I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C=\frac{Q}{V}$. What I don't understand, however, is why it is defined as coulomb per volt. Of course, the charge in the numerator makes sense but I don't get why capacitance would measured in relation to voltage. For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. Or, according to the equation $C=\frac{Q}{V}$, why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. Hope you can provide me with some intuition on this topic.

  • $\begingroup$ Because then it wouldn't be useful, and we'd define some other thing to be equal to $\frac{Q}{C}$ because that is useful. Imagine you're shopping for a capacitor. "Hey, this one is rated at .2C/m^3!" "Cool... how big is it?" If the capacitor's capacitance does depend on its volume, and the capacitor's volume is known by the manufacturer, why wouldn't the manufacturer just do the math for you? The manufacturer gives you a rating that tells you how it performs under conditions that are unknown to the manufacturer. $\endgroup$
    – Devsman
    Commented Apr 29, 2016 at 13:15
  • $\begingroup$ Historically, since thermodynamics seemed to have come before electrodynamics, $Q = CT$ where $Q$ is heat, $C$ is heat capacitance, and $T$ temperature was already known. Just as heat is viewed as something physical (which can be transferred), and temperature viewed more abstract / a characteristic for something more physical (although we think of temperature as "physical" intuitively - temperature tells you the rate of change of energy transfer. When you touch a "hot" object, it's not temperature that's transferring. Temperature tells you the rate). Likewise charge/electric field $\endgroup$
    – DWade64
    Commented Oct 21, 2018 at 20:51
  • $\begingroup$ are "physical" and electric potential is viewed more abstractly/ a characteristic for something more physical (though because the math is symmetric, what is deemed more "physical" is up for debate or personal taste). Anyways Q = constant * temperature is similar to Q (as charge) = constant * voltage $\endgroup$
    – DWade64
    Commented Oct 21, 2018 at 20:52

8 Answers 8


You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in)

How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored.

The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" is the cross section, in this picture. The broader, the more water you store with a given effort.

Now this is a direct analogy. The capacitor is never full (= the tube is very high), you can always store more charge, you just have to push harder.

Indeed, there is an upper rim of the tube, when the water flows out. That's when a spark jumps and partially discharges the capacitor. But this is another story.
A still more correct picture is two tubes in which a pump can create a difference in water level.

The question is: why $Q/V$ and not for example $Q/V^2$?
One answer: experiment shows, that a given capacitor will have a linear dependence of stored charge to applied voltage.
Another answer: The field produced by a charge is linearly proportional to $Q$ (Coulombs Law). And so will be the voltage (it's the integral of the field).

... You see, I can view the system from different perspectives, what the cause-effect direction is. You can say, that a high water column produces a high pressure, or you can say that a high pressure will push the water column high.
You may as well say, that a capacitor stores voltage instead of that it stores charge; both is right. The energy is given by the product, and this is what you really care about.

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    $\begingroup$ If the capacitor goes into full short circuit, then it is a greedy cup like water tube :) en.wikipedia.org/wiki/Pythagorean_cup $\endgroup$ Commented Apr 28, 2016 at 15:47
  • $\begingroup$ that's a good point! $\endgroup$
    – Ilja
    Commented Apr 28, 2016 at 15:49
  • 1
    $\begingroup$ A typical container for stuff will often require very little effort to put stuff in until it's nearly full, whereupon the effort required to store anything else will go up markedly. This is equivalent to a device whose capacitance is large at low voltages, but diminishes markedly near its operating-voltage limits. $\endgroup$
    – supercat
    Commented Apr 28, 2016 at 18:29
  • $\begingroup$ well, maybe the generic container in a mechanical model behaves like this. In this case the transition to the high-resistance-mode would be a good meaning for the concept of being full (that was misunderstood by the OP). And at once the concept of capacitance becomes meaningless, for such a device. $dQ/dV$ is certainly a meaningfull thing, but the total $Q/V$ is not any more. $\endgroup$
    – Ilja
    Commented Apr 28, 2016 at 21:03

Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage.

When you try to separate the charges, you unavoidably create electric fields ($\vec E$ from the positively charged part to the negatively charged part), and when you integrate this electric field $\vec E$ over $d\vec r$, the distance between the separated positive and negative charges, you obtain the voltage. (The electric field is measured in volts per meter.)

So the voltage is unavoidably induced by the separated positive and negative charges. And for a fixed capacitor – with a fixed geometry – there exists a proportionality law. The greater the charges $+Q$ and $-Q$ that we have separated are (note that the electric fields and forces are proportional to $Q$, e.g. by Coulomb's law), the greater are the electric fields $\vec E$ in the previous paragraph, and the greater the voltage (the integral of $\vec E$, basically $V=\vec E\cdot \Delta \vec r$), is, too. This proportionality boils down to the linearity of Maxwell's equations in the electric fields; and in the sources (and currents).

