Capacitance
A-Level Physics Topic 19 14:49 English narration · English + 中文 subtitles burned in
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Look closely at a camera flash.
仔细看看相机的闪光灯。
Before the picture, a tiny part hums and whines for a second — you might even hear it.
在拍照之前，有一个小部件会嗡嗡作响一秒钟——你甚至能听见。
It is charging a capacitor, storing energy slowly, sip by sip, from a small battery.
它正在给一个电容器充电，用一节小电池，一点一点地、慢慢地储存能量。
Then, the instant you press the button, all that stored energy pours out all at once — a burst of light far brighter than the little battery could ever give on its own.
然后，就在你按下按钮的一瞬间，所有储存的能量一下子全部倾泻而出—— 发出的光，比那节小电池自己所能给出的要亮得多。
Store slowly, release in a flash.
慢慢储存，瞬间释放。
That is what a capacitor does.
这就是电容器所做的事。
A capacitor stores charge, and stores energy — ready to release in an instant.
电容器储存电荷，也储存能量——随时准备在一瞬间释放。
Today: capacitance, combining capacitors, the energy stored, and how a capacitor discharges.
今天：电容、电容器的组合、 储存的能量，以及电容器如何放电。
Let's begin.
让我们开始吧。
A capacitor is beautifully simple: two parallel conductor plates — two metal plates — with an insulator between them.
电容器非常简单：两块金属板，中间夹一层绝缘体。
Connect it to a battery, and charge piles up — positive on one plate, negative on the other.
把它接到电池上，电荷就堆积起来—— 一块板带正电，另一块带负电。
How much it holds for each volt, we call its capacitance: the charge stored, divided by the voltage across it.
它每一伏特能存多少电荷，我们就叫它的电容： 储存的电荷除以两端的电压。
The unit is the farad.
单位是法拉。
But a farad is enormous, so real capacitors hold only tiny fractions of one.
但一法拉大得惊人， 所以真实的电容器只能存到它极小的一部分。
Double the charge, and the voltage doubles too — so the capacitance stays fixed.
电荷加倍，电压也加倍——所以电容保持不变。
The simplest capacitor is two parallel conducting plates with an insulator between them — a dielectric, or just vacuum or air.
最简单的电容器是两块平行的导体板，中间夹着绝缘体—— 可以是电介质，也可以只是真空或空气。
Connect it to a battery and charge plus Q builds up on one plate and minus Q on the other, with a potential difference V across the gap.
把它接到电池上， 一块板上聚集正 Q 的电荷，另一块上是负 Q，两板之间有电势差 V。
Notice the charges are EQUAL and OPPOSITE, so the capacitor as a whole is still neutral.
注意这两份电荷大小相等、符号相反，所以整个电容器仍然是中性的。
Nothing crosses the gap; the insulator makes sure of that.
没有电荷穿过中间的间隙；绝缘体保证了这一点。
What the capacitor stores is a separation of charge, held apart by the insulator, and that is what you get back when it discharges.
电容器储存的是被分开的电荷，由绝缘体把它们隔开， 而这正是放电时你能取回的东西。
Capacitance C equals Q over V — the charge stored per volt.
电容 C 等于 Q 除以 V——每伏特所储存的电荷。
The unit is the farad, one coulomb per volt.
单位是法拉，即每伏特一库仑。
A farad is enormous, so real components run from picofarads to millifarads.
一法拉非常巨大，所以实际元件的范围从皮法到毫法。
The definition is not limited to plates: an isolated sphere holding charge Q sits at potential Q over four pi epsilon-nought r, so its capacitance is four pi epsilon-nought r — set purely by its radius.
这个定义并不限于平行板：一个孤立的球体带电荷 Q 时， 其电势为 Q 除以四 pi epsilon 零 r，所以它的电容是四 pi epsilon 零 r—— 完全由它的半径决定。
And here is the point students most often miss.
下面是学生最常忽略的一点。
Capacitance is CONSTANT for a given capacitor.
对给定的电容器，电容是恒定的。
Double the charge and the voltage doubles too, so the RATIO does not move.
电荷加倍，电压也加倍，所以这个比值不变。
C is fixed by the geometry and the dielectric, not by how much charge you happen to have put on it.
C 由几何形状和电介质决定，而不是由你恰好放上去多少电荷决定。
A one hundred microfarad capacitor is charged to twelve volts.
