Gauss's Law
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| Gauss's law/ˈɡaʊsɪz lɔː/ | định luật Gauss |
| permittivity/ˌpɜːmɪˈtɪvəti/ | hằng số điện môi |
| Gaussian surface/ˈɡaʊʃn ˈsɜːfɪs/ | mặt Gauss |
Turn a nasty integral into one line
- Finding $E$ from a big charged object by integrating is hard work.
- Gauss's law 高斯定律 offers a shortcut — when the shape is symmetric.
- It links the flux through a closed surface to the charge trapped inside.
- With the right surface, you read $E$ off in a single step.
The law itself
- For any closed surface: $\oint \vec E \cdot d\vec A = \dfrac{Q_{\text{enc}}}{\varepsilon_0}$.
- The left side is the total flux; the right is the enclosed charge over $\varepsilon_0$.
- $\varepsilon_0 = 8.85\times10^{-12}$ is the permittivity 介电常数 of free space.
- Only charge inside the surface counts.
Flux from a charge
Watch the field lines from a charge spread out and cross any surface around it.
In Gauss's law, the flux through a closed surface equals:
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$\oint \vec E \cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$.
Outside charge cancels
- A charge outside the surface sends as much flux in as out.
- So its net contribution to the closed-surface flux is zero.
- This is why only $Q_{\text{enc}}$ appears on the right.
- The field on the surface still feels outside charge — but the flux doesn't.
A charge outside a closed surface adds to the net flux through it.
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False. Field lines from an outside charge enter and leave the surface equally, so the net flux is zero; only enclosed charge counts (Gauss's law).
A surface encloses $+6\ \text{nC}$ and $-2\ \text{nC}$. What net charge (in nC) sets its flux?
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Only enclosed charge counts: $6 + (-2) = 4\ \text{nC}$.
Pick a surface that matches the symmetry
- Sphere of charge → use a spherical Gaussian surface 高斯面.
- Line of charge → use a coaxial cylinder.
- Plane of charge → use a box (pillbox) through it.
- Choose it so $E$ is constant and either along or across each patch.
The imaginary closed surface you integrate over is called a ____ surface.
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It is a Gaussian surface, chosen to match the symmetry.
Select all true statements about Gauss's law.
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Always true, best with symmetry, driven by enclosed charge — over a closed surface.
Powerful results, fast
- Outside a charged sphere: $E = \dfrac{kQ}{r^2}$ — exactly like a point charge.
- Around an infinite line: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$.
- Near an infinite plane: $E = \dfrac{\sigma}{2\varepsilon_0}$ (constant).
- Each drops out of Gauss's law in a couple of lines.
Outside a uniformly charged sphere of charge $Q$, the field is:
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Gauss's law gives $E = kQ/r^2$ outside — identical to a point charge.
Find $E$ a distance $r$ outside a sphere holding charge $Q$.
- A spherical Gaussian surface gives $\oint E\,dA = E(4\pi r^2)$.
- Set it equal to $Q/\varepsilon_0$: $E = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{kQ}{r^2}$.
Gauss's law is always true, but only useful when symmetry makes $E$ constant on your surface. For a lopsided charge you can still write $\oint \vec E \cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$, but you can't pull $E$ out of the integral.
Gauss's law says $\oint \vec E \cdot d\vec A = Q_{\text{enc}}/\varepsilon_0$ — flux through a closed surface depends only on the enclosed charge. Choose a Gaussian surface matching the symmetry (sphere, cylinder, plane) to read $E$ off in one step.
Write from a shell field of 600000 newtons per coulomb
- A spherical shell holds 6 microcoulomb. A point outside is 0.30 m from the centre. Use k = 9 times 10 to the 9.
- Find the outside field. Say the field inside the empty shell.
A spherical shell holds 6 microcoulomb. Find the field 0.30 m from the centre, outside the shell, using k = 9 times 10 to the 9. What is the field inside the shell if the cavity is empty?
9 times 10 to the 9 times 6 times 10 to the minus 6, divided by 0.09, is 600000. Gauss's law gives 0 inside an empty shell.
What is the field outside and inside this shell?
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Outside, 54000 divided by 0.09 is 600000. Inside, enclosed charge is 0.