Because of this proportionality, there has to exist a law $Q=CV$ for the charge as a function of the voltage, for a fixed "design" of the capacitor, and the coefficient is simply known as the capacitance. We want to keep the voltage fixed, e.g. because the capacitor is a part of a circuit and its goal is to produce some fixed voltage in the circuit under some circumstances. The greater the capacitance is, the more charge the capacitor is able to separate at a fixed voltage.

The ratio of the charge and the volume isn't well-defined because the fixed capacitors can carry basically any charge – but the voltage will go up accordingly, according to the $V=Q/C$ law we derived above. So the charge per volume simply cannot be fixed for a given capacitor.

Even if there were a way to write the charge "per something else" (some function of the dimensions of the capacitor) that would be constant for a given capacitor, it wouldn't be terribly useful because the purpose of the capacitor is to influence the voltages and currents in the circuit. We're just not interested in how the capacitor achieves its job. We want to add this component according to what it is capable of doing.

The charge is basically $Q=I\cdot t$, the product of the current and time for which the capacitor may produce this current, and voltage is important in all circuits. We want to know how the current $I$ and the voltage $V$ are related because these are the two most important quantities in every circuit. Resistors have $U=RI$, Ohm's law, and capacitors have something similar, basically $$Q\equiv I\cdot t = C\cdot V$$ The current multiplied by the time for which the capacitor is capable of producing it is equal to the capacitance times the voltage at the beginning, before it gets discharged. We want to know how the components of the circuits influence currents and voltages because these are the basic quantities circuits work with. Currents go through wires and voltages are provided e.g. by batteries. Resistors affect the behavior of circuits according to their own rules and the constants $R,C$ describe how.

Inductance of inductors (coils etc.) is similar except that the time appears in the opposite way: $V=L \cdot dI/dt$. The voltage of the inductor is proportional to the time derivative of the current (the rate at which the current is changing with time), and the coefficient is known as inductance. So components of circuits have some effect on voltages and currents – the only major "intrinsically electromagnetic quantities" that are relevant in a current – and the circuits also operate in time which means that we may want to know how the currents or voltages are changing or how these changes are correlated with other things. A circuit achieves a certain job and capacitors and inductors (and especially transistors!) may be shrunk while the functionality of the circuit stays the same. That's why we need to know the relevant or required parameters to "keep the functionality the same".


We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality.

We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without changing its volume. Charge divided by voltage is constant.

I think the most important question you asked is:

Or, according to the equation $C=\frac{Q}{V}$, why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge.

I like this question because its slightly backwards, suggesting you're thinking about it in a different way. I like when people think about something backwards, because its show's they're really thinking, and willing to take a stab at trying to figure out what's going on!

The trick to this is that you will find you can't increase the voltage across the capacitor while keeping the charge constant, without doing some physical modifications to the capacitor itself. Reality simply wont let you. If you try to increase the voltage, you will find exactly enough charge will flow into the capacitor to balance the voltage out.

More interestingly, consider the case where you instantaneously change the voltage, say from 1V to 10V. In theory, that should "increase the voltage without increasing the charge," because there hasn't been any time for current to flow. You could draw this up in a circuit simulator, like PSPICE, and change the voltage at t=0. It looks like you have to be changing the capacitance.

In reality, we see a different effect. What we see is that, even though we increased the voltage over the system, the voltage across the capacitor will actually remain exactly the same! This makes sense from the equation, because we know the charge and capacitance didn't change, so voltage can't change. But now it looks like we have a broken circuit: somehow we have 10V on the input, but only 1V over the capacitor! Everyone knows that doesn't add up.

What we find happens in reality is that there are "parasitic resistances" in every device we use. The battery has a resistance, the capacitor has a resistance, even those wires you use to connect them have a resistance. So your real circuit isn't just a voltage source and a capacitor, it's a voltage source, a capacitor, and a bunch of small resistors.

In 99% of circumstances, we can ignore these resistors because they just don't change the circuit all that much. However, in this slightly pathological situation, they actually matter a lot. They are what "soak up" that extra voltage. You'll end up with 1V across the capacitor and 9V across the sum total of all of those resistors. Now the fun begins. because current through a resistor uses $V=IR$, we can calculate the current going through the system. The more ideal the wires and batteries were, the more current we're going to have to use to account for 9V. That current is a flow of charge. Where does it flow to? The capacitor. You will immediately start seeing the charge on the capacitor go up, as current flows through it, until eventually there's enough charge on the capacitor to generate 10V of potential across it. At that point, there's no more voltage to flow across the resistors, so the current drops to 0, and the circuit stays constant.

(Realistically there's some exponential terms in there, and it never technically gets to 10V exactly, but in realistic scenarios, we tend to get close enough to handwave away that set of extra complexities)

  • 1
    $\begingroup$ Another aspect: One can pull a charged capacitor apart, without changing the charge. But as different spacial dimensions mean different capacity, th eformula suggests that the voltag eshould increas by this, wven without any external power source connected - and it really does! (Actually, the power source is your muscles pulling the thing apart) $\endgroup$ Commented Apr 28, 2016 at 21:40

A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor.

