Electrostatics
Introduit| English | Sens |
|---|---|
| Gauss's law/ˈɡaʊsɪz lɔː/ | loi de Gauss |
| potential/pəˈtenʃl/ | The electric potential energy per unit charge at a point, measured in volts. |
| capacitance/kəˈpæsɪtəns/ | capacité |
| boundary conditions | Rules for how fields must match at the surface between two materials. |
The phone that feels a finger
- A touch screen knows where your finger is because the finger changes the electric field near the glass. Every capacitor, sensor and insulator is an electrostatics problem.
- For symmetric charges, one law turns the field into arithmetic. This lesson applies it to a charged sphere, inside and out.
Gauss's law
- Gauss's law 高斯定律 says the electric flux out of any closed surface equals the enclosed charge over $\varepsilon_{0}$: $\displaystyle\oint \mathbf{E} \cdot d\mathbf{A} = \dfrac{Q_{enc}}{\varepsilon_{0}}$.
- On a surface where the field is constant and normal, the integral is just $E \times A$. For a sphere of radius $r$: $E \cdot 4\pi r^{2} = \dfrac{Q_{enc}}{\varepsilon_{0}}$.
A 1 nC point charge. What is E at 30 cm, in V/m?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
$E = \dfrac{q}{4\pi\varepsilon_{0}r^{2}} = \dfrac{10^{-9} \times 8.99 \times 10^{9}}{0.09} = 99.9$ V/m.
Inside a uniformly charged sphere
- A sphere of radius $a$ and charge density $\rho$. Inside, at radius $r$, the enclosed charge grows as $r^{3}$ while the area grows only as $r^{2}$.
- So $E \cdot 4\pi r^{2} = \dfrac{\rho \cdot \tfrac{4}{3}\pi r^{3}}{\varepsilon_{0}}$, giving $E = \dfrac{\rho r}{3\varepsilon_{0}}$: zero at the centre, rising linearly to the surface.
The field at the centre of a uniformly charged sphere is zero.
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
$E = \dfrac{\rho r}{3\varepsilon_{0}}$ is zero at $r = 0$. By symmetry the pulls from all sides cancel there.
Outside: like a point charge
- Outside, all the charge is enclosed: $E = \dfrac{\rho a^{3}}{3\varepsilon_{0}r^{2}}$, the field of a point charge at the centre.
- At $r = a$ both formulas give $\dfrac{\rho a}{3\varepsilon_{0}}$, so the field is continuous at the surface.
The surface field of a charged sphere is 753 V/m. At twice the radius (r = 2a), the field is —
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
Outside, $E \propto \dfrac{1}{r^{2}}$, so doubling $r$ divides $E$ by 4: $\dfrac{753}{4} = 188$ V/m.
Worked example: the numbers
- $\rho = 1\,\mu\text{C/m}^{3}$, $a = 2\,\text{cm}$: at the surface, $E = \dfrac{10^{-6} \times 0.02}{3 \times 8.854 \times 10^{-12}} = 753\,\dfrac{\text{V}}{\text{m}}$.
- At $r = 4\,\text{cm}$, twice the radius, the field is a quarter: 188 V/m. At $r = 1\,\text{cm}$, inside, it is half: 377 V/m.
Same sphere (753 V/m at its surface). What is E at r = a/4, inside, in V/m?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
Inside, $E \propto r$: a quarter of the radius gives a quarter of the surface field, $\dfrac{753}{4} = 188$ V/m. The same value as at $2a$ outside.
Potential and capacitance
- The potential 电位 adds up the field from infinity: $V(r) = -\displaystyle\int_{\infty}^{r} \mathbf{E} \cdot d\mathbf{l}$. Capacitance 电容 is charge per volt, $C = \dfrac{Q}{V}$.
- An isolated sphere has $C = 4\pi\varepsilon_{0}a$, about 1.1 pF per centimetre of radius. That is why objects in open space barely couple.
The capacitance of an isolated sphere of radius 1 cm is about —
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
$C = 4\pi\varepsilon_{0}a = 4\pi \times 8.854 \times 10^{-12} \times 0.01 = 1.11$ pF. Capacitance depends on geometry, not on the charge.
A dielectric weakens the field
- Fill the space with a dielectric of relative permittivity $\varepsilon_{r}$ and the field drops by $\varepsilon_{r}$: $\varepsilon_{0}$ becomes $\varepsilon = \varepsilon_{0}\varepsilon_{r}$ in the law. With $\varepsilon_{r} = 4$ the surface value falls to 188 V/m.
- At a boundary between materials, the boundary conditions 边界条件 hold: tangential $E$ is continuous, and normal $D = \varepsilon E$ jumps only by free surface charge.
Embedding the sphere in a medium with εr = 4 changes the 753 V/m surface field to —
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
Only $\varepsilon$ changed in the law, so $E$ everywhere divides by 4. Bound charge in the dielectric partly screens the free charge.
Three common slips
- The wrong surface. A wire needs a cylinder, $2\pi rl$, not a sphere, $4\pi r^{2}$.
- Total charge inside the sphere. Gauss's law counts only the charge enclosed by the surface.
- Mixing $\varepsilon_{0}$ and $\varepsilon$. Choose one per problem.
Match each source to its Gauss's-law area.
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
Each area is the part of the Gaussian surface the field crosses; the ends of the cylinder and the sides of the pillbox carry no flux.
Recap
- Gauss's law with symmetry: $E \times A = \dfrac{Q_{enc}}{\varepsilon}$.
- Uniform sphere: $E \propto r$ inside, $E \propto \dfrac{1}{r^{2}}$ outside, equal at the surface.
- Isolated sphere: $C = 4\pi\varepsilon_{0}a$; a dielectric divides the field by $\varepsilon_{r}$.