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Electro-conceptual questions

  1. Feb 22, 2005 #1
    Physics Question:

    All right, well I have read over the textbook-type material behind the questions I am about to ask. So, I really don’t want equations thrown at me. My problem is mostly with conceptual-level information:

    A) Concerning Capacitance- If you had two capacitor plates that were not attached to a source of voltage, doubling the distance between them would theoretically double the value of capacitance (due to the definition of potential difference) at that immediate moment. That I am just fine with understanding. However, when attached to a source of voltage, the inverse relationship is present: greater distance means greater capacitance. This would imply that an electric field of constant magnitude would develop regardless of distance between the plates BECAUSE of the “manipulation” by the source of voltage in establishing that field. Charged plates by themselves, when being spread apart, obviously lose electric field strength as distance between them is increased. Can someone clear this up for me and set it in perspective?

    A Second Part to Part A- In my physics textbook, the value of electric field between two plates is expressed as a constant multiplied by surface charge density. I understand Gauss’s law, but area simply does not change correspondingly as you increase the distance between the plates. So what is up with this?

    B) A Fairly Simple Yet Odd Circuitry Question- With electromotive force aside, let’s look at the actual electrical forces, rather than the potential gained due to those forces. When I picture a simple squared off wire with a single source of voltage in the middle of a side of that wire, it would seem that certain points along that wire experience different forces in terms of direction of induced acceleration. Some flow of current seems to be against the electric force basically aimed towards the source of voltage at certain points along the wire. So how can charge flow remain so straightforwardly constant? When I think of a single electron in the wire, I imagine it getting stuck at certain points along the wire, not feeling favorable forces for the movement in which it must go in order to end up gaining energy in the end of things. Any clarity available?

    Thanks for your help!
  2. jcsd
  3. Feb 22, 2005 #2

    Doc Al

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    Not true. When the plates are separated, capacitance decreases. Applying a voltage will charge the capacitor, but won't change the capacitance, which is a function of geometry.

    Not sure what your question is. As long as you don't separate the plates too much, the field depends only on the surface charge density. If the charge remains constant, so will the field. (But the voltage will change!) If the voltage remains constant, the charge and the field will change.

    What makes you say that the force (due to the electric field) propelling the electrons is different at different points? Or that the current must travel in different directions?

    Welcome to PF, by the way!
  4. Feb 22, 2005 #3
    If you would like a conceptual approach to E and M, try Intro to Electrodynamics by David Grifiths.

    He gives the following explaination to your question b: Suppose, at the square kink, some charges were getting stuck. Picture the kink with positive charge flowing in to it that is not also flowing out; then there is a lump of positive charge built up in the kink. But notice this charge opposes the current flowing in, and increases the current flowing out. So the system is self correcting, over times on the order of 10^-12.

    For your first question, I think you are taking the formulae too literally. Yes, for a constant electric field, doubling the distance would double the voltage for a given amount of charge, but this never happens! It is the voltage in the battery which is set, and the distance between the capacitor plates and therefore the surface charge density i.e. electric field which is variable.

    What I am saying is, if you think of the charge on the plate being constant as you vary distance, then you are assuming an increase in voltage (but you are asking "why doesn't the voltage increase?").
  5. Feb 22, 2005 #4
    Well I understand the argument that the area of a capacitor plate does not influence electric field significantly. However, when it comes to distance, I just don't see where the argument is coming from. Two plates separated an infinite amount would have a potential difference of 0, considering that V = Q multipled by a constant divided by distance. Also, under a similiar argument, electric field should become zero. Really, I don't see how capacitors would be very different from just two point charges when it comes to distance considerations. The strength of electric field is dependent upon the distance to the mid-point between the plates. If charge remains the same, electric field strength should be decreased by a factor of 4 if the distance is doubled.
    Last edited: Feb 22, 2005
  6. Feb 22, 2005 #5
    Oh, that is a typical question. The reason E does not depend on distance is because the sheet of charge is "infinite", that is, we ignore end effects.

    The equation E =sigma/epsilon only applies up close to a conducting surface.

    Also remember, the only way you can increase distance and keep the charge constant is if you supply more voltage.

