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Homework Help: Electric potential and capaciters question

  1. Oct 17, 2006 #1
    I have a homework question where the correct answer is bit confussing to me.

    A parallel-plate capacitor is charged by connecting it to a battery.
    If the battery is disconnected and the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it?

    Answer: The charge remains fixed and the electric potential increases.

    I get that the charge remains fixed, althought i think negative charged dust in the air would start sapping away at the chaged plates.

    Why does electric potential increase...

    book doesn't give straight definition of electric potential, but my intrepretation is electric potential is the potential energy in an electric field...potential to move a charge in the field.

    so as the charged plates seperate some distance, it just makes sense to me that the electric potential decreases because the postitive plate can't influence the negative plate as much. or is it that which allows the potential energy to grow...hmm
  2. jcsd
  3. Oct 17, 2006 #2
    Assuming ideal conditions for the plate, the field will be uniform and the electric field is:
    [tex] \vec E = -\hat y \frac{V_{12}}{d} [/tex]

    Depending on your coordinate system and arrangement of the plates. Lets just drop the unit vector and look at the magnitude of the electric field (E) and the potential (V12).

    [tex] E = \frac{V_{12}}{d} [/tex]

    So what happens to V as you increase d?
  4. Oct 17, 2006 #3
    By the way, the potential is:

    [tex] V_{12} = -\int_2^1 \vec E \cdot d\vec l [/tex]
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