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In lab, my TA charged a large circular parallel plate capacitor to some voltage. She then disconnected the power supply and used a electrometer to read the voltage (about 10V). She then pulled the plates apart and to my surprise, I saw that the voltage increased with distance. Her explanation was that the work she did increased the potential energy that consequently, increases the voltage between the plates but the electric field remains constant. Although I tried to get more of a physical explanation out of her, she was unable to give me one. Can someone help me here?

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    $\begingroup$ I saw this demo also and was complete lost on the explanation. Great question. $\endgroup$ – Lisa Lee Oct 27 '13 at 15:20

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Toby, I agree that this is really counter intuitive and I was also quite surprised as well when I first saw this very demonstration. I am an undergraduate TA and this is how I explained it in my lab section. I hope this helps. I see two parts to a full explanation: (1) Why is the electric field constant and (2) why does the potential difference (or voltage) increase?

Why is the electric field constant as the plates are separated? The reason why the electric field is a constant is the same reason why an infinite charged plate’s field is a constant. Imagine yourself as a point charge looking at the positively charge plate. Your field-of-view will enclose a fixed density of field lines. As you move away from the circular plate, your field-of-view increases in size and simultaneously there is also an increase in the number of field lines such that the density of field lines remains constant. That is, the electric field remains constant. However, as you continue moving away your field-of-view will larger than the finite size of the circular plates. That is, the density of field lines decreases and therefore, the electric field decreases as well as the potential field.

enter image description here

To show this mathematically, the easiest way to show this for E = constant is using the relation between the electric and potential fields: $$E = -\frac{\Delta V}{\Delta d} \longrightarrow \Delta V =-E \Delta d$$ I would expect the voltage to increase linearly as long as the field is constant. When the electric field starts decreasing, the voltage also decreases and the fields behave as finite charged plates. Although I’ve only talked about one plate, this idea immediately applies to two plates as well.

Why does the work increase the electrical potential energy of the plates? One way to interpret why the voltage increases is to view the electric potential (not the electrical potential energy) in a completely different manner. I think of the potential function as representing the “landscape” that the source (of the field) sets up. Let me explain what the gravitational potential acts like when a ball is thrown upwards (of course, you know what happens in terms of the force of gravity or in the conservation of energy scenario). I claim that the potential function is related to the “gravitational landscape” that the earth sets up, which is derived from the potential energy and is equal to the potential energy per mass:

$$ {\Delta U = mg\Delta y} \longrightarrow \frac{\Delta U}{m} = \Delta V = g\Delta y$$

Plotting these functions, the constant gravity field sets up a gravitational potential ramp (linear behavior) that looks like

enter image description here

In terms of energy, the ball moves up this gravitational ramp were the ball is converting its kinetic energy into potential energy until the ball reaches it maximum height. However, the gravitational ramp exists whether the ball is thrown up or not. That is, gravity sets up a gravitational ramp (the landscape) and this is what the ball “sees” before it is thrown up.

If we now apply the above thinking to a constant electric field between the parallel plates, the electric potential function is derived in a similar manner:

$$ {\Delta U = qE\Delta r} \longrightarrow \frac{\Delta U}{q} = \Delta V = E\Delta r$$

If we look at the electric potential of the negative plate (it’s easier than the positive plate), it has a negative electrical ramp that starts at 0V.

enter image description here

So as your TA pulls the plates apart, the work she does moves the positive plate up the electrical ramp and increases the potential of the positive plate. So this interpretation of the electric potential is what you intuitively already think about in terms of mechanical situations like riding your bike up a hill. There is no difference in the electrical situation.

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    $\begingroup$ Thanks for taking the time to write such a long and detailed answer. Love your images. What software do you use to create these? $\endgroup$ – Toby Oct 27 '13 at 15:28
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    $\begingroup$ @Toby, I don't know if this is what Carlos used for this particular answer, but they look like they could have been done in Inkscape. $\endgroup$ – Colin McFaul Oct 27 '13 at 16:06
  • $\begingroup$ @Toby: Your welcome and I used Snagit by TechSmith to create those images. $\endgroup$ – Carlos Oct 27 '13 at 17:28
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    $\begingroup$ @Carlos - Your way of explaining why field E is constant and then it reduces is beautiful; thanks a lot. But your answer that potential would decrease as a result of decrease in electric field is incomplete, as potential is dependent on distance too and it is increasing. I think the appropriate way would be - "electric field reduces after a certain point; potential would reduce only after a point where the rate of decrease in electric field is larger than the rate of change of distance"...Correct me if I am wrong... $\endgroup$ – karthikeyan Nov 1 '13 at 19:25
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Here is my understanding:

When you increase the distance between electrodes - capacitance drops, but stored charge remains the same, as electrons have nowhere to go.

