Asymmetrically charged capacitor

[slow]

Hi. I put the simplest case. A capacitor formed by two flat metal plates faced in the vacuum, with charges of the same absolute value in both plates. In this case, the electric field of the capacitor, in a static situation, has an energy expressed in the following way.
$$E=\dfrac{q^2}{2 \ C}$$
Now my doubt. One of the plates is touched by something that changes the value of the charge on that plate. The charge of the other plate has not changed. Now the capacitor is charged asymmetrically. How is the energy of the electric field expressed in this case?

 

[kuruman]

The quickest way to do this is to find the new electric field between the plates and then use the energy density to find the total energy stored.

 

[Vanadium 50]

One of the plates is touched by something that changes the value of the charge on that plate. The charge of the other plate has not changed.

That doesn't happen. If it did, your capacitor would have a huge charge on it and blow itself apart.

 

[slow]

Thanks kuruman. To do it as you suggest, some doubts stop me. The plates are geometrically identical. You can wrap each one with a Gaussian surface. The electric field flow has a value on one plate and a different value on the other plate. If there were no vacuum between the plates, there would be several ways to have different flows. a) Curvature of the electric field lines. b) A permittivity gradient between the plates. c) Have scattered charges between the plates, of the same sign as the charge of lower value of the capacitor. Having empty, the permittivity has in all the points the value of the permittivity of the vacuum. Form b) is not possible. And in the vacuum there are no stable charged particles, so the form c) is not possible. Only the curvature of lines remains. And I do not know how to analyze that.

 

[slow]

Hello Vanadium 50. The capacitor explodes, I can understand that. I am simply interested in finding out what can be done in an ideal case, without thinking about the technical limitation of a common capacitor. Best regards.

 

[kuruman]

That doesn't happen. If it did, your capacitor would have a huge charge on it and blow itself apart.

Assuming that the plates are not connected to a battery, if one brings an uncharged conductor in contact with one of the plates, charge will flow from that plate to the conductor. Then removing the conductor will leave one plate with different absolute charge value than the other.

Piedo wrap each one with a Gaussian surface

I don't know what this means. In fact I do not understand what you are trying to say in post #4, parts a-c.

 

[slow]

Think of a delicate way of changing the charge value on a plate, for example, throwing an electron beam at the positive plate. Let's forget for now options a), b), c).

 

[Nik_2213]

http://demoweb.physics.ucla.edu/content/experiment-3-electrostatics

similar to 'Charging by induction' ??

 

[kuruman]

Regardless of how it's done, if one plate has surface charge density σ₁ and the other σ₂ (σ₁≠σ₂), the electric field between the plates in the "non-fringing field" approximation will be constant and given by $$E=\frac{\sigma_1}{2\epsilon_0}-\frac{\sigma_2}{2\epsilon_0}.$$ The energy stored in the capacitor will be $$U=\frac{1}{2}\epsilon_0 E^2Ad=\frac{1}{2\epsilon_0} \left( \frac{\sigma_1-\sigma_2}{2} \right)^2Ad$$
With $\sigma=Q/ A$, the equation becomes
$$U=\frac{1}{2\epsilon_0} \left( \frac{Q_1-Q_2}{2A} \right)^2Ad=\frac{1}{2} \left( \frac{Q_1-Q_2}{2} \right) ^2\frac{d}{\epsilon_0A}=\frac{1}{2C} \left( \frac{Q_1-Q_2}{2} \right) ^2.$$
Note that $Q$ is a signed quantity. For example, when the plates carry charges of equal magnitudes but opposite sign, we can set $Q_1=Q$ and $Q_2=-Q$ to get the well known expression $U=Q^2/(2C).$

 

[slow]

Thank you very much Kuruman. If possible, I still need help with some doubts, always accepting that the approach is approximate.The plates are geometrically identical, flat and mutually parallel. There is a vacuum between the plates. Then the permittivity is $\varepsilon_o$ in all points of the system. If the field is constant, uniform permittivity and geometrically equal plates, then Gauss's theorem gives the same absolute charge value in both plates. Instead of an asymmetrically charged capacitor we would have one symmetrically charged. Am I reasoning right or wrong?

 

[girts]

I wonder is it really the case when one can take a capacitor charge it up and then once it's charged disconnect it and using a conductor which is not attached to the opposite plate of the capacitor draw off charges?

Vanadium said earlier in a post hat this doesn't work that way , and I have doubts too can someone please clarify?
charges on a capacitor are attracted to the opposite charges in the opposite plate so I imagine that the only way to "bleed" them off is by connecting the two plates in a circuit, can I please ask for some elaboration?thank you.