一个一百微法的电容器被充到十二伏。
Find the charge stored.
求所储存的电荷。
Rearrange the definition to Q equals C V.
把定义变形为 Q 等于 C V。
Substituting gives one hundred times ten to the minus six, times twelve, which is one point two times ten to the minus three coulombs — one point two millicoulombs.
代入得到一百乘以十的负六次方，再乘以十二， 等于一点二乘以十的负三次方库仑——也就是一点二毫库仑。
The whole difficulty in this question is the prefix: the microfarads must become farads before you multiply, or your answer is out by a factor of a million.
这道题全部的难点就在词头：相乘之前必须先把微法换成法拉， 否则你的答案会差一百万倍。
Write the power of ten down explicitly rather than trying to hold it in your head.
把十的幂明确写下来， 不要试图只在脑子里记着它。
Here is what that range looks like in hardware.
这就是那个范围在实际元件上的样子。
The four large cans standing up are electrolytic capacitors — high capacitance, up to millifarads, used to smooth power supplies.
竖立着的四个大罐是电解电容器—— 电容很高，可达毫法级，用来对电源进行平滑滤波。
The small film and ceramic parts in the row below are picofarads to nanofarads, used for timing and filtering.
下面那一排小的薄膜电容和陶瓷电容是皮法到纳法级，用于定时和滤波。
Same physics, same C equals Q over V; the size difference is plate area and how thin the maker can get the dielectric.
物理是同一套，C 等于 Q 除以 V 也是同一个；尺寸上的差别在于极板面积， 以及厂家能把电介质做得多薄。
Larger area and a thinner gap both mean more capacitance, which is why the high-value ones are physically big.
面积更大和间隙更薄都意味着更大的电容， 这正是高容值的那些在体积上很大的原因。
Capacitors combine — but in a way that feels backwards.
电容器可以组合——但方式让人觉得反过来了。
Side by side, in parallel, they share the same voltage and their charges add, so the capacitances simply add up.
并排放着，也就是并联，它们共用同一个电压， 电荷相加，所以电容直接相加。
In a line, in series, they carry the same charge and their voltages add, so this time the reciprocals add — giving less than any single one.
排成一行，也就是串联，它们带着相同的电荷，电压相加， 所以这次是倒数相加——得到的比任何一个单独的都小。
Notice: this is exactly opposite to resistors.
注意：这和电阻恰好相反。
Parallel adds capacitance; series shrinks it.
并联让电容变大；串联让它变小。
Capacitors in parallel sit across the same two nodes, so they share the same potential difference V.
并联的电容器接在同样的两个节点之间，所以它们的电势差 V 相同。
Each stores its own charge, C one V and C two V, and the total charge is the sum.
每个各自储存自己的电荷，C 一 V 和 C 二 V，总电荷是两者之和。
Divide the total charge by the shared V and the combined capacitance is simply C one plus C two.
用总电荷除以共同的 V，合成电容就是 C 一加 C 二。
So a parallel combination is LARGER than either capacitor alone — which makes sense physically, because putting plates side by side is effectively one capacitor with more plate area.
所以并联组合比其中任何一个单独的电容器都要大—— 这在物理上是说得通的，因为把极板并排放在一起， 实际上就等于一个极板面积更大的电容器。
In series the charge is the same through both, because the isolated section of circuit between them can only redistribute the charge it already has.
串联时通过两者的电荷相同，因为它们之间那段孤立的电路 只能重新分配它本来就有的电荷。
So each capacitor carries Q, and the potential differences ADD: V total is Q over C one plus Q over C two.
所以每个电容器带的电荷都是 Q， 而电势差要相加：V 总等于 Q 除以 C 一，加上 Q 除以 C 二。
Divide through by Q and one over C series equals one over C one plus one over C two.
两边同除以 Q，就得到一除以 C 串联等于一除以 C 一加一除以 C 二。
A series combination is SMALLER than either one — adding capacitors in series is effectively increasing the plate separation.
串联组合比其中任何一个都要小——串联加电容器， 实际上相当于增大了极板间距。
These rules are the exact opposite of the resistor rules — resistors add in series, capacitors add in parallel — and mixing them up is a classic lost mark.
这些规则和电阻的规则正好相反——电阻是串联相加，电容是并联相加—— 而把它们弄混是一个经典的失分点。
Do not memorise two lists; look at where V sits in each definition.
