Uniform electric fields
| English | Chinese | Pinyin |
|---|---|---|
| uniform field | 匀强场 | yún qiáng chǎng |
| potential difference | 电势差 | diàn shì chà |
| plate separation | 板间距 | bǎn jiān jù |
| parabola | 抛物线 | pāo wù xiàn |
A field you can control
- Between two parallel charged plates, the field is uniform — same strength everywhere.
- It is a neat, controllable field.
- Old TVs used one to steer the electron beam across the screen.
Uniform electric field lab
Follow how a charge behaves between parallel plates.
Strength of a uniform field 匀强场
- $E = \dfrac{V}{d}$ — potential difference 电势差 divided by plate separation 板间距.
- It points from the + plate to the − plate. The unit $\dfrac{\text{V}}{\text{m}}$ equals $\dfrac{\text{N}}{\text{C}}$.
A charge entering a uniform field at right angles follows a parabolic path, like a projectile
Between parallel plates the field is uniform, E = V/d
Two plates $0.050\ \text{m}$ apart have $200\ \text{V}$ between them. What is the field strength?
$E = \dfrac{V}{d} = \dfrac{200}{0.050} = 4000\ \dfrac{\text{V}}{\text{m}}$.
A uniform field between two plates points:
The field runs from high potential (+) toward low potential (−).
The unit V/m is the same as N/C.
Both describe electric field strength; $\dfrac{\text{V}}{\text{m}} = \dfrac{\text{N}}{\text{C}}$.
A charge in the field
- A charge feels a constant force $F = qE$, so a constant acceleration $a = \dfrac{qE}{m}$.
- Just like a mass in a uniform gravitational field.
In a uniform field the potential falls steadily with distance, so the field strength V/d is constant
A charge in a uniform field has a constant acceleration.
The force $qE$ is constant, so $a = \dfrac{qE}{m}$ is constant — like free fall in gravity.
Parabolic path
- A charge entering at right angles follows a parabola 抛物线 — like a projectile under gravity.
- Constant sideways speed, steady acceleration across the field.
A charge entering a uniform field at right angles follows a:
Constant sideways speed plus steady acceleration across the field gives a parabolic path, like a projectile.
You've got it
- uniform field between plates: $E = \dfrac{V}{d}$, from + plate to − plate
- a charge feels a constant force $qE$ → constant acceleration
- entering sideways, it follows a parabola (like a projectile)