    Capacitance is the amount of charge for a given potential:

    C =Q/V

    You noticed C is zero if d is infinite. That just means that if the two plates are very far appart, wer have to supply lots of voltage for a tiny bit of charge. It makes sense.
    Last edited: Feb 22, 2005
  7. Feb 22, 2005 #6
    Hrmm, I see. I believe I am confused of the definition of electric potential difference in this situation. Initially, I just think of it as the amount of charge on a plate divided by distance between the plates. Hmm...

    So potential difference doubles when distance is doubled..what exactly does this mean in terms of energy gain if charge transfers between the plates?
  8. Feb 23, 2005 #7
    The potential difference only changes if the plates are not attached to a battery.. Q = CV, if they are not attached to a battery, they cannot transfer charge, and since the capacitance drops on increasing the distance, the voltage difference has to increase.. it also makes sense because the voltage difference is representative of the potential energy of the system.. to pull apart two oppositely charged plates, you have to do work on them and increase the energy of the system.

    If they are connected to a battery, and charge can transfer between the plates, then the battery will maintain the plates at a constant potential difference, that is Q/C will be constant. So as capacitance decreases as you increase the distance, the charge on each plate will also decrease.
  9. Feb 24, 2005 #8
    I'm still not quite seeing all there is to see here. Let's isolate the condition in which the plates are not connected to a battery. If C = Q/V is true, than in order for capacitance to follow these laws we discuss, the potential difference must be doubled when distance is doubled. This would imply that electric field remains constant between the plates no matter what the distance between them is. How can this be? Plates work much like point charges when it comes to distance considerations, and the electric field between two point charges varies drammatically as the distance between them changes.
    Last edited: Feb 24, 2005
  10. Feb 24, 2005 #9
    Go back and read my last post, plates are unlike charges, and the electric field does not depend on distance (because the plate has infinite area).

    Yes, if we are not connected to a battery and you double the distance, you double the voltage. How does the voltage double? Well, you did it by doing work to move the two plates apart.
  11. Feb 24, 2005 #10


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    We get away with calling the plates infinite because we assume the dimensions of the plate are much larger than the seperation d. When you talk about treating the plates as point charges, it seems like you are asking what happens when d gets very large. At this point, you can no longer treat the plates as infinite, and the formula for capacitance with the 1/d factor isn't valid.
  12. Feb 25, 2005 #11
    Ah, I see. Well perhaps I'm just confused over the meaning of the electric field between two capacitor plates. Even if the plates are considered infinite in area...well what is wrong with this understanding of the constant electric field:

    Let's say you have two plates (XXXXX) spread apart. Then the electric field magnitude would be about equal to the strength at the midpoint between the plates. I.E.








    If the electric field strength at D is equal to the constant value assigned to the magnitude of the constant electric field, then doubling distance should just about cut the electric field down to a quarter of its original strength.

    Seriously, this is the most confusing thing I've ever dealt with in physics. The textbooks I've used go into little detail about the concepts between capacitance. Thanks for the help so far.
  13. Feb 25, 2005 #12


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    Roughly speaking, as you get farther from a plane of charge, the test charges can "see" more of it. This field of view cone intersects the plane in a circle, so the charge they can see goes up as d2, which balances the decrease in strength from couloumbs law of 1/d2. That is why the field of an infinite plane of charge doesn't drop off with distance. Similarly, for a line of charge, the charge that a test charge sees goes up as d, which is why the field drops off as 1/d. This is all very qualitative, but it can be put on firm mathematical ground. As someone mentioned before, Griffith's book explains this all nicely.

    The field between the plates doesn't change with d, so the potential difference between the plates for a given Q goes up as d, since you integrate over this constant field from one plate to the other. That is why C=Q/V is proportional to 1/d.
  14. Feb 27, 2005 #13
    Oh! Okay then, that is what I needed to hear. Up until now, I had approached this problem much like this visually:

    [+++++] <- Capacitor plate

    - <- say a negative charge in the center between the plates

    [------] <- Capacitor plate

    In this simple view, the charge farthest away from the midpoint generates a strong upward electro-force. However, in realistic situations, there are MANY point charges spread along the capacitor surface for a much longer distance. As a result, many electro-forces would be felt by the midpoint charge at such an angle that most of the strength of these forces is canceled out by an electro-force pointed towards the opposite horizontal direction (at the horizontally opposite location along the capacitor surface). The very small vertical component of these forces becomes much more pronounced when the plates are spread apart. But then again, this would only work for an infinitely large area. Though, realistic areas would show this property fairly well for reasonable distances. Interesting.
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