So, basically you are storing charge on capacitors.

Let someone ask you how much charge you can store in your capacitor.What would you reply?

Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference you apply across capacitor. "

So, you need a standard to tell how much charge you can store at some universal condition.

The standard is 1V. Hence, the charge stored in capacitor at the standard of 1V is called capacitance of capacitor.

Why standard was 1V is because calculations become easy.

Why don't measure the ability to store something by the volume it takes so why not charge per unit volume. 

$ C = \epsilon°\frac{A}{d} = \epsilon°\frac{Ad}{d^2} = \epsilon°\frac{V}{d^2}$

So, there is relationship for volume too. But relation is not too direct . If you keep d constant and increase V charge you can store increases.

Instead if you keep A constant and then change V , it decreases.

Charge stored per unit volume, it can be actually given other names like charge density (or name it Smith :-) as you want).

This term may be useful to calculate size of capacitor required in any device. But , more direct use is of potential difference across capacitor.

Changing V for storing charge is much easier than changing volume of capacitors.

Or, according to the equation $C=\frac{Q}{V}$, why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge.

You are storing charge in capacitor. If you apply more PD, you can store more charge (I need not explain it).

If you can store more charge and hence more energy for same PD applied, won't it make you happy? So, capacitance is charge stored, and if you can store more charge for same PD of 1V, you say it has more capacitance.

  • $\begingroup$ I my self like my answer. +1 to me. $\endgroup$ Commented Apr 28, 2016 at 15:47

I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$

Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this:

If I have a basket with a capacity of 2 apples, then a basket with more capacity can hold more than 2 apples. So if I have a capacitor with a greater capacitance, it can hold more electric charge, right?

I see this misunderstanding come up in the context of electrical engineering quite a lot. Take this question, for example. Although the question doesn't come right out and say it, the way it's written suggests that the author suspects that if he can find a "big enough" capacitor ("big" meaning "high capacitance"), then this sufficiently big capacitor can hold enough energy for his needs.

But the fact is that an ideal capacitor is never "full", regardless of its capacitance. You can put as much charge or energy into it as you like. Just like you can stretch an "ideal" spring as far as you like. So any understanding of "capacitance" that's like "the capacity of an apple basket" is intuitively wrong and will never make sense.

Of course real springs will deform at some point, and real capacitors will fail at some point. But we are discussing the ideal case. And although I'm using the word "capacitor" as in the electrical device with two plates, the same applies to a single object and its self-capacitance. There's no limit to how charged a single object can be, in the ideal case.

Capacitance is simply how much the voltage will increase per unit charge. That is why a farad is equal to a coulomb per volt. One farad means that for every coulomb, there's a one volt change.

So you might think of capacitance as being analogous to the force constant of a spring. While the force constant tells you how much force is required to extend a spring, capacitance tells you how much voltage is required to charge a capacitor. A lower capacitance is like a stiffer spring.

With a little rearrangement, Hooke's law and the formula for capacitance are very similar:

$$ F = k X \\ Q = CV $$

For a higher force constant, it will take more force for a given change in extension. For a higher capacitance, it will take more charge for a given change in electric potential.


why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge.

(1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge $Q$ while the other plate has charge $-Q$ so there is no net electric charge stored.

(2) If you increase the voltage across a capacitor, the charge $Q$ must increase unless you're physically changing the arrangement of the the two plates in a particular manner.

For example, if the distance between the two plates of a parallel plate capacitor is increased (which decreases the capacitance) as the voltage across is increased (by, e.g., a variable voltage source), $Q$ can remain constant.

But keep in mind that it isn't the increase in voltage that changes the capacitance, it is this physical change of distance between the plates.

Now, there are devices which exhibit voltage dependent capacitance, e.g., varactor diodes but that is beyond the scope of this answer.


For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume.

There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it?

So here you have a capacitor and its charge per unit volume is $3 \;\text{C m}^{-3}$.

I ask "What happens to the charge per unit volume if you double the potential difference across the terminals?"
Your parameter does not help you produce an answer and you would need to use the idea that for your capacitor, charge is proportional to potential difference.

I ask "What happens if you halve the volume of the capacitor?"
Without any further information you would be unable to answer the question and to answer the question you will probably find it easier to go back to defining capacitance in the normal way.

"Charge per unit volume" might be useful in some applications but for the vast majority of instances $Q=CV$ is much more useful as a measure of charge (energy) storage capacity.


As I read your statements, I get the impression that the difference between capacity and capacitance is not clear to you. The capacity of a capacitor is defined by its "physical" construction (length, width, area, volume, material, etc. C = kA/d). However, capacitance is a measure of how difficult/easy it is for a capacitor to store charge (C = Q/V , similar to R = E/I). Although related, they are not the same thing. When you change the voltage, you change the capacitance, but not the capacity of the capacitor.
An analogy that might be useful is a dam. Its capacity to hold water is related to the height of the dam, but the actual amount of water it has, is related to the actual water height.


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