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Why does the capacitance of a capacitor increase when its plates are closer in distance to each other?

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Intuitive approach: if the distance wouldn't be a factor then you would be able to place the plates at an infinite distance apart and still have the same capacitance. That doesn't make sense. You would expect a zero capacitance then.
If the capacitor is charged to a certain voltage the two plates hold charge carriers of opposite charge. Opposite charges attract each other, creating an electric field,

enter image description here

and the attraction is stronger the closer they are. If the distance becomes too large the charges don't feel each other's presence anymore; the electric field is too weak.

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  • \$\begingroup\$ true, and nice graphic, but let's play devil's advocate: just because for a given charge Q, the electric field is stronger when the plates are closer doesn't give you any intuitive indication that the voltage is stronger or weaker (Q=CV so higher capacitance means lower voltage for fixed charge). I don't buy the infinite argument, either: infinitesimal electric fields integrated over an infinite distance gives an indeterminate voltage. \$\endgroup\$ – Jason S Jul 7 '11 at 22:36
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    \$\begingroup\$ -1, because conductors at an infinite distance actually have finite capacitance. Consider a single conductor sphere w/ radius R1, and charge Q. Outside the sphere, the field is Q/(4*pieps0*r^2), and if you integrate this from radius R1 to infinity, you get voltage V = Q/(4*pieps0*R1). If you superpose the electric fields of another sphere with voltage -Q of radius R2 infinitely far away, you get a total voltage between the spheres of Q/(4*pieps0)*(1/R1+1/R2) -- it's additive rather than subtractive (opposite signs of Q cancel the opposite path integral), so C = Q/V = 4*pieps0/(1/R1+1/R2) \$\endgroup\$ – Jason S Jul 8 '11 at 12:51
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    \$\begingroup\$ @Jason - parallel plate cap: \$C = \frac{\epsilon A}{d}\$. \$\epsilon\$ and A are finite, d is infinite, so C = 0. QED \$\endgroup\$ – stevenvh Jul 8 '11 at 12:56
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    \$\begingroup\$ Wrong. The equation only applies for d << the dimensions of the plate. \$\endgroup\$ – Jason S Jul 8 '11 at 12:57
  • \$\begingroup\$ For parallel discs of radius R and distance d, a closer approximation is \$C = \epsilon [\pi R^2/d + R ln (16\pi R/d - 1)]\$, but even that is still an approximation -- see santarosa.edu/~yataiiya/UNDER_GRAD_RESEARCH/… \$\endgroup\$ – Jason S Jul 8 '11 at 13:06
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FIG 1 to 4: Capacitor:

Capacitor diagram

It is obvious that as the distance between plates decreases, their ability to hold charges increases.

fig.1 = If there is unlimited distance between plates, even a single charge would repel further charges to enter the plate.

fig.2 = if distance bet plates decreases, they can hold more charges due to attraction from the opposite charged plate.

fig.4 = with minimum distance between the plates, the max attraction between them enables both to hold max amount of charges.

As Capacitance C = q/V, C varies with q if V remains the same (connected to a fixed potential elec source). So, with decreased distance q increases, and so C increases.

Remember, that for any parallel plate capacitor V is not affected by distance, because: V = W/q (work done per unit charge in bringing it from on plate to the other)

and W = F x d

and F = q x E

so, V = F x d /q = q x E x d/q

V = E x d So, if d (distance) bet plates increases, E (electric field strength) would drecrese and V would remain the same.

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  • \$\begingroup\$ Of course $V$ is affected by distance. You have $V = E \times d$ in your final sentence for example. And $V$ is an integral of $E$ over some distance so as $d$ increases we are adding up more of $E$ so $V$ should increase. \$\endgroup\$ – csss Jul 6 '16 at 12:20
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Capacitance is charge per EMF. Specifically Farads are Coulombs per volt. As you move the plates closer at the same applied voltage, the E field between them (Volts per meter) increases (Volts is the same, meters gets smaller). This stronger E field can hold more charges on the plates. Remember that the charges on the plates would otherwise repell each other. It takes a E field to keep them there, and the stronger the E field the more charges it can keep there. The higher charge at the same voltage means higher capacitance (more Coulombs at the same Volts).

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  • \$\begingroup\$ almost answers it... there's kind of a handwaving thing here about stronger E field implying more charge, but I'll give you a +1: linearity arguments (Q should be proportional to E) are probably good enough. \$\endgroup\$ – Jason S Jul 7 '11 at 23:11
  • \$\begingroup\$ @Jason, I was trying to keep it simple, because it actually is a pretty simple concept. It's hard to judge what level of detail the OP wants, so I don't know where to stop explaining and start waving the hands. Too far either way is bad. If you don't believe that, take a look at the mess Matt's answer has turned into. Without direction from the OP, I picked what I thought was a reasonable tradeoff he can ask more about if he wants to. \$\endgroup\$ – Olin Lathrop Jul 7 '11 at 23:27
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To get technical, you want to look at Coulomb's law. This states that

"The magnitude of the Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them." - Wikipedia

The formula for this is:

\$F = k_e \frac{q_1 q_2}{r^2}\$

Where \$F\$ is the electrostatic force between two charges, \$k_e\$ is a 'proportionality constant' (eg the dielelectric constant in a capacitor), and \$r\$ is the distance between the two charges \$q_1\$ and \$q_2\$.

There are other forms of the equation - such as this one specifically for an electric field:

\$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\$

Which tells us the force at a distance \$r\$ from the single point charge \$q\$.

