Are capacity and capacitance the same

capacitancedefinitionelectric-fieldselectrostaticsterminology

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?

Best Answer

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

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

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

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.

Yes your answer helps but please also include the answer/explanation for the question I wrote in title

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.

Related Solutions

[Physics] Capacitance of three plates

Consider the three-terminal device that is your stacked capacitor:

 A ----=============================================
                   (dielectric medium ɛ)
       =============================================---- B
                   (dielectric medium ɛ)
 C ----=============================================

All three plates have the same area A. It's pretty clear what happens when A is left "floating". (Defining floating: before operation we ground A so that $Q_A = 0$, then we disconnect it so that $A$ is not connected to any particular other component.) When we float A and measure the capacitance between B and C, or when we float C and measure the capacitance between A and B, we get an electric field $E = Q / (\epsilon A)$ and capacitance $C = \epsilon A / d$ where $A$ is the area of these plates, $Q$ is the charge on one terminal, and $d$ is the distance between them. The key observation here is that the capacitor has a very large area compared to its width, so the electric field outside of the capacitor tends to 0, so neither A nor C really "matters" when it's floating in that 0 electric field.

It is somewhat harder to think about what happens when we float B and then measure the capacitance between A and C. The electric field needs to be 0 inside B, because it is a perfect conductor and any electric field will cause current to flow. At the same time, the overall charge is 0 and the situation is still "capacitor-like" ($A$ is much much larger than $2d$ if it's much much larger than $d$) so the electric field outside of the parallel plates should tend to 0. Since the charge on the plate directly creates a discontinuity in the electric field, we have to come to this conclusion: the field in the dielectric between B and C is the same when we put $+Q$ on A, $-Q$ on C, as when we floated A, putting $+Q$ on B and $-Q$ on C. It has to be, because it's the same jump discontinuity from the same starting point ($E = 0$ outside the stack). The same must also be true between A and B. The field must be $Q / (\epsilon A)$ in both dielectrics.

The condition that $E = 0$ within the middle plate means that we induce a surface charge of $+Q$ on the BC side of B, and a surface charge of $-Q$ on the AB side of B. When you include those surface charges, it "looks exactly like" two capacitors in series, and you expect half of the capacitance.

And that's exactly what happens if you ignore B, too! If you ignore B, then you've got a constant field of $E$ over twice the distance, so $V = 2 E d$ for the same $Q$, so it takes twice as much voltage to get the same charge on each plate. So while B is doing "something", it's actually doing the nothingest something it can do. So you're right to intuit that it should just work like a single two-plate capacitor, if you ignore the thickness of the B plate in the calculation of how thick the capacitor is.

Now that we understand this, here comes situation 1. For situation 1, the easiest way to get an analogous result is to connect A and C with a wire, so they are at the same voltage, each plate holding charge $Q/2$ while the B-plate holds charge $-Q$. Then you are correct to intuit that this just looks like two capacitors in parallel. What happens? Well, the field is still 0 outside the system. The charge $Q/2$ therefore means that we have half the electric field inside the AB and BC dielectrics, which means that the same charge requires only half the voltage, so the capacitance doubles.

Now what if, as you say, we put charge +Q on plate A, +Q on plate C, and -Q on plate B? Well, we have a problem: the overall charge is no longer 0. Under the same "parallel plate" assumption that $A$ is much much larger than $d$ we find the fields by the principle of superposition:

                    E = + Q/(2 A ɛ)
 A ----=============================================
                    E = - Q/(2 A ɛ)
       =============================================---- B
                    E = + Q/(2 A ɛ)
 C ----=============================================
                    E = - Q/(2 A ɛ)

Now, we can't even define capacitance unless we choose two points to measure a voltage between. Suppose you want the points A and C: the voltage between these plates is 0, and the capacitance is infinite. Actually, since this voltage is 0, we wouldn't change the system fundamentally by connecting A to C. So then you can consider the voltage between A and B, and you get the same result as before, twice the capacitance as if C weren't there. The surplus charge on A and C, while it superimposes on the electric fields, doesn't affect the capacitances involved.

[Physics] Capacitors in a series circuit with dielectric

You have to deal with this in stages:

Adding a dielectric to $C1$ will increase its capacitance. With the same charge present, that means its voltage $V_{C1}$ will go down.
Since $V_{C1}$ goes down, the total of $V_{C1}$ and $V_{C2}$ will now be less than the battery voltage $V_B$. This will start current flowing through the circuit to compensate.
The current will add equal amounts of charge to both capacitors, and increase their voltages, until $V_{C1}+V_{C2}=V_B$.
Once stability returns, the following changes will have been made:
- The capacitance of $C1$ will have increased, but that of $C2$ will stay the same
- The charge on both capacitors will have increased, and by the same amount
- The voltage on $C1$ will have decreased, and that of $C2$ will have increased

Best Answer

Related Solutions

[Physics] Capacitance of three plates

[Physics] Capacitors in a series circuit with dielectric

Related Question