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Why is capacitance inversely proportional to the distance between the plates.

  1. Nov 1, 2012 #1
    I need help with something.I have problem to understand how the capacitance of a capacitor can be increased if we move the plates of the capacitor more close.
    Here's my reasoning:
    The capacitor is just a extended terminal of a battery,so the battery voltage must equal the capacitor voltage.Because the area of the plates of the capacitor is larger than,that of the battery, the plates of the capacitor can accumulate charge until the battery and the capacitor have the same voltage.This is logical for me and I can make a mental picture from it.

    I cannot understand how the capacitance can be increase, and voltage not if we move the plates more close.I understand that if we move the plates more close,the positive plate will affect the negative and more charge will accumulate on the negative plate,but if more charge accumulate on the same square are of the plates,that mean that we will have bigger voltage on the capacitor.

    For example if we have 1 farad capacitor and the voltage is one volt,we will have one coulomb of charge on the plates.But if we bring the plates more close, then the voltage would be the same but will will have more charge on the same area plates.If we have more charge on a plate with a same area then, the repulsive forces between the electrons should make the voltage bigger.

    Thank to anyone who can help me with this.I'm struggling to understand this.
     
    Last edited: Nov 1, 2012
  2. jcsd
  3. Nov 1, 2012 #2

    tiny-tim

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    welcome to pf!

    hi jonah! welcome to pf! :smile:
    no, the repulsive forces between point charges will be bigger … the field lines will spread out, covering an area 4πr2, so the force is proportional to 1/r2 (inverse square law)

    but between two plates the field lines are parallel, there is no spreading, and the force is the same for any value of r (provided it's small compared with the width of the plate)

    since the force is the same (per fixed charge), the charge must be increased when the distance between the capacitor plates is decreased, to maintain a "balancing" voltage equal to the applied voltage :wink:
     
  4. Nov 3, 2012 #3
    Hi tiny-tim.Thanks for your answer but I still don't understand the problem.

    I am considering only the charges of one plate.For example, if we take the negative plate,the negative plate will accumulate charge until the voltage across the capacitor is equal to that of the battery.Now if we bring the plates close together that mean that the negative plate will accumulate more charge on a same square area.{same goes for positive plate,but in a reverse manner}.If we have more electrons on a same square area{or a bigger shortage of electrons on the positive plate} that mean, that the repulsive force between the point charges would be bigger.More electrons,more close together on a same area makes a bigger potential.
     
  5. Nov 3, 2012 #4
    Re: welcome to pf!

    If we increase the battery voltage,we will increase the capacitance too.If this is correct then,increasing the capacitance by moving the plates more close should increase the voltage.
     
  6. Nov 3, 2012 #5

    tiny-tim

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    hi jonah! :smile:
    no

    if the applied voltage stays the same, and we bring the plates closer, then yes there are more electrons per area :smile:

    but that does not make a bigger potential, it makes a bigger force, proportional to 1/distance

    electric potential = potential energy per test charge = work done per test charge = force per test charge times distance

    so, if we change the distance, the force per test charge changes, and is proportional to 1/distance, so the electric potential (= applied voltage) stays the same :wink:
     
  7. Nov 6, 2012 #6
    Hi tiny-tim.Thank you very much for your help.I think I understand now what is going on inside a capacitor.If you want, please tell me if my reasoning is now correct.

    I still think that if we have more charge on a plate A the voltage would be bigger, but in this case because of the coulomb forces from the other plate B, that act on the plate A, we will have bigger charge but also no change in voltage.I think that because the other plate B have the opposite charge it brings more electrons on the plate A, and at the same time is lowering the repulsion force between them.

    If I am correct now,this must be true.If we disconnect the capacitor from the circuit and then we lower the distance between the plates then the voltage across the capacitor would be smaller, and if we increase the distance between the plates the voltage across the capacitor would be bigger.Tell me if this is true.

    Thank you again
     
  8. Nov 6, 2012 #7

    tiny-tim

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    hi jonah! :smile:
    yes, the charge on the capacitor plates is then the same …

    so the electric field is the same (E = Q/εA), so the voltage is proportional to the distance (V = dQ/εA) :smile:
    you've started by considering the voltage due to a single plate

    this makes no sense … voltage is relative, you can only talk about the voltage between two points (or plates)

    (and the repulsion force between electrons is affected by nothing but the distance, and the permittivity of the medium, between them)
     
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