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Let's say you have two infinitely large plates. I understand the formulas for capacitance. What I don't understand is the intuition behind the fact that when you increase the distance between the plates, capacitance decreases. I understand that as you increase distance, potential difference increases, and so based on one of the capacitance formulas, capacitance decreases. That's obvious. But what's the intuitive meaning / significance behind it? The textbook I'm using says "capacitance is a measure of the ability of a capacitor to store energy." That seems to me to go against the idea that as you increase distance, you decrease capacitance, because the further apart the plates are, the more energy is stored between them due the increasing potential difference. If you can answer my question, please do not use formulas. I understand the formulas. I'm really looking for just an intuitive understanding. Also, I think I would understand more if you don't provide an analogy (that's just me). Thank you in advance.

PS: Sorry I think it's important to distinguish between the case of infinitely large plates and not infinitely large. My question was addressing infinitely large plates. But I would appreciate an explanation on not infinitely large plates as well, and the differences between these two cases.

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  • $\begingroup$ The textbook is wrong. It is not the ability to store energy. A 100pf capacitor and a 1000μF capacitor are equally able to store energy. Capacitance is the amount of energy stored per volt of potential difference. Calling it “ability” leads to confusion and misunderstanding. $\endgroup$ – Martin Kochanski Jun 18 at 5:02
  • $\begingroup$ @Martin The textbook isn't wrong. The ability to store charge/energy (same thing) IS the ratio of charge a capacitor can store for a given voltage. The less voltage needed to store a given amount of Q, the better a capacitor is at storing charge/energy and thus a higher capacitance. It is a measure of efficiency and it is determined by the physical geometry of the capacitor which allows the charges to arrange themselves. Sometimes we want a lower Q and a higher V though so a higher capacitance doesn't automatically mean a more desirable capacitor. $\endgroup$ – Andrew Jun 18 at 16:11
  • $\begingroup$ It's no different than saying giving a ratio of steps I take to energy I use - a measure of my energetic efficiency to travel distance. $\endgroup$ – Andrew Jun 18 at 16:12
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Sorry I re-read your question and I misunderstood your confusion. It seems that you feel the equations lead to a contradiction with the definition of capacitance. This is because it seems you think a higher capacitance means more energy stored on the capacitor. From our equations for capacitance of an infinite capacitor:

$C = Q / V = Q / Ed$

$E$ is constant for infinite plates with charge uniformly distributed. If $E$ and $Q$ are constant and $d$ increases then it would appear that capacitance decreases so the capacitor's "ability to store energy" diminishes.

We know that energy for a capacitor is $U = \frac{1}{2} (QV) = \frac{1}{2}Q(Ed)$

Which would imply that as $d$ increases, $U$ increases which means the capacitor is actually storing MORE energy! This isn't a contradiction because while the energy stored by the capacitor went up, its ability to store energy is a different concept. $U$ and $C$ don't necessarily go up or down with each other. Its ability to store energy depends on its ability to store charge which depends on its geometry and distance to the other plate; not it's current level of energy storage. It's the capacitors size/shape that determines how much charge can be stored on it (its capacitance).

Here are some examples of expressions for capacitance of different shapes:

Parallel plate capacitor (for short distance $d$, area $A$ of plate): $C = \epsilon\frac{A}{d}$

Coaxial cable ($l$ is length of cable, $R_{1}$ and $R_{2}$ is inner and outer radius): $\frac{2\pi\epsilon l}{ln\frac{R_{2}}{R_{1}}}$

Each of these quantifies how much charge a capacitor can store (and thus how much energy it can store) and you'll notice they depend exclusively on the geometry of the capacitor AND the distance between the capacitors - $d$ for the parallel plate and $R_{1}$ and $R_{2}$ for the coaxial cable.

I think it's important to remember that infinitely sized plates don't exist. Plates are considered infinite either because they are so close as to appear infinite to each other or so big that moving further or closer makes no difference to their relative size to one another. But in reality, the "so big" infinite plate doesn't exist. We can get two objects so close to each other so that their electric field is constant in the region between them because the plates appear infinite in the region between them but we can't make an infinitely large plate.

Now, in the case of plates being so close as to appear infinite, when we move them further apart, they no longer appear infinite and $E$ doesn't remain constant; it decreases faster than $d$ increases because $E$ is an inverse square law (it is related to distance according to $\frac{1}{d^2}$). Which means $U$ and $C$ both decrease in that particular case.

Sorry for the hastily written answer. Let me know if you're still confused

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  • $\begingroup$ But what about the case of "infinitely large" plates? You mentioned that the electric field decreases as we move away. But if we're talking about this ideal case, how does this apply here? $\endgroup$ – user10796158 Jun 17 at 19:24
  • $\begingroup$ Or would the capacitance just be constant for this case no matter how far apart they are? $\endgroup$ – user10796158 Jun 17 at 19:26
  • $\begingroup$ I'll update my answer to include this, but if the plates are INFINITELY large, then there is no concept of distance between the two plates. The entire point of the plates being infinite is that they are either so close as to appear infinite to each other or so big that moving further or closer makes no difference to their relative size to one another which means their capacitance would remain constant regardless of "distance" for "infinite" plates $\endgroup$ – Andrew Jun 17 at 19:28
  • $\begingroup$ Thank you this is making more sense now $\endgroup$ – user10796158 Jun 17 at 19:31
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    $\begingroup$ @Andrew You are confusing the behavior of the electric field with that of the potential. The potential depends on the distance from the infinite plate. The potential difference between two infinite plates depends on the distance between the plates. Your answer contains some misleading/wrong statements. $\endgroup$ – nasu Jun 17 at 20:19
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Infinite plates have a constant electric field (at fixed charge density). Constant electric field means constant voltage gradient, so total voltage increases linearly with distance from the plate.

