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Electrical Field of a Parallel Plate Capacitor

  1. Aug 21, 2003 #1
    Just going through some review questions and wanted to clear up some questions I had. I figure you guys like showing off how smart you are.

    First Question:
    Consider the parallel plate capacitor classroom demonstration.
    A. Write a mathematical expression for Gauss' Law. Describe the meaning of Gauss' Law in words.
    I've already got this part.

    B. Using Gauss' Law: draw a picture of one plate, draw a suitable Gaussian surface, and calculate the electric field due to the charge density, [sig]C/m^2, on that plate.
    Pretty sure I've got this part but I just wanted to check. It specifies one plate, and I wound up with E = [sig]/(2e), where e is the permittivity of free space.

    C.What is the total electric field: external to the plates and inside the plates?
    Well, there is no electric field outside of the plates, correct? And inside the plates I would assume would be 2 * [sig]/(2e), but that's sort of an educated guess...

    D. Estimate the capacitance for the plates, based on your observations of the plate geometry and a calculation of the theoretical capacitance.
    This part throws me because on the practice sheet there is so much room to write. The answer I got was C=e*A/d, with d being the distance between the plates and A being the surface area of the plates.

    The second question is easier but the last section is giving me fits. I won't even bother to type the full question. The problem involves a single spherical shell capacitor, of radius a, connected with a wire to a resistor, R ohms, which is connected to the ground. The shell has a steady current i amps that is delivered to the sphere not using the wire. The section giving me problems is:

    If the capacitor has no charge on it to begin with, to what voltage Vf will it charge and roughly how long, T, will it take to reach that voltage. Answers should be in terms of i, C, and R.
    I got [del]V=iR, but I don't think that's what he's looking for...and as for the time, I assume it has to do with i=[del]Q/[del]T, but I can't seem to put it all together.

    Any help at all on these problems is greatly appreciated. Thanks.
  2. jcsd
  3. Aug 21, 2003 #2
    The main problem I'm having with that last problem is that it says the charge doesn't flow through the resistor. Would this matter? I think I managed to solve for [del]t anyway. [del]t=CR.
  4. Aug 21, 2003 #3

    Tom Mattson

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    That's right too. There should be a derivation in your book, so that will take the guesswork out of it.

    That is the correct expression for a capacitor when fringe effects can be neglected. Also, when they say "estimate the capcitance", I think they are looking for a number.


    How did you get that? To me, when it says the charge does not flow through the resistor, it means that all the charge is stored in the capacitor. That means that dq/dt=i, where q=the charge on the capacitor.
  5. Aug 22, 2003 #4
    Re: Re: Electrical Field of a Parallel Plate Capacitor

    I substituted different expressions into one another to get that result. I forgot to post that the question wanted the answer in terms of R, I, and C.

    My work is as follows:
    [del]V=iR; this is just a definition which is why I think that it's wrong.

    For the second part:
    i=[del]Q/[del]t ergo [del]t=[del]Q/i=C[del]V/i=CiR/i=CR
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