# Are capacity and capacitance the same?

Consider two parallel plate capacitors, one with a separation between plates of $$d_1$$ and other with $$d_2$$, where $$d_1,d_2\ll\sqrt A$$, and where $$A$$ is the area between the plates and both in the same medium.

Now let dielectric strength of material be $$E$$, therefore the maximum possible electric field that can be in the region containing the dielectric is $$E$$.

Both the capacitors can produce this electric field when charge $$Q=EA\epsilon_0$$ is given; therefore the maximum charge that both plates can hold, i.e. capacity of the plates is same.

If $$d_1>d_2$$, then the potential difference between the plates will be different, which will give higher capacitance of the plates with smaller separation.

Therefore, is the definition that capacitance is the ability to hold charge wrong? Since both plates can hold the same charge but have different capacitance? Should the correct definition be: capacitance is the ability to store charge per unit potential difference?

• You'll find that the first sentence of the Wikipedia article on capacitance gives exactly this definition.
– noah
Commented Dec 7, 2021 at 13:59
• I think dielectric strength in "...let dielectric strength of material be E..." should be $\epsilon_0$ not $E$. Commented Dec 7, 2021 at 14:00
• @BrendanDarrer I am not talking of dielectric constant( relative permittivity) but of dielectric strength( breakdown electric field) Commented Dec 7, 2021 at 14:01
• @BrendanDarrer No I am talking of en.wikipedia.org/wiki/Dielectric_strength Commented Dec 7, 2021 at 14:05
• "Should the correct definition be: capacitance is the ability to store charge per unit potential difference?" That is the definition of capacitance, $C=Q/V$. Commented Dec 7, 2021 at 16:27

Should the correct definition be: capacitance is the ability to store charge per unit potential difference?"

That is the electrical definition of capacitance, $$C=Q/V$$.

In the equation

$$Q=EA\epsilon_{o}$$

$$\epsilon_o$$ is the electrical permittivity of free space and $$E$$ is the magnitude of the electric field which, for a parallel plate capacitor, is related to voltage by

$$E=\frac{V}{d}$$

so

$$Q=\frac{V}{d}A\epsilon_o$$

The physical definition of parallel plate air gap capacitance is

$$C=\frac{\epsilon_{o}A}{d}$$

so increasing $$d$$ for the same dielectric and plate area decreases capacitance.

Substituting that into the previous equation gives

$$C=\frac{Q}{V}$$

So if the capacitor with larger plate separation holds the same charge as that with the smaller separation then the voltage on the capacitor with the larger separation has to be greater.

If by "capacity" you mean the amount of net charge on the plates, then obviously that's not the same as the capacitance of the capacitor which is the charge divided by the voltage. The capacitance of a capacitor is greater if the work required per unit charge to separate the charge on the plates (i.e., the voltage) is less.

Hope this helps.

• Yes your answer helps but please also include the answer/explanation for the question I wrote in title Commented Dec 7, 2021 at 16:50
• @LalitTolani I have updated my answer to respond. Commented Dec 7, 2021 at 17:02
• Thankyou for your help Commented Dec 12, 2021 at 4:45

Your thinking is correct. The material between the plates of any capacitor has a maximum electric field that it can tolerate before breaking down. And since the electric field between the plates is proportional to the charge on the plates, which is in turn proportional to the voltage applied, this means that most capacitors have a maximum voltage rating. Applying a voltage in excess of this rating can lead to dielectric breakdown of the material between the plates, which can lead to a more exciting day than you expected.

For many applications, though, we are more concerned with the proportionality between the voltage across the plates and the charge on them. (This is what's important in the analysis of AC circuits, for example.) Given two capacitors with the same voltage across them, the one with the higher capacitance will have more charge on it assuming that you don't exceed the voltage rating of either one. In this sense, capacitance can be thought of as "the capacity to hold charge"; a higher capacitance means more charge, all things being equal and assuming you haven't blown the dang thing up.

• a higher capacitance means more charge, all things being equal and assuming you haven't blown the dang thing up. This was the statement I was looking for. Commented Dec 9, 2021 at 4:37
• Thankyou for your help Commented Dec 12, 2021 at 4:45

The simple answer to your question is that the definition of capacitance is the charge accumulated on the plates per unit voltage - $$C = Q/V$$ - which is not the same as the maximum charge the plates could hold before dielectric breakdown occurs. The latter would be measured in units of charge, not charge per voltage, and would be given by $$Q_{max} = \epsilon_0 A E_{max}$$ as you say.

Giving this $$Q_{max}$$ the name "capacity" sounds like a marvelous way to invite confusion with the distinct idea of capacitance defined above. But that's just my opinion.

• Also note that "capacity" is in fact an archaic term for capacitance (I come across it from time to time in old EE books). And I think many languages use their equivalent of the word for capacity to refer to capacitance, too. Commented Dec 8, 2021 at 0:54
• Thankyou for your help Commented Dec 12, 2021 at 4:45

I think the confusion here is the overloaded word capacitance. When talking about a single charged object, you have what is termed self capacitance. When talking about two charged objects separated by distance, you have their mutual capacitance. To make things even more confusing, most EE texts (as well as most EEs) shorten mutual capacitance to simply capacitance, as they seldom, if ever, deal with charged objects in isolation.