Same charge in lower capacitance means higher voltage potential. Without that part of stored energy would just vanish :-)

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    $\begingroup$ I agree. Q = C.V The capacitance is inversely proportional to the plate spacing, so long as it is small compared to plate length and width (fringing fields), doubling the space, halves the capacitance, and doubles the Voltage, so long as the charge is unchanged. $\endgroup$ – user26165 Oct 28 '13 at 3:44
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I think as we know E = V/d, and the field is same, so for field remains constant between the plates of the capacitor, while increasing the distance the potential also increases. In the same manner as that of distance so that the ratio of V and D is same always.

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It is easy! We know that C=q/U, where C - capacitance, q - electrical charge, U - voltage between plates,

Charge (q) can not change, but when plates are separated, capacitance goes down so U goes up!

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When the two capacitors are charged, they are constantly trying to come closer due to electrostatic forcd between them, when you displace the plates away from each other there is a net displacement in opposite direction to that of force, hence - work is done by the capacitor system or in other words the energy of this system increases which gets stored as electrostatic potential.

Another explanation can be that a certain capacitor system is able to hold charges at lesser potentials than a single conductor can. This implies that for capacitors of lower capacitances you need more potential to store the same amount of charge, what your TA did was reduce the capacitance of the system so now to hold the same amount of charge the potential increases. You can also see that for large plates using approximations electric field comes out to be independent of distance, so when your TA pulls the plates apart thr electric field does not change; However potential depends directly on both electric field and distance. Sos even when electric field remains constant, the increment in length between the plates increases the potential difference.

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as you know that inside a capacitor electric field remains same. If you increase the distance between the two plates electric field does not change just because electric field= surface charge density/ epsilon. so E=V/D gives increment in V as D increses so that electric field remain same.

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The explanation is simple.

Start with a definition of voltage: the work done moving 1 coulomb of charge from point $a$ to point $b$. In this case $a$ to $b$ is one plate of the capacitor to the other, since we are talking about the voltage across the capacitor.

And now a definition of the work done: it's $\text{force} \times \text{distance}$.

A capacitor has an even electric field between the plates of strength $E$ (units: force per coulomb). So the voltage is going to be $E \times \text{distance between the plates}$. Therefore increasing the distance increases the voltage.

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I see it from a vector addition perspective. Using a single test charge in the directly between the plates, the force vectors created by individual charges on the plate begin to point more "horizontal" as the plate is moved away from the test charge. So although the individual vectors decrease in length, the horizontal components of forces from charges at the edge of the plates increase maintaining a constant overall field strength. I'm sure there is a mathematical proof, but this is how I'll explain it to my HS physics students. Does this make sense?

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Opposite charges attract. Pulling them apart requires external energy supply that would be stored in charge system. As charge remains constant, per charge energy increases as well (that is potential difference).

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I think I may have a good explanation for you. I've been an electrical engineer for nearly 3 decades and I teach circuit theory as well. I need to explain how a capacitor works all the time. Let's start with a metaphor:

  • You pick up a rock and put it atop building 30 feet tall. It now has potential energy stored relative the ground, stored in the gravitational field (which we assume is constant at these elevations). That energy came from a crane that put it up on the top. We could get that energy back by dropping the rock and converting it back to kinetic energy at the bottom. Think about this - initially, the ball had ZERO kinetic energy (when sitting on the ground) but now it has kinetic energy when it comes back to the ground. Why? An external force gave it to the rock (the crane) in the form of potential energy and we retrieve it in the form of kinetic (which ultimately gets lost as heat as the rock comes to rest).

  • Now, take that rock sitting atop the 30 foot building. It's the same rock as before. And instead of pushing it off the roof, the crane picks it up and lifts it another roof another 30ft above, so now the rock is 60 feet above the ground. Now push it off the higher roof - when it hits the ground, it is moving even FASTER than before - it has DOUBLE the kinetic energy it had when it was pushed off the 30 foot building! How did that happen? How does it now have DOUBLE the potential energy? Well, by using the crane you stored DOUBLE the potential energy in the gravitation field.

  • How do we know we "stored it in the gravitational field?" Because if the gravitational field were to suddenly vanish, the rock would just sit there, even pushed off the roof, and not drop. Now where is the energy? Is it gone? Well, yes, if the gravitational field simply "went away" we wouldn't have needed it the first place.

So what does this all mean? Well, the exact same logic applies to capacitors.

1) Instead of a rock, let's say the rock is an electron. 2) Instead of a crane, let's say the crane is a battery that can move electrons. 3) Instead of a graviational field, let's say it's an electric field.

So....

We take a pair of metal plates and form a parallel plate capacitor. And we make sure the distance between the plates is REALLY REALLY THIN relative to the area of the plates. This means that any electric field between the plates will be constant - just like the gravity is constant close to the earth (it is, really, trust me!).

