# Does the equation CV = q break down when V is small or a wire is very long?

I'm trying to understand how the equation $CV=q$ relating the voltage difference and charge stored on a charged capacitor breaks down under extreme circumstances. In most problem statements that I'm seeing, they might begin with a capacitor attached to a battery via a circuit and an open switch. They'll say that when the switch is open, the voltage difference between the two ends of the battery is $120 V$. Then, when the switch is closed and when the capacitor is fully charged the stored charge on each plate will have magnitude $|q| = 5 C$. Finally, the problem will ask me to calculate the capacitance of the capacitor.

From what I understand, these problems assume that the potential difference between the two plates is the same as the original potential difference between the two ends of the battery. Is this a truly accurate assumption? Technically speaking, couldn't the redistribution of charge along the circuit change the electric field within and outside of the circuit and thus couldn't it also change the potential difference between the two conductors? (The two conductors being the two separately charged batteries and their corresponding wires/capacitor plates).

If the closing of the switch and the charging of the capacitor can change the potential difference, when is this change in potential difference significant? Is it ever significant to an electrical engineer (even if such a circumstance is an outlier)? Or is it only detectable in the most extreme of circumstances, ones that no electrical engineer would ever have to deal with? (i.e. when V is very small or there is a very long wire over which the excess charge can redistribute itself?)

When you ask questions about things "in the limit", the answer is almost always "It depends". In this case, the answer is "it depends".

The equation $Q=CV$ assumes linear behavior of the capacitor - in reality the dielectric of most capacitors has hysteresis as well as a nonlinear component, so as you increase the voltage, the capacitance will change. This effect is particularly strong (and useful) in certain PIN diodes that can be used, when reverse polarized, as voltage-dependent capacitors (handy for electronically tuning RF circuits, for example).

The equation also assumes that $Q$ is only the charge on the capacitor - that no charge accumulates elsewhere. Now the capacitance of a pair of 16 AWG wires separated by 1 cm is almost exactly 10 pF/m - see this handy calculator. So if your capacitor is attached to the battery with a sufficiently long wire, the wire itself will act as a non-negligible capacitor. Again, at RF frequencies (especially when you get up into the GHz range) these "parasitics" really start to add up, and in fact they dominate the behavior of circuits. They are in fact the basis for transmission lines, and definitely are considered important for electrical engineering.

Back to your original question: if you have a battery connected to a capacitor with a pair of wires and a switch, then we usually consider the following simplifications:

1. The battery voltage is the same before I close the switch and after
2. The only charge of interest is on the capacitor
3. No current flows in the circuit except at the instant that the switch is closed

At high enough precision, none of these assumptions are true. Let's take them one at a time:

1. When charge flows from a battery, the chemical reaction that takes place will cause the voltage to drop slightly. In your example, you used 5 C of charge - that is 1 A for 5 s, which is a not inconsiderable amount of current for a battery. A 120 V battery might consist of 80 1.5 V cells in series; if each cell drops 1 mV because of this transient load, the voltage of the battery would drop by 80 mV.

2. As I showed above, the wires will collect some charge; how much charge will depend on the geometry. If the value $Q$ of interest is only the charge on the capacitor (rather than the charge that left the battery) this difference doesn't matter - but that was not made explicit in the question.

3. A capacitor has finite resistance - charge will slowly bleed off. How quickly it bleeds depends on many factors: most notably the dielectric material used, and the electric field across it. Making a capacitor with high capacitance per unit volume requires a very thin dielectric and thus a high field and possibly high leakage. Precision capacitors tend to be low-value, and bulky.

