Uniform electric fields · 匀强电场
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| uniform field/ˈjuːnɪfɔːm fiːld/ | 匀强场 | yún qiáng chǎng |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | 电势差 | diàn shì chà |
| parabola/pəˈræbələ/ | 抛物线 | pāo wù xiàn |
| projectile/prəˈdʒektaɪl/ | 抛体 | pāo tǐ |
The oil drop that hung in mid-air
- In 1909 Robert Millikan sprayed tiny oil drops between two horizontal charged plates and adjusted the voltage until a drop hung perfectly still.
- Still meant the electric force exactly balanced the weight, which turned the drop into a scale for its own charge. Millikan measured hundreds, and every charge was a whole-number multiple of one tiny value.
- That value was $e$. A uniform field between two plates was enough to prove that charge comes in indivisible lumps.
- This lesson is the uniform field 匀强场 between parallel plates, what it does to a charged particle, and the two situations the exam sets: the balanced drop and the deflected beam.
悬在半空中的那滴油
- 1909 年,Robert Millikan 把微小的油滴喷进两块水平带电板之间,调节电压直到某一滴完全静止地悬着。
- 静止意味着电场力恰好平衡了重力,这就把油滴变成了称量它自身电荷的天平。Millikan 测了几百滴,而每一个电荷都是同一个微小数值的整数倍。
- 那个数值就是 $e$。两块板之间的一个匀强场,就足以证明电荷是一份一份、不可分割的。
- 这一课讲平行板之间的匀强场(uniform field)、它对带电粒子做了什么,以及考试常设的两种情形:平衡的油滴和被偏转的束流。
Strength of a uniform field
- Between two parallel plates a distance $d$ apart with potential difference 电势差 $V$ between them, the field is uniform apart from edge effects:
- It points from the higher-potential plate to the lower one. The unit $\text{V/m}$ comes straight from this equation and equals $\text{N/C}$.
- Only two things change the field: increase the p.d., or move the plates closer. A resistor in series with the supply changes nothing, since no current flows once the plates are charged, so there is no p.d. across it.
Evenly spaced parallel lines: the same strength everywhere between the plates
匀强场的强度
- 在相距 $d$、其间电势差(potential difference)为 $V$ 的两块平行板之间,除边缘效应外场是均匀的:
- 它由高电势板指向低电势板。单位 $\text{V/m}$ 直接来自这个式子,并且等于 $\text{N/C}$。
- 只有两样东西能改变这个场:增大电势差,或把两板移得更近。与电源串联的电阻毫无影响,因为两板充好电后没有电流,所以电阻上没有电压。
等间距的平行线:两板之间处处强度相同
Two plates $0.050\ \text{m}$ apart have $200\ \text{V}$ between them. What is the field strength? · 两块极板相距 $0.050\ \text{m}$，其间电势差为 $200\ \text{V}$。场强是多少？
$E = \dfrac{V}{d} = \dfrac{200}{0.050} = 4000\ \dfrac{\text{V}}{\text{m}}$.
A uniform field between two plates points: · 两块极板之间的匀强电场方向指向：
The field runs from high potential (+) toward low potential (−). · 电场从高电势 (+) 指向低电势 (−)。
The unit V/m is the same as N/C. · 单位 V/m 等同于 N/C。
Both describe electric field strength; $\dfrac{\text{V}}{\text{m}} = \dfrac{\text{N}}{\text{C}}$. · 两者均描述电场强度；$\dfrac{\text{V}}{\text{m}} = \dfrac{\text{N}}{\text{C}}$。
Which changes increase the field strength between two charged parallel plates? Select all · 所有 that apply. · 哪些改变会增加带电平行极板间的电场强度？选择 所有 适用项。
E = V/d depends only on those two. No current flows once the plates are charged, so a series resistor drops no voltage and changes nothing. · E = V/d 仅取决于这两个量。极板充电后无电流流过，因此串联电阻不会分压，也不会改变任何参数。
A charged particle in the field
- The force $F = qE$ is the same everywhere, so the acceleration $a = qE/m$ is constant. This is exactly a mass in a uniform gravitational field, with $qE$ in place of $mg$.
