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Homework Help: Physics behind capacitance

  1. Mar 26, 2008 #1
    Hi, I'm doing a piece of (A-level) coursework into the factors that affect the capacitance of a parallel plate capacitor. After searching the internet I've come up with the following explanation for why plate separation affects capacitance:

    "Capacitance is inversely proportional to plate separation because the magnitude of the electric field between the plates is inversely proportional to there distance apart as in the equation (E = electric field strength, V = potential difference,
    d = distance between plates):

    E = V/d

    The amount of charge that a capacitor can store on its plates depends on the electric field between its plates. When a capacitor is charged the negative charges on the negative plate repel each other due to the electric field created by the negative charge in the plate. This field is cancelled out by the electric field created by the positive plate, so the negative charge remains on the plate (also the negative plate has the same effect on the positive plate). When the negative charge on the plate is increased, the electric field strength in the plate is is increased, the negative charges repel each other more so it is more difficult to add more negative charge. To increase the amount of charge the negative plate can hold, the electric field that cancels out the field due to the negative charges must be increased. Therefore the electric field strength must be increased (as electric field strength is proportional to force produced by an electric field.) Electric field strength is increased in this experiment by reducing the separation between the plates (keeping potential difference constant) as:

    E proportional 1/d

    Electric field strength relates to the charge on capacitor plates by the equations:

    E = σ/e0 (σ = charge density)

    Charge density is the amount of charge per area of the plate:

    σ = Q/A (Q = charge on the plate, A = area of plate)


    Q proportional σ prop. E prop. 1/d

    So Q prop. 1/d, charge stored on the plates is proportional to 1/d. if there is a constant potential difference and charge on the plates is increasing, then capacitance increases as:

    C = Q /V


    C prop. 1/d"

    However, im now thinking this explanation to be complete rubbish, especially in the light of the fact that placing a material with a higher dielectric constant in the capacitor has the effect of reducing the electric field between the plates but increases capacitance. Can anyone explain to me what the real reason that capacitance is proportional to 1/ separation is and explain the equations behind it? (Also need help understanding why capacitance is proportional to plate area and dielectric constant)

  2. jcsd
  3. Mar 26, 2008 #2

    Chi Meson

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    The bold part is an incorrect interpretation of the formulae.

    Notice how this statement makes no mention of the distance.

    You have some of the reasoning right, however sloppy in your presentation. As distance decreases, the electric FORCE that pulls the negative charges to the positive plate increases. This allows a tighter "packing" of the electrons on the negative plate allowing a higher charge density.

    The reason why dialectrics increase capacitance is that they allow the plates to get closer together. If a plate was too close, electrons could jump the empty gap across them. Without a dialectric between the plates, a voltage of 6V could cause electrons to jump a gap of half a mm or so. Dialectrics can be micrometers thin, yet prevent a current from several hundred volts.
  4. Mar 27, 2008 #3
    Dielectrics can also increase the capacitance for a fixed distance of the plates. They can get polarized, the atoms in it will become dipoles with the positive charge displaced towards the negative plate and the negative charge towards the positive plate. This will reduce the field strength in the capacitor for a given charge, Or allow more charge on the plates for a given potential difference
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