Dielectrics
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| dielectric/ˌdaɪɪˈlektrɪk/ | điện môi |
| polarised/ˈpəʊləraɪzd/ | phân cực |
| dielectric constant/ˌdaɪɪˈlektrɪk ˈkɒnstənt/ | hằng số điện môi |
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.
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.
A dielectric polarises and weakens the field between the plates.
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Its aligned charges oppose the plates' field, weakening the net field.
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.
Add a dielectric
See how storing more charge at the same voltage means a larger capacitance.
Inserting a dielectric of constant $\kappa$ changes the capacitance to:
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$C = \kappa C_0$, and $\kappa > 1$, so capacitance rises.
A $5\ \mu\text{F}$ capacitor gets a $\kappa = 4$ dielectric. Find the new capacitance (in μF).
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$C = \kappa C_0 = 4 \times 5 = 20\ \mu\text{F}$.
For any real dielectric, the constant $\kappa$ is always greater than ____.
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$\kappa > 1$ for all dielectrics, so they always raise $C$.
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.
Select all true effects of adding a dielectric.
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A dielectric raises C, weakens the field, and raises breakdown voltage.
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 $Q$.
A dielectric is inserted while the battery stays connected ($V$ fixed). The stored charge:
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With $V$ fixed and $C$ up, $Q = CV$ increases.
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.
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.
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).
Write from a capacitance of 24 microfarad
- A capacitor has C0 = 6 microfarad. A dielectric with kappa = 4 fills it. It is disconnected from the battery, and the voltage was 12 volts.
- Find the new capacitance and the new voltage.
A 6 microfarad capacitor receives a dielectric of kappa 4 while disconnected, after being charged to 12 volts. What are the new capacitance and the new voltage?
4 times 6 is 24 microfarad. At fixed charge, voltage is divided by 4, so 12 volts becomes 3 volts.
What happens when this dielectric is inserted while disconnected?
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kappa times 6 microfarad is 24 microfarad, and 12 volts divided by 4 is 3 volts.