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Potential/voltage, and charging up one side of a capacitor

  1. May 29, 2005 #1
    I'm an electrical engineer, so I'm used to circuits and components and know them pretty well. What I don't know is "broken circuits"; the physics of a charged object hanging out in space by itself, discharging into other objects, etc. So I'd like to ask a few things that have been confusing me lately:

    1. This site talks about the voltage on a single charged sphere floating in space. But what is that voltage measured relative to? Isn't voltage by definition a "potential difference"? A difference between two points? (And what does "potential" mean, anyway?) So wouldn't this sphere's voltage have to be measured relative to another point? Is that "other point" infinity? If so, what does that mean exactly? We explored how to calculate stuff like this in electromagnetics class, but I guess I never really learned some of the concepts behind our calculations...

    2. If I take a metal plate and hang it in the air by insulators, scuff my feet along the ground, and touch it, I can charge it. Now what if I put another metal plate very close to it with a dielectric between them? (A floating capacitor.) If I do the same thing, can I "fit" more charge on the plate that I touched, because the oppositely charged particles from the other plate crowd up next to it and "condense" more charge into a smaller area/volume? (Hence the word "condenser". (Does anyone know who invented the word "capacitor" and why it replaced condenser in English but not in any other language?))

    3. If I charge up one side of a floating capacitor, as described above, can I connect the other plate of that capacitor to a neutral conductor and "extract" charge from it? (Because there should be a charge imbalance if I am thinking of this correctly.)

    4. If I do the same thing again, but this time the other side of the cap is connected to a large conductor (like the Earth), can I fit even more charge than in the case of #2?

    This is all related to an experiment I've been looking at, which I will start another thread for. :-)
  2. jcsd
  3. May 29, 2005 #2
    Well for the first, the voltage difference is measured between the center of the sphere and the edge which is shown in the integral limits a and b. The radii a and b are labeled in the diagram.
  4. May 29, 2005 #3
    See "isolated sphere capacitor" below.
  5. May 29, 2005 #4
    They give you both points, they are measuring the potential at the radius in comparison to infinitely far away (0).
  6. May 29, 2005 #5
    Ok, but that's a pretty abstract notion. Don't really understand it.
  7. May 29, 2005 #6
    Do you understand potential energy?
  8. May 29, 2005 #7
    sure, i do. is "electric potential" a form of energy?
  9. May 29, 2005 #8
    The energy that a charge gains falling through a potential gap is proportional to the electric potential. So yes, voltage is essentially energy.

    Imagine two positively charged sphere that are near eachother. If left to themselves they will repel eachother untill the distance between them is unbounded. Even as we let the distance become infinite, calculations show that the charges will not gain an infinite amount of energy (they will gain some certain energy, proportional to the electric potential between where the charge started and infinity).
  10. May 29, 2005 #9
    Here is a better description of my main question. See attached image.

    In case A, a metal sphere is charged up and brought in contact with another, neutral, identical sphere. The excess charge redistributes itself so that each sphere now has 1/2 the excess charge, right?

    In B, the neutral sphere is now 1/2 the mass/volume/whichever is important here. This time only 1/3 of the charge is distributed to the smaller sphere, because there is "less room" for it to fit.

    In C, the exact same thing as B, except the smaller sphere has been hammered out into a thin plate. The physical shape of the second object has no effect on how much charge will be transferred, does it? Edit: I think this is wrong. The charges are trying to get as far apart as possible, so a very thin very wide sheet would collect more of the charges than a sphere of the same mass/volume.

    In D, the same exact plate as B, except now a thin insulating plate has been stuck to the metal plate, and another conductive plate placed very close. How much charge gets transferred to the plate now??

    Attached Files:

    Last edited: May 29, 2005
  11. May 29, 2005 #10
    well it's defined as joules/coulomb, so it's more like "energy density".

    they will gain energy?? if you let a ball drop to the earth it is losing energy. converting potential into kinetic. if you let two unequally charged spheres fall into each other, they are also losing energy, potential into kinetic. i would think the same thing would be the case for repulsion...
  12. May 30, 2005 #11


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    The potential at the surface of the sphere corresponds to the energy you would have to impart to an unit charge of the same sign to move it from infinity to the sphere.

    Theorectially you can put any charge you want in the plate. The more charge you add, the greater the electrical potential. If the plate is hanging in space, the potential will be relative to infinity. If you have a parallel plate capacitor, the potential difference will be relative to the other plate.
    Of course, the voltage between the two plates will be limited by the effectiveness of your dielectric.
    About the name capacitor, it is the same in Portuguese (once it was condensador).

    The conductor would be part of the capacitor. The total charge would be the same.

