Capacitors
| English | Chinese | Pinyin |
|---|---|---|
| capacitor | 电容器 | diàn róng qì |
| capacitance | 电容 | diàn róng |
| farad | 法拉 | fǎ lā |
The device that saves up charge and dumps it in a flash
- A camera flash charges quietly, then releases its energy in a blink.
- The part that stores that charge is a capacitor 电容器.
- It is just two conductors holding equal and opposite charge.
- Capacitors show up in every circuit, timer, and power supply.
Capacitance = charge per volt
- Capacitance 电容 $C$ measures how much charge a capacitor holds per volt.
- $C = \dfrac{Q}{V}$, in farads 法拉 (1 F $= 1$ C/V).
- More charge for the same voltage means a bigger $C$.
- The two plates always carry $+Q$ and $-Q$ — equal and opposite.
Charge a capacitor
Change the voltage and plate spacing and watch the stored charge and field respond.
Capacitance is defined as:
$C = Q/V$, charge stored per volt, in farads.
The SI unit of capacitance is the ____.
1 farad $= 1$ C/V.
Geometry sets the capacitance
- For parallel plates: $C = \dfrac{\varepsilon_0 A}{d}$.
- Bigger plates (more area $A$) hold more charge — larger $C$.
- Closer plates (smaller gap $d$) hold more charge — larger $C$.
- $C$ depends on shape and size only, not on $Q$ or $V$.
To increase a parallel-plate capacitor's capacitance you can:
$C = \varepsilon_0 A/d$: a smaller gap $d$ raises $C$. Voltage does not change $C$.
The capacitance of a fixed pair of plates does not change when you raise the voltage.
$C$ depends on geometry only; higher $V$ just stores more $Q$.
Energy stored
- Charging a capacitor stores energy in its field: $U = \tfrac12 C V^2$.
- Equivalent forms: $U = \tfrac12 QV = \dfrac{Q^2}{2C}$.
- The flash dumps this energy fast to make a bright pulse.
- Double the voltage and you store four times the energy.
A $2\ \text{F}$ capacitor is charged to $3\ \text{V}$. Find the stored energy (in J).
$U = \tfrac12 CV^2 = \tfrac12 (2)(3^2) = 9\ \text{J}$.
The field between the plates
- Between the plates the field is uniform: $E = \dfrac{V}{d}$.
- Push the plates closer (smaller $d$) and the same $V$ gives a stronger field.
- Too strong and the gap sparks — the capacitor's voltage limit.
- This uniform field is the parallel-plate picture from the fields unit.
Select all true statements about capacitors.
Equal/opposite charge, geometric C, energy ½CV². Raising V does not change C.
Two plates of area $0.02\ \text{m}^2$ sit $1\ \text{mm}$ apart. Find $C$. ($\varepsilon_0 = 8.85\times10^{-12}$.)
- $C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.85\times10^{-12})(0.02)}{0.001}$.
- $C \approx 1.8\times10^{-10}\ \text{F} = 180\ \text{pF}$.
$C$ is set by geometry alone — area and gap. Raising the voltage does not change $C$; it just pushes proportionally more charge onto the same capacitor ($Q = CV$).
A capacitor stores equal and opposite charge on two plates: capacitance $C = Q/V$ (farads). For parallel plates $C = \varepsilon_0 A/d$ — geometry only. It stores energy $U = \tfrac12 CV^2$, with a uniform field $E = V/d$ between the plates.