Electrostatics
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| Gauss's law/ˈɡaʊsɪz lɔː/ | Định luật Gauss |
| bound charge | bound charge |
| capacitance/kəˈpæsɪtəns/ | điện dung |
One law, two regions
- Gauss's law: $\oint E \cdot dA = Q_{enc}/\varepsilon_{0}$. For a sphere of charge, symmetry turns it into $E \cdot 4\pi r^{2} = Q_{enc}/\varepsilon_{0}$ — inside $E \propto r$, outside $E \propto 1/r^{2}$, equal at the surface.
$\rho = 1\ \mu$C/m³, $a = 2$ cm: at the surface $E = \rho a/3\varepsilon_{0} = 753$ V/m from BOTH formulas ✓; at $r = 4$ cm the outside form gives 188 V/m.
{exercise:uni-ee0721-sphere}
At twice the sphere radius (r = 2a), the field is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Outside the sphere E ∝ 1/r², so doubling r divides E by 4: 753/4 = 188 V/m.
Dielectrics and boundaries
- A medium with $\varepsilon_{r}$ divides the field by exactly $\varepsilon_{r}$; the missing lines end on bound charge $Q_{bound} = Q(1 - 1/\varepsilon_{r})$.
- At an interface: tangential E continuous, normal D jumps by free surface charge. The isolated sphere's capacitance $C = 4\pi\varepsilon_{0}a$ (0.1 pF/cm).
{exercise:uni-ee0722-dielectric}
Embedding the sphere in a medium with εr = 4 changes the outside field to —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Only ε changed in the law, so E everywhere divides by 4; Q_bound = Q(1 − 1/4) = 0.75Q hides just under the surface.
The capacitance of an isolated sphere of radius 1 cm is about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
C = 4πε0 a = 4π × 8.854e-12 × 0.01 = 1.11 pF — capacitance is geometry, not charge.