不要去背两张表； 看 V 在各自定义式中所处的位置。
Resistance is R equals V over I, with V on TOP.
电阻是 R 等于 V 除以 I，V 在上面。
Capacitance is C equals Q over V, with V on the BOTTOM.
电容是 C 等于 Q 除以 V，V 在下面。
Flipping V from top to bottom flips which arrangement adds.
V 从上面翻到下面， 哪种连接方式相加也就跟着翻了过来。
And you can always check it against the physics: parallel plates side by side give more area, so more capacitance — that alone tells you parallel is the one that adds.
而且你随时可以用物理来检验： 极板并排放置给出更大的面积，因而电容更大—— 单凭这一点你就知道，相加的那种是并联。
A capacitor stores energy, and charging one takes work — because every extra bit of charge must be pushed against the voltage already sitting there.
电容器储存能量，而给它充电是要做功的——因为每多存一点电荷， 都必须顶着已经存在的电压把它推上去。
Plot voltage against charge, and you get a straight line.
把电压对电荷作图，你会得到一条直线。
The energy stored is the area underneath — a triangle.
储存的能量就是它下面的面积——一个三角形。
So the energy is one half the charge times the voltage, or equally, one half the capacitance times the voltage squared.
所以能量等于电荷乘电压的一半， 或者同样地，电容乘电压平方的一半。
That factor of one half is there because the voltage climbs from zero — on average, you push against only half the final voltage.
那个二分之一的因子之所以出现，是因为电压从零开始爬升—— 平均来说，你只顶着最终电压的一半在推。
Charging a capacitor takes work, because each extra bit of charge has to be pushed against the p.d. that is ALREADY there.
给电容器充电需要做功，因为每一份额外的电荷都必须顶着已经存在的电势差被推上去。
When the charge so far is little q, the p.d. is q over C, so adding d q costs V d q.
当已充的电荷为小 q 时，电势差是 q 除以 C，所以再加上 d q 要花费 V d q 的功。
Integrate that from zero to Q and you get Q squared over two C.
把它从零积到 Q，就得到 Q 平方除以二 C。
Using V equals Q over C, the same energy has three equivalent forms — a half Q V, a half C V squared, and Q squared over two C.
利用 V 等于 Q 除以 C， 同一个能量有三种等价的形式——二分之一 Q V、二分之一 C V 平方， 以及 Q 平方除以二 C。
Pick whichever matches the quantities you were given.
哪一个与题目给你的量相符，就用哪一个。
And a worked example: one hundred microfarads at twelve volts stores a half times one hundred microfarads times twelve squared, which is seven point two millijoules.
再看一个例题：一百微法充到十二伏，储存的能量是二分之一乘以一百微法 乘以十二的平方，等于七点二毫焦。
Plot V against Q and you get a straight line through the origin — gradient one over C, since V equals Q over C.
把 V 对 Q 作图，你得到一条过原点的直线——斜率是一除以 C， 因为 V 等于 Q 除以 C。
The energy stored is the AREA under that line up to the charge you reached, and for a straight line through the origin that area is a triangle, a half Q V.
所储存的能量是这条线下方、直到你所达到电荷处的面积， 而对一条过原点的直线，那块面积是一个三角形，等于二分之一 Q V。
This is where the one-half comes from, and now you can say WHY it is there: the p.d. grew from zero up to V during charging, so the AVERAGE p.d. that the charge was pushed against is V over two, not V.
这就是那个二分之一的来源，而且现在你能说出它为什么在那里： 充电过程中电势差是从零一直长到 V 的，所以电荷被顶着推上去时 所面对的平均电势差是 V 的一半，而不是 V。
That reasoning is worth a mark on its own.
光是这个推理本身就值一分。
Here is a result that surprises people.
下面这个结果会让人意外。
Connect a capacitor C straight across an ideal battery of e.m.f. V.
把一个电容器 C 直接接到电动势为 V 的理想电池上。
The battery pushes charge Q equals C V through a p.d. of V, so it gives out C V squared of energy.
电池把电荷 Q 等于 C V 推过 V 的电势差，所以它输出的能量是 C V 平方。
But the capacitor ends up storing only a half C V squared.
可是电容器最终只储存了二分之一 C V 平方。
Exactly half the energy is missing, and it is lost as heat in the connecting wire.