If you want to start getting really technical then you need to start reading up on quantum mechanics and the interactions between particles and the energies involved in it.

When two particles (say electrons in this case) interact they send quantum particles between them (photons). These, like the rats in the basement, require energy to move. The greater the distance the higher the energy. The higher the energy taken to move the photons the lower the charge left between the two plates.

That's a very simplistic view of it and there is one helluva lot more detail in there to be discovered - such things as Quantum Tunneling, Leptons, Fermions, Bosons, etc. It's fascinating reading if you have the time. I'd recommend Steven Hawking's A Brief History of Time as a good starting point. Follow that up with F. David Peat's Superstrings and the Search for the Theory of Everything and you won't go far wrong. While both these books are getting a bit long in the tooth now and the theories are all still evolving, they give good insights into the workings of the universe at a subatomic level.

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    \$\begingroup\$ You spell out the formulas which show the relationship to distance, but I got the impression OP already knows that. He doesn't ask if distance affects capacitance, but why it does. if (nitpicking) then say_sorry; \$\endgroup\$ – stevenvh Jul 7 '11 at 8:29
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    \$\begingroup\$ @stevenvh The why is what the formulas demonstrate - we're getting into quantum mechanics here. Is there a difference between what and why, and even where and when? Oh, and it should be if(nitpicking) { say_sorry(); } ;) \$\endgroup\$ – Majenko Jul 7 '11 at 8:49
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    \$\begingroup\$ Yeah, I was a difficult guy in college. I often asked why and the professor would always point to formula, which left me frustrated, because I didn't find that satisfying. There had always to be an intuitive explanation :-). And my code is pseudo-code, so it compiles correctly! ;-) \$\endgroup\$ – stevenvh Jul 7 '11 at 9:07
  • \$\begingroup\$ I'm sorry, but it segfaults on my core - it must be an incompatibility in firmware. For more into the "Why" you want to read "A Brief History of Time" (Steven Hawking) followed by "Superstrings and the search for the theory of everything" (F David Peat) and you'll have a lot more knowledge, but still be non the wiser ;) \$\endgroup\$ – Majenko Jul 7 '11 at 9:22
  • \$\begingroup\$ @stevenvh - Your code compiles fine with Delphi & FreePascal :o} \$\endgroup\$ – MikeJ-UK Jul 7 '11 at 10:17
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A key thing to understand is that if a plate has more electrons coming in than going out, it's going to build up a negative charge which will serve to repel any more electrons from coming in (likewise for a plate with more electrons leaving than arriving). It wouldn't take very many electrons coming into an isolated plate for the charge to build up to millions of volts. If, however, there is a positively-charged plate near the negatively-charged one, the positively-charged plate would try to pull electrons toward itself and consequently toward the negative plate (likewise the negatively-charged plate would try to push electrons away from itself and consequently away from the positive plate). The force from the positive plate trying to draw in electrons cannot completely counter-balance the force of the negative plate trying to push them away, but it if the plates are close together it can counterbalance it significantly. Unfortunately, if the plates are too close, the plates won't be able to build up too much of a charge before electrons start hopping from one plate to the other.

It turns out there's trick to ease this problem. Some materials allow electrons to move about within them, but they don't allow electrons to enter or leave. Placing such a material (called a dielectric) between the two plates can greatly improve the performance of a capacitor. What happens, essentially, is that the charge difference between the negative and positive plates moves the electrons in the dielectric toward the positive one. The side of the electric toward the negative plate thus has a relative shortage of electrons, drawing electrons toward the negative plate, while the side toward the positive plate has a surplus of electrons, pushing electrons away from the positive plate. This behavior can improve the performance of a capacitor by many orders of magnitude.

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    \$\begingroup\$ -1: you're talking about dielectric strength, but you don't make any mention, either quantitatively or qualitatively, about the capacitor's capacitance. \$\endgroup\$ – Jason S Jul 7 '11 at 23:13
  • \$\begingroup\$ @Jason S: Capacitance is the ratio of the amount of charge imbalance to the amount of electromagnetic force required to hold that level of charge imbalance. Perhaps I should have defined capacitance in terms of coulombs per volt, but I believe the first paragraph pretty well answers the question that was asked. The second question was intended to make clear that it's not just the electrons on the plates that play a role in capacitor behavior; those in the dielectric are often very important as well. \$\endgroup\$ – supercat Jul 7 '11 at 23:21
  • \$\begingroup\$ @supercat: It's not electromagnetic force. Magnetism has nothing to do with capacitors. It's stricly about EMF (ElectroMotive Force). This is the physical property often measured in Volts. \$\endgroup\$ – Olin Lathrop Jul 7 '11 at 23:31
  • \$\begingroup\$ @Orin Lathrop: Sorry, my terminology in the comment was wrong, though I don't use the term "electromagnetic force" in the answer. I think the key point I was trying to put forth in my answer was that electrons can flow into the negative plate, despite the charge imbalance, because they are attracted toward the positive plate. Without the attraction from the positive plate, one could push a few electrons into the negative plate, but not a whole lot. \$\endgroup\$ – supercat Jul 8 '11 at 1:29
  • \$\begingroup\$ @supercat: I still see nothing in your answer or comments to explain why capacitance increases when the plates are closer. Why doesn't capacitance decrease when the plates are closer? Why doesn't it stay the same? The quantitative/qualitative behavior of capacitance as a function of plate distance is different from (but related to) the quantitative/qualitative behavior of charge or of electric field. \$\endgroup\$ – Jason S Jul 8 '11 at 12:33

protected by Dave Tweed Apr 22 '15 at 10:28

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