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Homework Help: Multiple dielectrics in one capacitor

  1. Sep 14, 2008 #1
    Hi guys, I'm absolutely, completely and utterly stumped with this one question I've been given..

    Here is the exact question:

    Given that you possess the following materials:
    (i) two sheets of copper of uniform thickness;
    (ii) one sheet of mica of uniform thickness 0.10 mm and dielectric constant κ= 6;
    (iii) one sheet of glass of uniform thickness 2.0 mm and dielectric constant κ = 7;
    (iv) one sheet of paraffin wax of uniform thickness 1.0 cm and dielectric constant
    κ = 2;
    determine which sheet, or combination of sheets, will produce a parallel plate capacitor of
    the largest capacitance when placed between the copper sheets. Assume that all sheets
    have the same shape.

    Relevant equations that I have are:
    but there are probably plenty more. I have equations for potential energy and such, but I'm not sure if they are necessary for the question.

    attempt at a solution

    Well as I said, I am completely stumped. I can calculate the capacitance of each dielectric on its own in the capacitor, but the combination ones are the ones I'm stuck with. This is what I've got so far {probably way off track}.

    C[mica] = 6(epsilon)(area)/0.1x10^-3
    C[glass] = 7(epsilon)(area)/2x10^-3
    C[wax] = 2(epsilon)(area)/0.01

    Can I ignore epsilon and area because they are common in all the equations?

    for the dielectrics next to each other;

    C[effective] = C[no dielectric]*k[1]*k[2]

    This is most likely wrong. I know that for one dielectric, you can simply say that C[effective] = C[no dielectric]*k
    where k = kappa

    But I simply don't know how to derive a formula for multiple dielectrics.

    Also, I do have knowledge of the theory of how dielectrics work, ie. how they generate opposing electric fields to that of the capacitor, but I am still 100% stuck with this question..

    Thanks in advance
  2. jcsd
  3. Sep 14, 2008 #2
    If you have multiple sheets of dielectric stacked on each other, you can imagine that each sheet is a distinct capacitor and these capacitors are connected in series. The total capacitance C is then

    [tex]\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}[/tex]
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