恰好有一半的能量不见了， 它以热的形式损耗在连接导线上。
The striking part is the last clause: this is true WHATEVER the wire's resistance.
最引人注目的是最后这一句： 无论导线的电阻是多少，这都成立。
A lower resistance just means a bigger current for a shorter time, and the heat comes out the same.
电阻更小只不过意味着电流更大、 持续时间更短，放出的热量还是一样。
Now let the capacitor empty out through a resistor.
现在让电容器通过一个电阻器放电。
At first, fully charged, it drives a strong current.
一开始，充满电时，它驱动很强的电流。
But as charge leaves, the voltage drops, so the current falls too — which means the charge now leaves more slowly.
但随着电荷流走，电压下降，电流也随之减小——这意味着电荷现在流走得更慢了。
Less charge, less push, slower still.
电荷越少，推力越小，就越慢。
The result is a smooth exponential decay: the charge, the voltage, and the current all fade away along the same curve, never quite reaching zero.
结果是一条平滑的指数衰减曲线：电荷、电压和电流， 全都沿着同一条曲线渐渐消退，却永远到不了零。
This is the standard circuit.
这是标准电路。
A two-way switch puts the capacitor either across the supply, at position A, where it charges to V nought, or across the resistor, at position B, where it discharges through R.
一个双掷开关把电容器要么接到电源上，也就是位置 A， 在那里它充电到 V 零；要么接到电阻上，也就是位置 B，在那里它通过 R 放电。
The two-way switch matters: it means the capacitor is never connected to both at once, so the discharge really is just C and R alone.
双掷开关很重要：它意味着电容器绝不会同时接到两边， 所以放电过程确实只有 C 和 R 参与。
That is the circuit every exam question on this topic describes, so recognise it from the shape.
这道题目的每一道考题描述的都是这个电路，所以要从图形认出它来。
The exponential is not something to memorise — it falls out of the circuit.
这个指数函数不是要去背的东西——它是从电路中自然得出的。
Kirchhoff's second law round the loop says the capacitor's p.d. equals the resistor's p.d.
对回路用基尔霍夫第二定律，电容器上的电势差等于电阻上的电势差。
Put in V C equals Q over C, V R equals I R, and I equals minus d Q by d t, the minus because the charge is falling.
代入 V C 等于 Q 除以 C，V R 等于 I R，以及 I 等于负的 d Q 比 d t， 取负号是因为电荷在减少。
That gives Q over C equals minus R d Q by d t.
于是得到 Q 除以 C 等于负 R d Q 比 d t。
Read what that equation SAYS: the rate of fall is proportional to how much charge is left.
读一读这个方程在说什么：下降的速率与剩下多少电荷成正比。
A quantity whose rate of fall is proportional to itself decays exponentially — that is the definition of exponential decay, and it is why the same shape shows up in radioactive decay too.
一个下降速率与自身成正比的量，就按指数衰减—— 这正是指数衰减的定义，也是同样的形状在放射性衰变中也出现的原因。
Solving that equation gives Q equals Q nought e to the minus t over R C.
解这个方程得到 Q 等于 Q 零乘以 e 的负 t 除以 R C 次方。
And because V equals Q over C and I equals V over R, the p.d. and the current follow exactly the same law with exactly the same time constant — V equals V nought e to the minus t over R C, I equals I nought e to the minus t over R C, with I nought equal to V nought over R.
而由于 V 等于 Q 除以 C，I 等于 V 除以 R， 电势差和电流遵循完全相同的规律，时间常数也完全相同—— V 等于 V 零乘以 e 的负 t 除以 R C 次方，I 等于 I 零乘以 e 的负 t 除以 R C 次方， 其中 I 零等于 V 零除以 R。
So one curve shape serves all three; only the axis label changes.
所以一种曲线形状可以服务于全部三个量； 改变的只是坐标轴的标签。
On the graph, after a time R C the charge has fallen to Q nought over e.
在图上，经过时间 R C 之后， 电荷已经降到 Q 零除以 e。
The time constant tau equals R C.
时间常数 tau 等于 R C。
Check its units and you get ohms times farads, which really does come out in seconds.
检查它的单位，得到欧姆乘以法拉， 而这确实就等于秒。
Tau is the time to fall to one over e, about thirty-seven percent, of the starting value.
tau 是降到起始值的一除以 e、 也就是约百分之三十七所需要的时间。
After two tau you are at about thirteen and a half percent; after five tau, below one percent — which is the usual working definition of fully discharged.