In your example, a charge of 5 C and a voltage of 120 V is possible with a capacitance of $c=\frac{5}{120} F = 42 mF. This is big, but not too unwieldy. I found one online that was 40 mF and 160 V rating; it was 76 mm diameter, by 150 mm long. As I said - quite big. I looked at the data sheet for this family of capacitors but could not find any information on leakage current. However, since this is an electrolytic capacitor, the leakage will be quite large. For example, I found a generic article on the leakage of aluminum electrolytic capacitors which included the following plot as figure 3: As you can see, there is significant current flowing in large electrolytic capacitors - this was for a 1 mF, 35 V device so you can imagine it will be quite a bit worse for 44 mF, 120 V. These differences are often much more important than the voltage drop of the battery (which your question was asking about). Long wires plus leakage currents could, in principle, result in a sustained voltage drop across the wires. But as a rule of thumb, electrical engineers will be careful when designing precision circuits (ones where a few mV on 120 V might matter) with short wires; and if you need long runs with a precise voltage at the end of it, you might use local regulation at the far end, or a 3 or 4 wire solution which allows sensing at the remote end so that the voltage source can compensate. Batteries are not capacitors. Closing the switch doesn't redistribute some fixed amount of charge in the circuit. Instead, the batteries can create new charge through chemical reactions. The redistribution would slightly lower the potential, allowing the chemical reaction to proceed until it comes back to the resting voltage. While real batteries are not ideal (the voltage they provide is not constant), such effects are not usually modeled for this type of problem. A designer would simply make sure the circuit was valid over a range of voltages. The drop in voltage under current is going to overwhelm any changes in the zero-current voltage. • That clarifies many things for me, but I understand that a battery is not a capacitor. Do batteries themselves not have voltage differences or something? Is it the capacitor, the battery, or both that the voltage difference is attributed to? May 19, 2015 at 2:41 • What do you mean by "differences"? Differences between batteries? A voltage difference between the electrodes? The battery is normally used as a voltage source in the circuit, a device that maintains a (near) constant voltage difference between its contacts. May 19, 2015 at 4:22 • I don't know. That's kind of my confusion that leads me to ask that question. My textbook has always just given a value of "V" without specifying (unless I missed the specification) what the potential difference is between (the battery electrodes, capacitor plates, or something else). This is off topic from my OP, though, so I probably should not have asked that question in a comment May 19, 2015 at 5:16 • (cont) In practice I understand that this distinction does not often matter because the circuit that the capacitor is attached to will have an approximately uniform potential, but I wished to know what standard terminology referred to when it talks about the "potential difference" of a capacitor. May 19, 2015 at 5:27 • Except for certain types of "textbook problems", any real (DC) circuit is going to have a voltage source, and the capacitor will collect charge until it matches the attached voltage. May 19, 2015 at 5:31 First, I'm a bit confused by your statement about two batteries. The capacitance of a two-sided capacitor is defined to be the ratio of the magnitude of separated charge (commmonly, the magnitude of charge on each plate) to the resultant potential difference (aka, voltage) between the plates. It's technically a "what if" formula. The actual capacitance depends on the geometrical and material arrangement of the plates. We often turn this definition,$C=\frac{q}{V}$, into an operational equation such as$q=CV.$And this is okay, because, at least to first order (and likely to 2nd and 3rd order) in an ordinary capacitor, the capacitance is independent of the voltage up to the dielectric strength of the gap material. Whatever the voltage across the capacitor, operationally,$q=CV$where$V$is the voltage from one plate to the other. Precisely how this relates to a battery in the circuit depends on what other elements are in the circuit. In a series circuit with a real battery and real wires with real resistance, the voltage across the plates could be slightly less than the emf of the battery while the capacitor is charging. If the capacitor is leaky, then a current will continue to flow and the voltage will continually be slightly less than the emf. If we put two capacitors of different capacitances in parallel, they will accumulate different charges. If we put two capacitors of different capacitances in series with each other, they will have different voltages but the same charge. Bottom line: the voltage in$q=CV\$ is the voltage across the capacitor only. It may happen to be the same value as the voltage across another component simply because of the way the circuit is connected. As the capacitor gaining charge after your switch is closed, its voltage changes at a rate determined by the actual resistances in the wires and the battery. If that resistance is small, the charging happens very quickly.

• Whoops, fixed the two batteries comment. Just meant two ends of the battery May 19, 2015 at 3:28