Capacitance is charge (which is fixed) per volts (which increases with distance); hence: capacitance decreases with distance between the plates.

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I just want to dispel with the idea that a capacitor has to have "infinitely large" plates. Obviously real capacitors don't have "infinitely large" plates.

What should be said is that any dimension of the plates should be much greater than the distance between the plates (thickness of the dielectric, $d$) so that the electric field $E$ can be considered constant between the plates (neglecting edge effects) and is $E=\frac{V}{D}$.

For example, given a thickness of a plastic film dielectric (e.g. polyester, polypropylene or polycarbonate) on the order of 10 x $10^{-6}$ meters used in a 25 mm dia capacitor, the ratio of the diameter to the plate separation is 2500:1.

Hope this helps.

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Let's say you have two infinitely large plates. I understand the formulas for capacitance. What I don't understand is the intuition behind the fact that when you increase the distance between the plates, capacitance decreases.

The capacitance of two infinitely large plates separated by a distance $d$ is infinite for any finite $d$. I don't think you want to say "infinitely large plates" for this problem. You might try arbitrarily large plates such that $A \ggg d$ and then $C$ and $Q$ are finite while $C = \epsilon A / d$ holds to arbitrary precision.


The textbook I'm using says "capacitance is a measure of the ability of a capacitor to store energy." That seems to me to go against the idea that as you increase distance, you decrease capacitance, because the further apart the plates are, the more energy is stored between them due the increasing potential difference.

If I read this correctly, you're expecting that the amount of energy stored must decrease if the capacitance decreases due to the increasing distance between the plates. But it doesn't follow that the energy should decrease.

For argument's sake, stipulate that the distance $d$ is doubled so that the capacitance $C$ is halved and, assuming $Q$ is held constant, it follows that the voltage $V$ across is doubled.

The energy stored before $d$ is doubled is

$$W = \frac{1}{2}CV^2$$

while the energy stored after $d$ is doubled is

$$W = \frac{1}{2}\frac{C}{2}(2V)^2 = CV^2$$

So the stored energy is doubled even the though the capacitance $C$ is halved. However, if we had doubled the voltage across the capacitor without doubling the distance, the stored energy would be

$$W = \frac{1}{2}C(2V)^2 = 2CV^2$$

Which is twice as much for the same voltage across. That is, after doubling the distance $d$, the capacitor stores half as much energy, for a given voltage across, as it did before despite the fact that there is more energy stored after $d$ is doubled (the additional energy stored by moving the plates apart is equal to the work done by whatever force moved the plates apart).

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  • $\begingroup$ Thank you for your answer. So this is kind of answering what I’m getting at, but not quite yet. I’m going to ask a related followup question if that’s okay. So capacitance tells us how much charge we can store for a given voltage. Say you have a capacitor, and you change the (small) distance between its (very large) plates. The capacitance will change. So really what’s the point of talking about ratio of charge to voltage for a given capacitor if it changes with distance? What is this change telling us about the capacitor? $\endgroup$ – user10796158 Jun 18 at 3:08
  • $\begingroup$ @user10796158, (1) to be clear, a capacitor doesn't store electric charge. The charge $Q$ is the charge on one of the plates while the other plate has charge $-Q$ so a charged capacitor is not electrically charged, it is electrically neutral. However, a charged capacitor has stored energy (analogous to a charged battery). (2) Ordinarily, the capacitance of a capacitor is effectively constant so that we can write $i_C = \frac{dQ}{dt} = C\frac{dv_C}{dt}$ for electric circuits. There's a paper I read a while back that addresses some of the problems of taking this derivative when $C = C(t)$. $\endgroup$ – Hal Hollis Jun 18 at 22:04
  • $\begingroup$ @user10796158, I'll see if I can find it, and I'll post a link if I do. $\endgroup$ – Hal Hollis Jun 18 at 22:05
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I will try explaining it by analogy.

We have 2 rectangular rooms (plates) - M and W. The room M is full of young men (particles with minus charge), and the room W is full of young women (particles with plus charge).

enter image description here The both rooms are overcrowded, so no other young man and no other young woman is able to enter the appropriate room, because neither the young men, nor the young women have a will to reduce the distance between their peers (identically charged particles repel one other).

But all the young men, and all the young women are so nice, that every young man is attracted to every young woman, and every young woman is attracted to every young man, and moreover both have a glass wall toward the other room (the dashed lines in the picture).

Rooms are in very large distance one from other, let's say 400 meters - long enough for recognizing individual persons, so the attraction doesn't work.

Now move the rooms closer one to other, let's say to 100 m distance:

enter image description here In this new situation the attraction begins to work - young men begin to crowd together in the direction to the glass wall, and young women will do the same. Consequently, some new young men will be able to enter room M, and some new young women will be able to enter room W. The capacity of the both room becomes larger.

  • The shorter distance, the stronger attraction, and - consequently - the larger capacity.

  • The longer glass wall, the more young men / young women may be closer one to other, and - consequently - the stronger attraction, the larger compression, and the larger capacity.

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