Dielectrics · 电介质
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| dielectric/ˌdaɪɪˈlektrɪk/ | 电介质 | diàn jiè zhì |
| polarised/ˈpəʊləraɪzd/ | 极化 | jí huà |
| dielectric constant/ˌdaɪɪˈlektrɪk ˈkɒnstənt/ | 介电常数 | jiè diàn cháng shù |
Slide an insulator between the plates and it holds even more
- A real capacitor rarely has plain air between its plates.
- Slip an insulating slab — a dielectric 电介质 — into the gap.
- Suddenly the same capacitor stores more charge at the same voltage.
- One cheap slab boosts capacitance and lets it survive higher voltages.
把绝缘体滑进两板之间,它持有的更多
- 真实电容器的两板之间很少是纯空气。
- 把一块绝缘板——一个电介质——塞进间隙里。
- 突然,同一个电容器在相同电压下储存更多电荷。
- 一块便宜的板既提升电容,又让它耐受更高的电压。
The dielectric polarises
- The plates' field tugs on the slab's molecules, lining them up.
- Each molecule becomes a tiny $+/-$ pair — the slab is polarised 极化.
- These lined-up charges make their own field against the plates' field.
- The net field between the plates gets weaker.
电介质被极化
- 两板的场拉扯板里的分子,把它们排列整齐。
- 每个分子变成一个微小的 $+/-$ 对——这块板被极化。
- 这些排齐的电荷产生自己的场,对抗两板的场。
- 两板之间的净场变得更弱。
A dielectric polarises and weakens the field between the plates. · 电介质极化并减弱极板间的电场。
Its aligned charges oppose the plates' field, weakening the net field. · 其排列的电荷对抗极板电场，削弱净电场。
Capacitance goes up by κ
- A weaker field means a smaller voltage for the same charge — so $C$ rises.
- $C = \kappa C_0$, where $\kappa$ is the dielectric constant 介电常数 ($\kappa > 1$).
- Air is about $1$; plastics are $2$–$5$; some ceramics reach into the hundreds.
- Insert a $\kappa = 3$ slab and the capacitance triples.
电容增大 κ 倍
- 更弱的场意味着相同电荷对应更小的电压——于是 $C$ 上升。
- $C = \kappa C_0$,其中 $\kappa$ 是相对介电常数($\kappa > 1$)。
- 空气约为 $1$;塑料是 $2$–$5$;某些陶瓷高达几百。
- 插入一块 $\kappa = 3$ 的板,电容就变三倍。
Add a dielectric · 插入电介质
See how storing more charge at the same voltage means a larger capacitance. · 观察在同一电压下储存更多电荷意味着更大的电容。
Inserting a dielectric of constant $\kappa$ changes the capacitance to: · 插入介电常数为 $\kappa$ 的电介质会将电容改变为：
$C = \kappa C_0$, and $\kappa > 1$, so capacitance rises. · $C = \kappa C_0$，且 $\kappa > 1$，因此电容增大。
A $5\ \mu\text{F}$ capacitor gets a $\kappa = 4$ dielectric. Find the new capacitance (in μF). · 一个 $5\ \mu\text{F}$ 电容器放入 $\kappa = 4$ 电介质。求新电容（单位：μF）。
$C = \kappa C_0 = 4 \times 5 = 20\ \mu\text{F}$.
For any real dielectric, the constant $\kappa$ is always greater than ____. · 对于任何真实电介质，常数 $\kappa$ 总是大于 ____。
$\kappa > 1$ for all dielectrics, so they always raise $C$. · $\kappa > 1$ 适用于所有电介质，所以它们总是提高 $C$。
Three helpful effects
- Bigger $C$ — more charge stored per volt.
- Weaker field inside for the same charge — less chance of sparking.
- Higher breakdown voltage — the capacitor tolerates more before it arcs.
- The slab also physically holds the plates apart.
三个有用的效果
- 更大的 $C$——每伏储存更多电荷。
- 相同电荷下内部场更弱——更不容易打火。
- 更高的击穿电压——电容器在起弧前能承受更多。
- 这块板还在物理上把两板隔开。
Select all · 所有 true effects of adding a dielectric. · 选择添加电介质的 所有 正确效应。
A dielectric raises C, weakens the field, and raises breakdown voltage. · 电介质提高 C，减弱电场，并提高击穿电压。
Two cases: battery on or off
- Battery connected ($V$ fixed): $C$ rises, so charge $Q = CV$ increases.
- Battery removed ($Q$ fixed): $C$ rises, so voltage $V = Q/C$ drops.
- Ask which quantity is held fixed before predicting the change.
- Same slab, opposite effects on $V$ and $Q$.
两种情形:电池接着或断开
- 电池接着($V$ 固定):$C$ 上升,所以电荷 $Q = CV$ 增大。
- 电池断开($Q$ 固定):$C$ 上升,所以电压 $V = Q/C$ 下降。
- 在预测变化之前,先问哪个量被固定。
- 同一块板,对 $V$ 和 $Q$ 有相反的效果。
A dielectric is inserted while the battery stays connected ($V$ fixed). The stored charge: · 在电池保持连接（$V$ 固定）的情况下插入电介质。储存的电荷：
With $V$ fixed and $C$ up, $Q = CV$ increases. · 随着$V$固定且$C$增加，$Q = CV$增加。
A capacitor has $C_0 = 4\ \mu\text{F}$. A dielectric of $\kappa = 3$ is inserted.
- New capacitance: $C = \kappa C_0 = 3 \times 4 = 12\ \mu\text{F}$.
- If still connected to the battery, the stored charge triples too.
一个电容器 $C_0 = 4\ \mu\text{F}$。插入一块 $\kappa = 3$ 的电介质。
- 新电容:$C = \kappa C_0 = 3 \times 4 = 12\ \mu\text{F}$。
- 若仍接着电池,储存的电荷也变三倍。
Before predicting $V$ or $Q$, decide whether the battery is still connected. Battery on → $V$ fixed, $Q$ rises. Battery off → $Q$ fixed, $V$ falls. Mixing these two cases is the classic dielectric mistake.
在预测 $V$ 或 $Q$ 之前,先判断电池是否仍接着。电池接着 → $V$ 固定,$Q$ 上升。电池断开 → $Q$ 固定,$V$ 下降。混淆这两种情形是经典的电介质错误。
A dielectric slab polarises, weakening the field, so capacitance rises: $C = \kappa C_0$ with dielectric constant $\kappa > 1$. It also raises the breakdown voltage. With the battery connected $Q$ rises ($V$ fixed); disconnected $V$ falls ($Q$ fixed).
一块电介质板发生极化,削弱场,于是电容上升:$C = \kappa C_0$,相对介电常数 $\kappa > 1$。它还提高击穿电压。电池接着时 $Q$ 上升($V$ 固定);断开时 $V$ 下降($Q$ 固定)。