1. Not finding help here? Sign up for a free 30min 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!

Electric potential HELP

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data
    An empty capacitor is connected to a 11.7-V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (κ = 3.1) is inserted between the plates. Find the magnitude of the amount by which the potential difference across the plates changes.

    2. Relevant equations
    if the charge is 11.7-V and k= 3.1, how is k used to find V?

    3. The attempt at a solution
    q = C*V
  2. jcsd
  3. Jan 29, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Inserting a dielectric causes a polarization of the charges inside the dielectric. This internal electric field in turn causes the E field between the terminals to decrease and thus decreases the potential difference. The capacitance C turns into k*C. Q must remain unchanged. How does this change V?
  4. Jan 29, 2008 #3
    Hmm, do you know about the displacement field? Basically it is a convenient way of redefining Maxwell's equations for materials. In the case you would be concerned about

    [tex] \nabla \cdot \mathbf{D} = \rho_f[/tex]

    which will translate to the Gauss's law everyone is used to

    [tex]\iint \mathbf{D} \cdot \hat{n}da = Q_{free}[/tex]

    Where the free charge density and total free charge are the quantities on the right side, they would be charges on your capacitors. We have to distinguish between the free charges and other (bound) charges because the dielectric will have all these dipoles floating around that we wouldn't care about for this case. Then you can use the relation that

    [tex]\mathbf{D} = \epsilon \mathbf{E}[/tex]

    where epsilon would be, in your convention

    [tex]\epsilon_0(1+k) = \epsilon[/tex]

    So almost everything is the same that you would do to find a regular capacitor, except the constants have changed.
    Last edited: Jan 29, 2008
  5. Jan 29, 2008 #4
    would V increase by "k" also? (V*k)??
  6. Jan 29, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    Dead wrong. Q is constant. Actually, Mindscrape might know this stuff better than me. I'll turn you over to him.
  7. Jan 30, 2008 #6
    I'm not really sure what the k the OP is using, but I assume it is the dielectric constant. It could be epsilon, though I highly doubt it because that would make no sense at all. Regardless, maybe we should rename the constant out front to be Y because Y is never used. I'm 99% positive you really want the epsilon I defined before to use as Y.

    So from gauss's law we would get

    [tex]DA = \sigma_f A = \sigma A[/tex]

    [tex]D = \sigma[/tex]

    which would bring us to E like so

    [tex]E = \frac{D}{Y} = \frac{\sigma}{Y}[/tex]

    So then V is the integral of E over the length

    [tex]V = E*L[/tex]

    At this point you should have everything you would ever need, within the limits of notation differences.
    Last edited: Jan 30, 2008
  8. Jan 30, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    I think it is just the dielectric constant. A multiple of the vacuum permittivity. I.e. k=1 is vacuum. Or am I wrong? It's been a long time. So C becomes C*k when the material is inserted?
  9. Jan 30, 2008 #8
    Yeah, that is the way I interpreted it, as the dielectric constant. I am used to [itex]\chi[/tex] being the dielectric constant, and the OP just put in some constant open to interpretation, so I don't know if the k is supposed to be dielectric constant or not, but I'm 99% sure it is supposed to be.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric potential HELP