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Electric Potential Intuition Needed

  1. Feb 21, 2014 #1
    I'm trying to understand why the voltage across parallel circuit elements is the same. I keep finding the answer "because they are both connected to the same voltage source."

    Why is it the case, in a physical sense, that voltage MUST be the same across, say, capacitors that are connected in parallel to the same voltage source? How does this physically dictate the same potential difference existing across each capacitor in parallel?

    I have other questions about voltage and circuits, but I will hold off on them and see where this example takes us.

    Thanks for any help!
  2. jcsd
  3. Feb 21, 2014 #2


    Staff: Mentor

    Hi LikesIntuition, welcome to PF!

    Parallel circuit elements are each connected between the same pair of conductors, and each conductor has a single voltage. So the voltage across any element is V2-V1. Since V2-V1=V2-V1 the voltages are the same.
  4. Feb 21, 2014 #3
    Ok thanks! So, how exactly does that voltage come about in those circuit elements? What happens at the moment when they're connected to the conductors that have a voltage?

    Edit: Also, when I put a voltage across a resistor, is a uniform field set up in that resistor? If so, why doesn't the field drop off over distance like it would in a vacuum?
    Last edited: Feb 21, 2014
  5. Feb 22, 2014 #4
    The point is not the vacuum or lack of it. The field of a point charge drops off with distance, but here we're not dealing with a single point charge, are we? We're dealing with a large collection of charges spread over the conductor
  6. Feb 22, 2014 #5


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    Hi LikesIntuition! :smile:
    Voltage is (electric) potential energy per charge.

    Difference in voltage is difference in (electric) potential energy per charge.

    The difference in potential energy depends only on where you start and finish …

    it's like gravitational potential energy, which also depends only on where you start and finish: in a playground if two slides connect the same points, the difference in potential energy between them is the same whichever slide you go down.

    It doesn't matter whether it's two capacitors or any other components: if they're connected to the same two points, they have the same potential energy difference (per charge), ie the same voltage difference. :wink:
  7. Feb 22, 2014 #6


    Staff: Mentor

    You didn't specify the voltage source, so I cannot say. Different voltage sources will have different mechanisms for establishing the voltage.

    You didn't specify the types of circuit elements connected, so I cannot say. Different circuit elements will respond differently (i.e. with different currents) to an applied voltage step.

    Yes, approximately.

    You can have uniform fields in vacuum too. Fields don't necessarily drop off over distance, it all depends on the boundary conditions.
  8. Feb 22, 2014 #7
    When I ask how voltage is set up in the circuit elements, I mean: physically, what are the electrons in the circuit doing?
    Could you also explain a bit more about boundary conditions?

    And thanks, tiny-tim! That explanation makes a lot of sense. So, as I cross a circuit element, I will go through some sort of change in electric potential, and that has to do with what voltage source the ends of the circuit element are connected to?
  9. Feb 22, 2014 #8


    Staff: Mentor

    As I already told you, you have to tell me the specific circuit element or voltage source before your question can be answered with any physical detail. I cannot read your mind. You can continue to ask the same question and get the same answer if you like, but a more productive approach would be to actually provide some details about the specific voltage sources and circuit elements that you are interested in.

    Laws of physics are differential equations, in the case of circuits the corresponding differential equations are Maxwell's equations and the Lorentz force law. When you solve a differential equation you don't get a single answer, you get a whole family of answers. For example, if you solve the differential equation for projectile motion you get a family of parabolas. In order to narrow down that family to a single answer you have to provide boundary conditions. For example, the initial velocity and position of your projectile.
  10. Feb 22, 2014 #9
    I'm sorry, I didn't understand what you meant completely the first time, and your answer wasn't the type of answer I had expected, so I thought maybe trying to reword my question might give you more insight into what I was asking. As you say, we can't read each other's mind!

    So, how about a typical AA battery setting up a field in a wire that has resistance? Would this be a case of a uniform field being set up? If so, what is happening to set it up? My professor talks a good deal about electrons trying to arrange themselves to eliminate the field imposed on them, like in a conductor. Is this what's happening, but the resistance doesn't allow them to eliminate the field?

    Thanks for your patience!
  11. Feb 22, 2014 #10


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    If you want a uniform field, you need a pair of large plates, parallel to each other with spacing x. The field between them, away from the edges will be a pretty good V/x volts per metre. If you want to consider the fields inside the wire, that is an entirely different matter.
  12. Feb 22, 2014 #11


    Staff: Mentor

    So a battery establishes a voltage by making a microscopic push or pull on an individual electron. On the side where oxidation occurs the molecules literally push an electron onto the electrode, and on the side where reduction occurs the molecules literally pull an electron off of the electrode. The pushing and pulling do work on the electrons which is physically how the voltage is established (remember voltage is work or energy per unit charge).

    A resistor responds to a voltage by allowing a current through it which is proportional to the applied voltage. Technically, any conductive material responds that way, and the difference between a resistor and a conductor is just the amount of current for a given voltage.

    Yes, approximately.

    The field is the negative gradient of the voltage. So where you have a voltage which changes over distance you have a field. So saying that you have set up a spatially changing voltage is the same thing as saying that you have set up a field.

    That "eliminate the field thing" only applies for electrostatic situations. When there is a current you are not in an electrostatic situation any more and you can have fields which are not eliminated in a conductor.
  13. Feb 22, 2014 #12
    That makes a lot of sense! I've never thought of it that way. So if voltage changes spatially, it means that the work done on a charge would change over that region of space, and if work is being done we have a force. Am I looking at this accurately?
  14. Feb 23, 2014 #13


    Staff: Mentor

    Yes, that is correct. I am glad I could help!
  15. Feb 23, 2014 #14
    Here is an interesting explanation of the electric field inside a wire.
    http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf [Broken]
    And about the uniformity of the electric field - take a look at the generalized form of Ohms's law.
    J = σE
    That tells you that as long as the current density and the conductivity/resisitivity are the same, the field strength will be the same.
    Last edited by a moderator: May 6, 2017
  16. Feb 23, 2014 #15


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    sorry, i don't like that description at all

    it depends not only on the source but also on everything else in the circuit

    any particular circuit element only gets the voltage difference that is left over after the voltage difference across everything else in series has been subtracted

    you should think of a voltage (a potential) as existing at every point, and then the difference in voltage between two points tells you the work done per charge moving between them along any path

    (of course, often there's only one path)

    yes, but i'm not sure that that's helpful in most cases

    yes you do need the spatial change of voltage to find the voltage across eg the gap in a capacitor (you are using work done = force times distance)

    but no in most parts of most circuits the spatial change is unused and unhelpful (you are using work done per time = volts times charge per time = volts times current, with no mention of force)who cares what the electric field (force per charge) is? … you can find everything using voltage current etc!

    (this is different from the gravitational comparison …

    in gravity, W = Fd only,which involves the potential

    but in electricity, W = Fd and W = Vq, and the potential is only involved in the latter, which is not explicitly spatial)​
    Last edited: Feb 23, 2014
  17. Feb 23, 2014 #16
    Thanks! The diagrams in the pdf were exactly the type of thing I was looking for!
    Last edited by a moderator: May 6, 2017
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