    The Earth would be the second plate of the capacitor.
    Last edited by a moderator: May 30, 2005
  13. May 30, 2005 #12


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    It's nothing like a density. Just because something is a ratio, doesn't mean it's like a density. It's potential; a very fundamental concept. Your are also mixing Units and Concepts. Using your statement, one could equally (and erroneously) well say that Energy is defined as a Volt-coulomb.
  14. May 30, 2005 #13


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    You are correct to think you were wrong. (Make sense?) The capacitance of an object depends upon its shape.
    You can calculate the capacitance of the new plate. It does depend upon the dielectric between the plates, so the amount of transferred charge is different. But there is also another thing. The added plate is isolated, so has no way to develop a net charge. Nevertheless, it does have a non-neutral charge distribution. Think of the added plate as 2 surfaces: the surface facing the first plate and the surface facing away. If there is charge Q on the old plate, then there is charge -Q on the surface of the new plate facing this old plate, and charge +Q on the surface of the new plate facing away from the old plate.
  15. May 30, 2005 #14
    Ok. That makes sense.

    Well, you could measure the potential of each plate relative to infinity, too. But here we are measuring "potential" relative to something, which means a "potential difference", which is the definition of a voltage. Is there such a thing as a "lone potential" measurement, or is it implicitly measured relative to infinity?

    Also, isn't there a "voltage function" in the space between the two plates? A linearly changing voltage? What relevance does this have to real life measurements? (Obviously you cannot measure a voltage in air between two plates of a cap with a voltmeter.)

    Ah. It seems that most languages use some variant of "condenser" (French condensateur or the German kondensator). Portuguese now uses "capacitor"?

    Let me explain a little better. I have a neutral (with respect to other objects and with respect to itself; same amount of protons and electrons on each plate) capacitor with two plates, A and B. If I add charge to plate A (without removing any from plate B; a regular static electrical charge), wouldn't the oppositely charged particles on plate B be attracted to the dielectric, while the similarly charged particles would move to the end of the B lead? Then if I connected plate B to a neutral object, some of the non-neutral charge distribution would try to reach equilibrium and move onto that other object, leaving a net opposite charge on plate B when the large object was removed.

    The Earth would be connected to the second plate of the capacitor. I am still using a regular capacitor, but with one lead wired to the Earth.

    It is defined as the amount of potential energy per unit charge. A 1000 uF cap charged to a potential difference of 10 V has a lot more energy than a 100 pF cap charged to 10 V. Voltage is not energy.

    An electronvolt is a unit of energy, no?

    Yeah. My mistake.

    THAT is what I was wondering! The amount is larger, correct?

    Is the Q charge on the "old plate" uniformly distributed? Or is it all "gathered around" the dielectric, with a neutral charge at the end of the lead?
  16. May 30, 2005 #15


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    You could say that each plate has a potential relative to infinity. The voltage between the plates is the potential difference.

    Yes, you would have equipotential surfaces between the plates. Each equipotential surface has a different potential. In the case of a parallel plate capacitor, the equipotential surfaces are planes. And yes, the voltage will vary linearly.
    Of course, this is only true if the plates are of infinite dimensions. For limited plates the shape of the equipotential surfaces will depart from planes at the edges.


    You are correct.
  17. May 30, 2005 #16
    So a "potential" can be thought of as a "voltage relative to infinity"?

    But these can't be measured with a voltmeter. What can they be measured with?

    I am correct that connecting one lead of a capacitor to the earth allows a creater amount of charge to be built up on the opposite plate by a given potential than if the cap were not connected to earth?
  18. May 30, 2005 #17


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    If you consider that infinity corresponds to zero potential, yes.

    You can't!

    No, you are wrong! You can put any amount of charge in a capacitor, regardless if there is a single plate or two plates and if the second plate is or not connected to anything.
    Remember that Q = CV. So, if you put more charge in a capacitor, it's voltage will increase.
    The energy needed to add an amount ΔQ of charge to the capacitor is
    ΔQxV. So, the greater the voltage, the harder it will be to increase the charge. That is what limits the amount of charge you can put in a capacitor (I am supposing that the dielectric will not be ionized by the growing voltage).
  19. May 30, 2005 #18
    What other things might I consider?

    What does an electrometer measure, then?

    I said "by a given potential".

    (Plus there is, technically, a limit to the amount of charge you can put on a metal object, but it isn't at all feasible.)

    But will connecting plate B of a capacitor to ground increase the capacitance of plate A?
  20. May 31, 2005 #19


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    You can attribute the zero reference anywhere you want.

    By a given potential V and a given capacitance C, the charge will be Q = CV.

    Prior to saturate the metal object with charges, the voltage will build up to such great values that anything near the object will be ionized by the electric field. Besides, as I said before, as the potential grows you need more energy to take more charges to the object.

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