Conservation of Electric Energy · 电能守恒
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| kinetic energy/kɪˈnetɪk ˈenədʒi/ | 动能 | dòng néng |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | 电势差 | diàn shì chà |
| electron-volt/ɪˈlektrɒn vəʊlt/ | 电子伏特 | diàn zi fú tè |
Let a charge go and stored energy becomes speed
- Release a charge in a field and it accelerates on its own.
- Its electric potential energy falls while its kinetic energy 动能 rises.
- The total energy stays the same — nothing is lost.
- This one idea solves problems that would be hard with forces alone.
松开一个电荷,储存的能量变成速度
- 在场中释放一个电荷,它会自己加速。
- 它的电势能下降,而它的动能上升。
- 总能量保持不变——没有任何损失。
- 仅凭这一个想法就能解决单靠力很难的问题。
Energy is conserved
- With only the electric force acting: $\Delta KE + \Delta U = 0$.
- Energy just changes form, from potential to kinetic (or back).
- So a gain in speed is paid for by a drop in potential energy.
- No need to track the changing force along the path — just the endpoints.
能量守恒
- 只有电力作用时:$\Delta KE + \Delta U = 0$。
- 能量只是改变形式,从势能变成动能(或反过来)。
- 所以速度的增加由势能的下降来支付。
- 不必追踪沿途变化的力——只看两端点。
A charge is released in a field (only the electric force acts). As it speeds up, its potential energy: · 一个电荷在电场中被释放（仅受电场力作用）。当它加速时，其电势能：
$\Delta KE + \Delta U = 0$: as KE rises, $U$ falls. · $\Delta KE + \Delta U = 0$：随着动能增加，$U$ 下降。
Charge falling through a voltage
- Moving a charge $q$ through a potential difference 电势差 $\Delta V$ changes its energy by $q\,\Delta V$.
- If that is the only force: $q\,\Delta V = \Delta KE = \tfrac12 m v^2$ (from rest).
- A $+$ charge speeds up moving to lower potential; a $-$ charge speeds up moving to higher.
- The sign of the charge decides which way is "downhill".
电荷穿过电压下降
- 把电荷 $q$ 移过一个电势差 $\Delta V$,它的能量改变 $q\,\Delta V$。
- 若这是唯一的力:$q\,\Delta V = \Delta KE = \tfrac12 m v^2$(从静止)。
- $+$ 电荷移向更低电势时加速;$-$ 电荷移向更高电势时加速。
- 电荷的符号决定哪个方向是"下坡"。
A $2\ \text{C}$ charge falls through $\Delta V = 5\ \text{V}$. How much kinetic energy (in J) does it gain? · 一个 $2\ \text{C}$ 电荷穿过 $\Delta V = 5\ \text{V}$ 下落。它获得了多少动能（单位：J）？
$KE = q\,\Delta V = 2 \times 5 = 10\ \text{J}$.
An electron speeds up when it moves toward: · 电子向以下哪个方向运动时会加速：
A negative charge gains KE moving to higher $V$ (its $q\Delta V < 0$ means $\Delta U < 0$). · 负电荷移动到更高 $V$ 时获得动能（其 $q\Delta V < 0$ 意味着 $\Delta U < 0$）。
A handy unit: the electron-volt
- One electron-volt 电子伏特 (eV) is the energy an electron gains crossing $1\ \text{V}$.
- $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$ — tiny, but perfect for particles.
- Accelerators quote energies in keV, MeV, and GeV.
- It saves writing awkward powers of ten for single charges.
一个方便的单位:电子伏特
- 一个电子伏特(eV)是一个电子越过 $1\ \text{V}$ 获得的能量。
- $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$——很小,但对粒子刚刚好。
- 加速器用 keV、MeV、GeV 来报能量。
- 它省去为单个电荷写别扭的十次幂。
Speeding up or slowing down? · 加速还是减速？
As a charge moves through a field, energy converts between kinetic and potential. Sort each case. · 电荷在电场中运动时，能量在动能和势能之间转换。对每种情况进行分类。
The energy an electron gains crossing $1\ \text{V}$ is one ____. · 电子跨越 $1\ \text{V}$ 所获得的能量称为一个 ____。
That energy is $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$. · 该能量为 $1\ \text{eV} = 1.6\times10^{-19}\ \text{J}$。
Path doesn't matter
- The energy change depends only on the start and end potentials.
- Any route between them gives the same $q\,\Delta V$.
- The electric force is conservative — like gravity.
- That is exactly why a potential (a single number per point) can exist.
路径无关紧要
- 能量变化只取决于起点和终点的电势。
- 它们之间的任何路线都给出相同的 $q\,\Delta V$。
- 电力是保守的——像引力一样。
- 这正是为什么能存在一个电势(每点一个数)。
The energy change $q\Delta V$ depends only on the start and end potentials, not the path. · 能量变化 $q\Delta V$ 仅取决于起点和终点的电势，与路径无关。
The electric force is conservative — path-independent. · 电场力是保守力——与路径无关。
Select all · 所有 true statements about electric energy conservation. · 选择关于静电能守恒的 所有 正确陈述。
Total energy is conserved, the force is conservative, eV is energy. Path does not matter. · 总能量守恒，力是保守力，eV 是能量单位。路径无关紧要。
An electron (charge $e$) is accelerated from rest through $100\ \text{V}$. Find its kinetic energy.
- $KE = q\,\Delta V = e \times 100\ \text{V} = 100\ \text{eV}$.
- In joules: $100 \times 1.6\times10^{-19} = 1.6\times10^{-17}\ \text{J}$.
一个电子(电荷 $e$)从静止经 $100\ \text{V}$ 加速。求它的动能。
- $KE = q\,\Delta V = e \times 100\ \text{V} = 100\ \text{eV}$。
- 换成焦耳:$100 \times 1.6\times10^{-19} = 1.6\times10^{-17}\ \text{J}$。
Watch the sign of the charge. A positive charge speeds up going to lower potential, but a negative charge (like an electron) speeds up going to higher potential. The energy $q\,\Delta V$ carries that sign automatically.
注意电荷的符号。正电荷移向更低电势时加速,但负电荷(如电子)移向更高电势时加速。能量 $q\,\Delta V$ 自动带上这个符号。
With only the electric force, energy is conserved: $\Delta KE + \Delta U = 0$. A charge through $\Delta V$ gains $q\,\Delta V$ of energy, so $q\,\Delta V = \tfrac12 m v^2$ from rest. The force is conservative (path-independent); a handy unit is the electron-volt.
只有电力时,能量守恒:$\Delta KE + \Delta U = 0$。电荷经过 $\Delta V$ 获得 $q\,\Delta V$ 的能量,所以从静止 $q\,\Delta V = \tfrac12 m v^2$。这个力是保守的(与路径无关);一个方便的单位是电子伏特。