- Released at rest, the particle accelerates in a straight line and gains kinetic energy. The energy method is usually quickest: $E_{\text{k}} = qV$, so $v = \sqrt{2qV/m}$, with no kinematics at all.
- Entering at right angles, it keeps its constant sideways speed while accelerating across the field, so it follows a parabola 抛物线, exactly like a projectile 抛体.
Constant velocity one way, constant acceleration the other
场中的带电粒子
- 力 $F = qE$ 处处相同,所以加速度 $a = qE/m$ 是恒定的。这正是匀强重力场中的一个质量,只是把 $mg$ 换成了 $qE$。
- 从静止释放时,粒子沿直线加速并获得动能。用能量法通常最快:$E_{\text{k}} = qV$,于是 $v = \sqrt{2qV/m}$,完全不需要运动学。
- 垂直进入时,它保持恒定的侧向速度,同时横越场加速,所以走抛物线(parabola),与抛体(projectile)完全一样。
一个方向匀速,另一个方向匀加速
Uniform electric field lab · 匀强电场实验
Follow how a charge behaves between parallel plates. · 观察电荷在平行极板间的运动行为。
A charge in a uniform field has a constant acceleration. · 电荷在匀强电场中具有恒定的加速度。
The force $qE$ is constant, so $a = \dfrac{qE}{m}$ is constant — like free fall in gravity. · 力 $qE$ 是恒定的，因此 $a = \dfrac{qE}{m}$ 也是恒定的——类似于重力作用下的自由落体。
A charge entering a uniform field at right angles follows a: · 垂直进入匀强电场的电荷其轨迹为：
Constant sideways speed plus steady acceleration across the field gives a parabolic path, like a projectile. · 恒定的侧向速度加上垂直于电场方向的恒定加速度产生抛物线路径，类似于抛射体运动。
Worked example: an electron crossing the gap
- Two plates in a vacuum are $0.041\ \text{m}$ apart with $250\ \text{V}$ between them. An electron is released from rest at the negative plate. Find the field, the acceleration, the time to cross, and the kinetic energy on arrival.
- $E = V/d = 250/0.041 = 6.1 \times 10^3\ \text{V/m}$, and $F = eE = 9.8 \times 10^{-16}\ \text{N}$, so $a = F/m_{\text{e}} = 1.07 \times 10^{15}\ \text{m/s}^2$.
- From rest with constant acceleration, $d = \tfrac12 at^2$, so $t = \sqrt{2d/a} = 8.8 \times 10^{-9}\ \text{s}$.
- Kinetic energy: use energy, not kinematics. $E_{\text{k}} = qV = (1.60 \times 10^{-19})(250) = 4.0 \times 10^{-17}\ \text{J}$, giving $v = 9.4 \times 10^6\ \text{m/s}$, which the kinematic route confirms.
例题:穿过间隙的电子
- 真空中两块板相距 $0.041\ \text{m}$,其间电压 $250\ \text{V}$。一个电子从负极板由静止释放。求场强、加速度、穿过所需时间,以及到达时的动能。
- $E = V/d = 250/0.041 = 6.1 \times 10^3\ \text{V/m}$,$F = eE = 9.8 \times 10^{-16}\ \text{N}$,所以 $a = F/m_{\text{e}} = 1.07 \times 10^{15}\ \text{m/s}^2$。
- 从静止作匀加速,$d = \tfrac12 at^2$,所以 $t = \sqrt{2d/a} = 8.8 \times 10^{-9}\ \text{s}$。
- 动能:用能量,不用运动学。$E_{\text{k}} = qV = (1.60 \times 10^{-19})(250) = 4.0 \times 10^{-17}\ \text{J}$,给出 $v = 9.4 \times 10^6\ \text{m/s}$,与运动学的路径结果一致。
Worked example: the balanced oil drop
- An oil drop of mass $2.6 \times 10^{-15}\ \text{kg}$ carrying $-4.8 \times 10^{-19}\ \text{C}$ hangs stationary between horizontal plates $2.0\ \text{cm}$ apart. Find the p.d., and say which plate is positive.
