1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proofing Capacitance

  1. Feb 1, 2015 #1
    Hello everyone, I'm an engineer, and though I highly doubt I'd use this in my mechanic courses...I do truly want to understand what it is I'm putting into application here:

    So I'm going to explain my understanding of capacitance, be prepared for an extensive reading. I'm doing this because I feel like if I'm explaining things to people, I'll understand the underlying concept to an extent that doing problems alone will not. Here it goes (please correct me in any area that I'm wrong in, this is a very basic and humble understanding of my knowledge in electrostatics), also it would be helpful if you wrote along what I'm saying to see where I'm mistaken:

    Assume parallel-plated capacitor (PPC) and in a vacuum -

    Capacitance (C) is the ratio of charge (Q) to the electric potential (V) of the two conductors in a capacitor, where C is measured in one Farad (Coulomb/Volt). Capacitance is the ability of a capacitor to store charge, and therefore store energy.

    Proofing that Electric field (E) of PPC is E = Q / (epsilon-nought x Area of plate):
    In a single charged plate, we find through using a Gaussian surface that flux will be equal to 2EA. Setting 2EA equal to Q enclosed over epsilon-nought (Gauss's law) where Q is equal to sigma times Area of Gaussian surface, we find that the electric field for one charged plate is E = sigma / 2epsilon-nought.

    Now, in a capacitor, with two charged plates (one negatively charged/one positively charged) the Electric field will be twice that of E = sigma / 2epsilon-nought since the electric field of the negative plate and positive plate are traveling in the same direction (From positive plate to negative plate). So, our Electric field equation for a PPC will be E = Sigma / epsilon-nought. Where Sigma = Charge per unit Area, we also derive E = Q / epsilon-nought x A.

    Because the electric field is uniform and given a distance (d) between the two charged plates, we can find the electric potential of the capacitor. From the negative plate to the positive plate, the Electric potential (V) is measured as electric field (E) times the distance between the plates (d). From the previous paragraph, we found E of a PPC is E = Q / epsilon-nought x A. Plugging in, we then get the electric potential (V) = Qd/ (epsilon-nought x A).

    Having obtained the Electric potential (V), we then plug into the equation C = Q/V and get:
    C = (epsilon-nought x A) / d

    Looking at this equation, I conclude that the capacitance of a capacitor is therefore independent of the charges placed on each conductors and is dependent on the sizes of the conductors.

    This is what I got out after piecing bits of information on capacitance, and was written through by analysis. Just wanted to confirm if my thinking process was correct in figuring out the capacitance this way, and if there's anything wrong, please let me know. This basic understanding of capacitance will help me understand way more things that I'm doing right now for a course...
  2. jcsd
  3. Feb 1, 2015 #2


    User Avatar
    Gold Member

    Yes, it is. That's like saying that the length of a ruler doesn't depend on what it measures or the resistance of an ideal wire is independent of the amount of current it is carrying. The charge you put on a capacitor is just something that you do TO the capacitor and does not change its fundamental properties. It has capacitance whether it carries any charge or not.
  4. Feb 1, 2015 #3


    User Avatar

    Staff: Mentor

    Not εo. This will be ε which is short-hand for the product εoεr.

    .... dependent on the geometry of the plates (area of plates and their separation), as well as the dielectric.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook