qenc enclosed by the surface if E = (3.00yj) N/C, with y in meters
- F
- qenc if E = [– 4.00i + (6.00 + 3.00y)j] N/C
The Correct Answer and Explanation is:
To solve this problem, we need to use Gauss’s Law, which states: ΦE=qencε0\Phi_E = \frac{q_{\text{enc}}}{\varepsilon_0}
Where:
- ΦE\Phi_E is the net electric flux through a closed surface,
- qencq_{\text{enc}} is the net charge enclosed by the surface,
- ε0=8.854×10−12 C2/N\cdotpm2\varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2/\text{N·m}^2 is the permittivity of free space.
Part 1: E = (3.00y) j N/C
This is a non-uniform electric field in the y-direction that increases linearly with yy. The cube has edge length L=1.40 mL = 1.40\ \text{m}, and we assume its sides are aligned with the axes and it’s centered in space.
Only the top and bottom faces contribute to the net flux since the electric field is in the y-direction. The field through side faces in xx and zz directions contributes zero net flux.
- Bottom face (y = 0): Ebottom=0E_{\text{bottom}} = 0
- Top face (y = 1.40 m): Etop=3.00×1.40=4.20 N/CE_{\text{top}} = 3.00 \times 1.40 = 4.20\ \text{N/C}
Each face has area: A=(1.40)2=1.96 m2A = (1.40)^2 = 1.96\ \text{m}^2
So, net flux: Φ=Etop⋅A−Ebottom⋅A=4.20⋅1.96−0=8.232 N\cdotpm2/C\Phi = E_{\text{top}} \cdot A – E_{\text{bottom}} \cdot A = 4.20 \cdot 1.96 – 0 = 8.232\ \text{N·m}^2/\text{C}
Net enclosed charge: qenc=Φ⋅ε0=8.232⋅8.854×10−12≈7.29×10−11 Cq_{\text{enc}} = \Phi \cdot \varepsilon_0 = 8.232 \cdot 8.854 \times 10^{-12} \approx \boxed{7.29 \times 10^{-11}\ \text{C}}
Part 2: E = [–4.00 i + (6.00 + 3.00y) j] N/C
This field has x- and y-components. However:
- The x-component is constant (−4.00 N/C)(-4.00\ \text{N/C}), so the flux entering and leaving the sides cancel.
- The y-component is variable: Ey=6.00+3.00yE_y = 6.00 + 3.00y
As before, compute flux only through top and bottom faces:
- At bottom (y = 0): Ebottom=6.00E_{\text{bottom}} = 6.00
- At top (y = 1.40): Etop=6.00+3.00⋅1.40=10.20E_{\text{top}} = 6.00 + 3.00 \cdot 1.40 = 10.20
So: Φ=Etop⋅A−Ebottom⋅A=(10.20−6.00)⋅1.96=4.20⋅1.96=8.232 N\cdotpm2/C\Phi = E_{\text{top}} \cdot A – E_{\text{bottom}} \cdot A = (10.20 – 6.00) \cdot 1.96 = 4.20 \cdot 1.96 = 8.232\ \text{N·m}^2/\text{C}
Same as before. qenc=8.232⋅8.854×10−12=7.29×10−11 Cq_{\text{enc}} = 8.232 \cdot 8.854 \times 10^{-12} = \boxed{7.29 \times 10^{-11}\ \text{C}}
Summary Answers
| Case | Net Flux ΦE\Phi_E (N·m²/C) | Enclosed Charge qencq_{\text{enc}} (C) |
|---|---|---|
| 1: E=3.00yj^E = 3.00y \hat{j} | 8.23\boxed{8.23} | 7.29×10−11\boxed{7.29 \times 10^{-11}} |
| 2: E=−4.00i^+(6.00+3.00y)j^E = -4.00\hat{i} + (6.00 + 3.00y)\hat{j} | 8.23\boxed{8.23} | 7.29×10−11\boxed{7.29 \times 10^{-11}} |
In both cases, the net flux and charge enclosed are the same, because only the y-component of the field varies within the cube and contributes to net flux.