- Two forces act, not one: the electric force and the weight. "Electric force only" is the standard wrong answer. Stationary means they are equal and opposite:
- The drop is negative, so the force on it is opposite to the field. For that force to be upwards, the field must point down, so the top plate is positive.
- The charge here is $3e$: three excess electrons. That whole-number multiple is Millikan's result.
例题:平衡的油滴
- 一滴质量 $2.6 \times 10^{-15}\ \text{kg}$、带电 $-4.8 \times 10^{-19}\ \text{C}$ 的油滴,静止悬在相距 $2.0\ \text{cm}$ 的两块水平板之间。求电势差,并说出哪块板是正的。
- 有两个力,不是一个:电场力和重力。"只有电场力"是标准的错误答案。静止意味着两者大小相等、方向相反:
- 油滴带负电,所以它受的力与场方向相反。要让那个力朝上,场必须朝下,所以上板是正的。
- 这里的电荷是 $3e$:三个多余的电子。那个整数倍正是 Millikan 的结果。
An electron is released from rest and crosses a p.d. of 250 V. What is its kinetic energy on arrival, in units of 10^-17 J? (e = 1.60e-19 C) · 电子从静止释放并穿过 250 V 的电势差。到达时的动能是多少（单位为 10^-17 J）？(e = 1.60e-19 C)
Ek = qV = 1.60e-19 x 250 = 4.0e-17 J, with no kinematics needed. The energy route is almost always quicker than finding a, then t, then v. · Ek = qV = 1.60e-19 x 250 = 4.0e-17 J，无需运动学公式。能量法通常比求 a、再求 t、最后求 v 要快得多。
Two ways to change a uniform field
- Between parallel plates the field is $E = V/d$, so the field depends on both the p.d. and the separation. What happens when you move the plates apart depends on what is held fixed, and the exam asks both versions.
- Connected to a supply, so $V$ is fixed: doubling $d$ halves $E$. The charge on the plates falls to match.
- Isolated after charging, so the charge is fixed: the field between the plates is unchanged, and it is $V$ that doubles instead.
- The second case surprises people every year. With the charge fixed, the field is set by the charge per unit area on the plates, and moving them apart does not change that.
- Read which case the question describes before writing anything. The two give opposite answers to "what happens to the field".
改变匀强场的两种方式
- 平行板之间的场是 $E = V/d$,所以场同时取决于电压和间距。把板拉开会发生什么,取决于什么被固定住,而考试两种版本都问。
- 接在电源上,即 $V$ 固定:$d$ 加倍则 $E$ 减半。板上的电荷也随之减少。
- 充电后断开,即电荷固定:板间的场不变,改为加倍的是 $V$。
- 第二种情形年年让人意外。电荷固定时,场由板上单位面积的电荷量决定,把板拉开并不改变这一点。
- 动笔之前先读清题目说的是哪一种。对"场会怎样"这个问题,两者给出相反的答案。
A charged oil drop hangs stationary between two horizontal plates. Which forces act on it? · 带电油滴悬浮在两块水平极板之间保持静止。作用在它上面的力有哪些？
"Electric force only" is the standard distractor. Stationary means the two forces balance, which is what makes the drop a measuring device for its own charge. · “仅受电场力”是标准干扰项。静止意味着两个力平衡，正是这种平衡使油滴成为测量自身电荷的工具。
Two parallel plates are moved further apart. Match each case to what happens to the field between them. · 两块平行极板被拉开距离。将每种情况与极板间电场的变化匹配。
With the charge fixed the field is set by the charge per unit area, which moving the plates does not change. Read which case the question describes before answering. · 电荷固定时，电场由面电荷密度决定，移动极板不会改变这一点。答题前请仔细阅读题目描述的是哪种情况。
Worked example: deflecting a beam
- An electron travelling horizontally at $2.0 \times 10^7\ \text{m/s}$ enters the field between horizontal plates $18\ \text{mm}$ apart with $400\ \text{V}$ across them. The plates are $30\ \text{mm}$ long. Find the deflection as it leaves.