 

[anorlunda]

High voltage charges can leak charge into the air by emitting ions. In power transmission we call it corona. But that is only at very high voltages.
In a sense that just fools us. We are really making a closed circuit, but partially though invisible conduction paths in air and ground.

Similarly, you could direct a beam of charged particles at the plate. Hit it with only a single charged particle and the charge changes. But it is better to thin of the beam as an invisible wire.

There are limited ways to build and manipulate electrostatic charges. If we talk about devices like capacitors in circuits, those methods don't apply.

 

[girts]

But as you say the air is the invisible wire, I guess it just means that when there is high humidity, high voltage power lines tend to "short out" a little along the way between phases and neutral and also I assume ground?

But in a sense taking a say 200v capacitor and charge it with some 150v DC and then disconnecting it and then just poking a neutral conductor at one of it's disconnected leads won't cause any significant amount of charge to bleed off.If I may ask just one more thing, but what would happen if this capacitor gets charged and then while being disconnected , its positive terminal becomes say grounded to Earth via a lightning rod or something similar, or a DC ground from a circuit, but its other plate "negative" one is still not connected to anything, then the capacitor should bleed I guess, I assume I'm actually asking whether the electrons that would go to neutralize the positive plate have to be the ones coming from the other plate or from somewhere else as in theory they are all the same.
I think this is confusing in the case of a capacitor because if the positive plate is neutralized with other electrons, then what happens to the negative charge in the opposite plate, I assume this is the reason one can't fully discharge a capacitor by other means than only connecting its two plates together.

 

[anorlunda]

@kuruman gave you an excellent answer. You are ignoring the most important part of his answer.

Regardless of how it's done

 

[kuruman]

... the electrons that would go to neutralize the positive plate have to be the ones coming from the other plate or from somewhere else as in theory they are all the same.

They don't have to be. Say you have two plates close to each other, one with charge +Q and one with -Q and you ground the negative plate that has excess electrons. Grounding means that you are connecting the plate to the Earth which is an infinite reservoir of electrons. This means that the Earth can accept or provide as many electrons as you wish without changing its electric potential that is always zero. This is similar to the the idea that the ocean is an an infinite reservoir of water drops; you can add or remove as many drops as you wish without changing its level. So now if you connect the negative plate to the Earth, some but not all excess electrons will move into the Earth because the positive charges (deficit of electrons) on the positive plate hold in place some electrons on the negative plate. If you wish, the electric field lines starting at the positive plate have to end at negative charges. Some of these negative charges are on the Earth, some are on the negative plate. Only if you remove the positive plate out to infinity with the negative plate still grounded will the excess electrons move into the Earth.

If, instead, you connect the two plates with a conducting wire, there is initially an electric field inside the wire. Electrons will experience a force qE so they will move (and keep on moving) until the electric field in the plates + conducting wire becomes zero. The total charge on the plates plus conducting wire will be $Q_{Total}=Q_1+Q_2$.

 

[slow]

Hello, I will try to take up the issue that motivated this thread. I have a doubt. In post #9 we have the following expression for energy.
$$U=\frac{1}{2C} \left( \frac{Q_1-Q_2}{2} \right) ^2$$
As far as I thought I learned, the energy of a capacitor corresponds to the electrostatic interaction between the charges housed in the plates. If a plate is neutral, this type of interaction between the plates is not established. That means, for example, that with $Q_1=0$ the equation should give $U=0$ . Am I reasoning right or wrong?

 

[anorlunda]

[QUOTE="slow, post: 5928352, member: 639237"
As far as I thought I learned, the energy of a capacitor corresponds to the electrostatic interaction between the charges housed in the plates. If a plate is neutral, this type of interaction between the plates is not established. That means, for example, that with $Q_1=0$ the equation should give $U=0$ . Am I reasoning right or wrong?[/QUOTE]

That is a misconception, and it may be confusing you. Consider for example the capacitance between a power transmission line and the ground. Earth ground is considered neutral, but the capacitance is very genuine.

 

[kuruman]

Am I reasoning right or wrong?

Your reasoning is incorrect. According to the equation in post #9, you must have $Q_1=Q_2$ for $U$ to be zero.

 

[anorlunda]

Your reasoning is incorrect. According to the equation in post #9, you must have $Q_1=Q_2$ for $U$ to be zero.

I didn't say U=0. I said that the capacitance of the line to ground is nonzero.