经过两个 tau 大约剩百分之十三点五； 经过五个 tau，就低于百分之一了——这是通常所说的"完全放电"的实用定义。
Worked example: a one hundred microfarad capacitor charged to twelve volts, discharged through forty-seven kilohms.
例题：一个一百微法的电容器充到十二伏，通过四十七千欧放电。
Tau is forty-seven thousand times one hundred microfarads, which is four point seven seconds.
tau 等于四万七千乘以一百微法，等于四点七秒。
And after one time constant the voltage is twelve times zero point three seven, about four point four volts.
而经过一个时间常数之后，电压是十二乘以零点三七，约为四点四伏。
To get tau from a measured curve you have two routes.
要从实测曲线求 tau，你有两条路。
The quick one: read off the time taken to fall to one over e of the start.
快的那条：读出降到起始值一除以 e 所用的时间。
The better one, and the one practical questions want, is to take natural logs.
更好的那条，也是实验题想要的那条，是取自然对数。
Log both sides of the decay law and you get l n V equals l n V nought minus t over R C.
对衰减公式两边取对数，得到 l n V 等于 l n V 零减去 t 除以 R C。
So plotting l n V against t gives a STRAIGHT line, with gradient minus one over R C.
所以把 l n V 对 t 作图会得到一条直线，斜率是负的一除以 R C。
A straight line is far easier to fit by eye than a curve, and it uses every data point rather than just one, so it is the more accurate method — say that if the question asks you to justify it.
直线比曲线容易用眼睛拟合得多，而且它用到了每一个数据点， 而不是只用一个，所以这是更精确的方法—— 如果题目要你说明理由，就这样答。
Two more graphs turn up, and both are straight lines, not curves.
还有两种图会出现，而且两种都是直线，不是曲线。
Q against V C is a straight line through the origin of gradient C, because Q equals C V holds at every instant — during discharge the point just slides down that line towards the origin.
Q 对 V C 是一条过原点、斜率为 C 的直线，因为 Q 等于 C V 在每一时刻都成立—— 放电过程中，那个点只是沿着这条直线滑向原点。
And I against V C is a straight line of gradient one over R, because I equals V C over R.
而 I 对 V C 是一条斜率为一除以 R 的直线，因为 I 等于 V C 除以 R。
So those graphs are how you MEASURE C and R.
所以这些图正是你测量 C 和 R 的手段。
Finally, the standard question routine: read V nought at t equals zero, find the time to fall to V nought over e to get tau, then use C equals tau over R, or R equals tau over C, and predict any later value from the exponential.
最后是标准的解题流程： 在 t 等于零处读出 V 零，求出降到 V 零除以 e 所需的时间得到 tau， 再用 C 等于 tau 除以 R，或者 R 等于 tau 除以 C， 然后用指数公式预测任何较晚时刻的值。
How fast does it fade?
它消退得有多快？
That is set by a single number: the time constant, the resistance times the capacitance.
这由一个数字决定：时间常数，也就是电阻乘以电容。
After one time constant, the charge has fallen to about thirty-seven percent of its start.
经过一个时间常数，电荷降到起始值的大约百分之三十七。
After two, to about thirteen.
经过两个，降到大约百分之十三。
After five, it is all but gone.
经过五个，就几乎没有了。
A bigger resistor, or a bigger capacitor, means a longer, slower decay.
更大的电阻，或更大的电容，意味着更长、更慢的衰减。
Read the time to reach thirty-seven percent off any discharge graph, and you have found the time constant.
从任何一条放电曲线上，读出降到百分之三十七所需的时间，你就找到了时间常数。
Three marks to secure.
三个要拿稳的分。
First, capacitance is charge over voltage, and the energy stored is one half charge times voltage — the area under the graph.
第一，电容等于电荷除以电压，储存的能量是电荷乘电压的一半—— 也就是图线下的面积。
Second, combine capacitors the opposite way to resistors: parallel add, series reciprocal.
第二，电容器的组合和电阻恰好相反：并联相加，串联取倒数。
Third, discharge is exponential, with time constant resistance times capacitance — the time to fall to thirty-seven percent.
第三，放电是指数式的，时间常数等于电阻乘电容——也就是降到百分之三十七所需的时间。
Master these, and capacitors are yours.
掌握这些，电容器就是你的了。