- Treat the two directions separately, as with any projectile. Horizontally the motion is unaffected, so the time in the field is $t = 0.030/(2.0 \times 10^7) = 1.5 \times 10^{-9}\ \text{s}$.
- Vertically, $a = eV/(m_{\text{e}}d) = 3.9 \times 10^{15}\ \text{m/s}^2$, so the deflection is $y = \tfrac12 at^2 = 4.4\ \text{mm}$.
- The horizontal speed sets the time, and the field sets the acceleration. Mixing them is the usual error.
例题:偏转一束电子
- 一个电子以 $2.0 \times 10^7\ \text{m/s}$ 水平飞入相距 $18\ \text{mm}$、电压 $400\ \text{V}$ 的水平板之间的场。板长 $30\ \text{mm}$。求它离开时的偏转量。
- 像处理任何抛体一样,把两个方向分开处理。水平方向不受影响,所以在场中的时间是 $t = 0.030/(2.0 \times 10^7) = 1.5 \times 10^{-9}\ \text{s}$。
- 竖直方向,$a = eV/(m_{\text{e}}d) = 3.9 \times 10^{15}\ \text{m/s}^2$,所以偏转量是 $y = \tfrac12 at^2 = 4.4\ \text{mm}$。
- 水平速度决定时间,场决定加速度。把两者混在一起是常见的错误。
Put the steps of finding the deflection of an electron crossing a uniform field in order. · 将电子穿过匀强电场发生偏转的步骤按顺序排列。
The horizontal speed sets the time and the field sets the acceleration. Treating it as a projectile is the whole method. · 水平速度决定了时间，电场决定了加速度。将其视为抛射体处理即为整个方法。
Marks that slip away
- A stationary charged drop has two forces on it, the electric force and its weight. Say both.
- For a negative charge, the force is opposite to the field, which is what decides which plate is positive.
- The field between plates depends only on $V$ and $d$. A series resistor changes nothing, since no current flows.
- For a deflection, split the motion: constant velocity along, constant acceleration across, and the time comes from the horizontal distance.
容易丢掉的分
- 静止的带电油滴上有两个力:电场力和它的重力。两个都要说。
- 对负电荷,力与场方向相反,而这正是决定哪块板为正的依据。
- 两板间的场只取决于 $V$ 和 $d$。串联电阻毫无影响,因为没有电流。
- 求偏转时要分解运动:沿向匀速、横向匀加速,而时间由水平距离决定。
You've got it
- between parallel plates the field is uniform, $E = V/d$, directed from the higher-potential plate to the lower; only changing $V$ or $d$ changes it
- a charge feels a constant force $qE$ and so a constant acceleration; released from rest it gains $E_{\text{k}} = qV$
- entering at right angles it follows a parabola: constant speed along, constant acceleration across, with the time set by the horizontal distance
- a stationary drop balances the electric force against its weight, $qV/d = mg$, and the sign of the charge decides which plate is positive
你掌握了
- 平行板之间的场是均匀的,$E = V/d$,由高电势板指向低电势板;只有改变 $V$ 或 $d$ 才能改变它
- 电荷受恒力 $qE$,因而有恒定加速度;从静止释放时获得 $E_{\text{k}} = qV$
- 垂直进入时走抛物线:沿向匀速、横向匀加速,时间由水平距离决定
- 静止的油滴让电场力与重力平衡,$qV/d = mg$,而电荷的正负决定哪块板为正