# Why does it seem that the potential difference dependence of capacitance and total energy stored in a parallel-plate capacitor are contradictory?

Consider a parallel-plate capacitor. Charge is stored physically on electrodes ("plates") which are flat and parallel to one another. If one electrode has charge $+Q$ and the other electrode has charge $-Q$, and $V$ is the potential difference between the electrodes, then the capacitance $C$ is $$C = \frac{Q}{V}$$

(This definition of $C$ is given in, for example, Introduction to Electrodynamics by David J. Griffiths.)

But, now, let's think about the energy stored in the electric field between the electrodes of this parallel-plate capacitor. As stated in Griffiths on page 105, "How much work $W$ does it take to charge the capacitor up to a final amount $Q$?" It turns out that $W$ is $$W = \frac{1}{2} CV^2$$

So:

(i) the capacitor's capacitance $C$ goes like $\frac{1}{V}$; and

(ii) the energy $W$ stored in the electric field goes like $V^2$.

Are statements (i) and (ii) at odds with one another? I am sure that they cannot be. But conceptually I am having difficulty.

We desire high capacitance -- we want to put as much charge on the electrodes as possible, because if we accomplish this, then I think that will increase the energy density of the system. But is what I just said true?

If we manage to increase $Q$, then by $V = \frac{Q}{C}$, the potential difference $V$ between the plates will also increase. This, I think, is why capacitor electrodes are separated by a material (such as a polarizable dielectric material like a slab of plastic); otherwise $V$ will become too large and the breakdown voltage will be reached, generating a spark.

But, now, the equation $W = \frac{1}{2} CV^2$ (where I think that $W$ can be conceptualized as the energy stored in the electric field between the electrodes) seems to say that as $V$ increases, so does the energy $W$, quadratically.

So, my question is, do we want a capacitor to have a large potential difference $V$ or a small potential difference $V$? If $V$ is large, then $W$ is large (which we want), but $C$ is small (which we do not want).

Am I somehow thinking of two different potential differences $V$ and confusing them?

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You somehow think that $C$ is dependent on $V$. It isn't. –  C.R. Jul 30 '12 at 6:34

As a summary of what other answers have already stated, in essence:

• Capacitance is a function of the geometry of the capacitor (directly proportional to the overlap area of the plates and inversely proportional to it's separation) and of the relative permittivity of the dielectric employed. Once this constructive parameters have been fixed, it's capacitance gets uniquely defined, and so it's constant when the capacitor gets charged and discharged (assuming linear response, as it's the norm in standard circuits).
• The relation between voltage between the plates of the capacitor, the accumulated charge and it's capacitance is just the definition of the latter. That capacitance is a property of the system which is a function only of geometry and dielectric's permittivity, is a fact that can be deduced from this definition.
• So, once you have picked up a capacitance (by fixing the parameters involved as explained above) maximization of stored energy in the electric field generated gets down to increasing voltage between the plates as much as possible. Nevertheless, increasing the electric field's strenght has a limit (the dieletric strength), which corresponds to the breakdown voltage of the dielectric, so that is the maximum voltage safely and technically attainable. So:

$W_{max}=\displaystyle\frac{1}{2}CV^{2}_{max}\ ;\\V_{max}=E_{max}\cdot d\ ;\\ C=\displaystyle\varepsilon_{0}\varepsilon_{r}\frac{A}{d}\Longrightarrow W_{max}=\displaystyle\frac{1}{2}\varepsilon_{0}\varepsilon_{r}E_{max}^{2}\cdot A\cdot d=\displaystyle\frac{1}{2}\varepsilon_{0}\varepsilon_{r}E_{max}^{2}\cdot Vol_{\ dielectric}$

Corollary:

• The amount of energy stored in a fully charged capacitor is just obliquely related to it's capacitance. Nevertheless, even though the energy of the electric field is directly proportional to the volume of dielectric between the plates (the product of the plate's area and their separation), for a given amount of dielectric material, the preferred geometry implies a large area and as little separation as possible, because that arrangement allows more compact designs. That's why a larger capacitances leads to larger energy densities in practical applications.
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First of: the energy W does not increase exponentially it increases quadratic!

Second:

What do you want?

Energy density!

So to get good high energy density you need high C because you can (as you already stated) only go to a certain amount of voltage until you get a breakdown. The problem is, C does not change easily. If you take a simple Plate - Plate capacitor $C \propto \frac{A}{d}$ where A is the Area of the plates and d the distance between them. The smaller the distance the higher the electric field gets between the plates gets relative to Voltage ($|\vec E| \propto \frac{U}{d}$) the faster you get a spark. So the only thing you can actually increase safely is the area. That of course increases the wight/size of you capacitor, reducing the Energy density. This generally applies to all capacitors

Wikipedia does have an excellent article about capacitors:

http://en.wikipedia.org/wiki/Capacitor

Third:

A spring is a very good analog to your problem:

A newtonian spring has a certain flexibility, describing the force you have to pull it relative to the expansion of the spring:

$F=k \times x$

Where F is the Force k is the flexibility and x the way. So the capacitance is actually k in this analog. But F is not the Energy W.

$E = \int_0 ^x F(x) dx$

So the more you pull the harder it gets to pull because the force from the spring increases. So to get the energy you can not just calculate $E = F \times x$ but you rather have to integrate the way.

That of course gets you to

$E = \frac{1}{2} k x^2$

Back to the Capacitor:

The more electrons you put in it, the stronger becomes the electric field between the plates of the capacitor. So you have to increase the voltage. The capacitance just tells you how high your voltage has to be (how hard you have to push) to put more electrons on the plate.

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Thanks! Still, though, do we want large $V$ or small $V$? My difficulty is still that "What do you want? Energy density!" seems to imply that we want both large $C$ and large $W$. Both $C$ and $W$ depend on $V$, but in very different ways: $\propto \frac{1}{V}$ versus $\propto V^2$, and I still find this confusing. –  Andrew Jul 29 '12 at 17:40
think about what you can change! C does not only depend on V, it also depends of Q. You can not decrease the voltage on one and the same capacitor without changing Q as well. Think about diminishing the force of a spring without changing the expansion of it. You would need another spring. –  miceterminator Jul 29 '12 at 17:51

Capacitance depends on the geometry of the conductors holding the charge. To maximise the energy stored, you maximise the capacitance by changing the geometry of the conductors, and maximise V by transferring as much charge as possible.

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Thanks! I understand that we need to "maximise the capacitance," since the total energy $W$ stored is proportional to the capacitance $C$, by $W = \frac{1}{2} CV^2$. But I don't understand why we need to "maximise $V$" if $C = \frac{Q}{V}$, where increasing $V$ causes a decrease in $C$. –  Andrew Jul 29 '12 at 18:23
@Andrew You misunderstand the use of $C = Q/V$. That is a prescription about how to measure the capacitance. Once you have chosen a geometry the capacitance does not change. Again, the capacitance is a function of the geometry and materials chosen. It is not a function of the potential because the charge Q and the potential V are linked to one another. –  dmckee Jul 29 '12 at 18:41