Next, we take the battery, which can rip electrons off of one plate and "stick" them onto the other. How? Chemistry - but that's not important right now. Let's just know that it took ENERGY of some kind and we added it to the electron, and now it is on the other plate.

What just happened? The pulled electron leaves an unmatched proton on the lower plate. So there is now an ELECTRIC FIELD established between the plates. We have STORED the energy from the battery in the electric field. How do I know this? Because if the field could magically "vanish" the electron would just float around happily on the other plate and not be pulled by the proton (again, this was if the electric field could suddenly vanish - which it can't - just like gravitational field just can't "vanish").

Since voltage is the amount of energy that gets stored per unit charge (Joules/coulomb is how Voltage is measured - it's a derived quantity!). We must be able to get that energy back. How? KINETICALLY!!!

Suppose you disconnect the battery, and the plates are 1mm separated. And let's then assume you have really really really tiny tweezers and you could yank the electron off the upper plate, put it next to the plate, and let it go. What would happen? It would go whizzing back to the proton from where it came from in the first place and essentially "crash" into it with kintetic energy.

So think about it. I take an electron off of the lower plate. It initially had no kinetic energy. A battery "sticks it on" the upper plate by using its chemical energy - giving the electron "potential energy" which gets stored in the ELECTRIC FIELD between the plates.

NEXT: Let's suppose instead of a battery, i use my hands and now separate the plates from 1mm to 2mm. That electron that we put on the top plate, is now still on the top plate, but guess what, our HANDS ADDED ENERGY to pull the plates apart a little bit, so little, we didn't even realize we did it - but we did! We added just a LITTLE BIT more potential energy to the system.

This extra energy is now stored in the electric field. How do I know? Because again, if this electric field suddenly vanished - the electron would just sit there, not getting pulled towards its matching proton.

Now, let's suppose I pluck off that lone electron and drop it next to the plate. It will now whiz back to the proton, but since the distance was double, it would have TWICE the kinetic energy it did before. It crashes onto the bottom plate, and loses this kinetic energy to HEAT.

The mechanical and electrical systems described herein are metaphorically identical.

The key learning here is that we store mechanical energy in a GRAVITATIONAL field.

WE store electrical energy in an ELECTRIC field.

The real question is not "why did the voltage go up" but rather, "why does a gravitation field or electrical field allow us to store energy within it."

And that is where the real mystery continues to lie. We still don't know. We don't know how a positive charge "pulls" on a negative charge, just like we don't know how two masses pull on each other.

All of what is written above holds true so long as you assume the gravitational field or the electric field remain CONSTANT while raising the rock or pulling the plates apart. And this is a very real behavior so long as the scales are small.

Gravity is essentially constant for hundreds of miles above the surface of the earth until the field weakens farther out in space. The principles still apply, but now you have to use calculus to figure things out.

Electric fields between two parallel plates are essentially constant when the plates are very close together. Once the distance between the place gets large, relative to the surface area of the plates, then again, the principles still apply, but you can't assume the field is constant.

Oh, and one final thought about this: We generally use a resistor as the path for the electrons to return to the lower plate. And what gets created as this occurs - HEAT is generated. Why? Because as the electron attempts to return to the bottom plate at near the speed of light, it gets slowed way down by the resistor which gets heated up in doing so, absorbing the kinetic energy along the way. That way, when the electron gets back to the bottom plate, guess what, it has no more kinetic energy, just the way it started when the battery pulled it to the top plate in the first place.

Hope this helps!

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    $\begingroup$ You go a little overboard on the ALL CAPS. Maybe tone it back a bit? $\endgroup$ – Chris Jan 20 '18 at 2:29
  • $\begingroup$ Chris, thank you - I removed the over-the-top caps. I was trying to emphasize not offer negative tone. I appreciate the feedback. I love this stuff, and I'm passionate about it. $\endgroup$ – Allan Jan 20 '18 at 3:10
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    $\begingroup$ It's not a negative tone, just a bit of an eyesore. Welcome to the site, by the way :) $\endgroup$ – Chris Jan 20 '18 at 3:27
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Capacitance increases as the voltage applied is increased because they have a direct relation with each other according to the formula $C=Q/V$. Capacitance decreases as the distance between the plates is increased because capacitance is inversely proportional to distance between the plates according to a relationship $C \propto \frac{1}{d}$.

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  • $\begingroup$ The capacitance is (approximately) constant with respect to the voltage. That is the whole point of the capacitance: It describes how much charge is stored on a capacitor in dependence of the voltage (in other words: the charge increases when the voltage increases). $\endgroup$ – Sebastian Riese Jun 1 '18 at 17:27

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