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!

Capacitance value between two conducting plates.

  1. Apr 1, 2013 #1
    1. The problem statement, all variables and given/known data

    We have two conducting plates. One is moveable, the other is not moveable. The moveable has the charge of +1V. While the other plate has the charge 0V and is part of a enclosing envelopment. See fig:

    http://tinypic.com/view.php?pic=wb2onm&s=6

    But when we move the upper plate horizontaly shown in the figure below the capacitance gets lower.

    http://tinypic.com/view.php?pic=4hcrbb&s=6

    In the figure the E-field and potential is plotted. How am I to motivate why the capacitance gets lower?

    2. Relevant equations

    From the relevant equations we got:

    Q=CV (1)

    and

    [tex]V_{12}=\int{E\cdot{dl}} (2)[/tex]

    [tex]C=\epsilon\frac{S}{d} (3)[/tex]

    3. The attempt at a solution

    I tried to motivate that the E-field gets weaker because the distance is greater in the horizontal axis. Then i suppose that the value of V gets smaller when you calculate the potential between the metal plates by equation (2).

    But this doesn't seem quite correct to me since the capacitance C is independent of V and Q. We see by equation 3 that the capacitance is dependent of the surface area S and the distance d.

    But the surface is the same, the only thing that changes is the distance. But the vertical distance is still the same.

    So how do i motivate that the value of the capacitance decreaces?
     
  2. jcsd
  3. Apr 1, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    0V and 1V are not charges.

    There are two ways to describe the motion:
    A) The distance between the plates increases. As the charge is the same, the electric field strength is the same, and the voltage increases. Using Q=CV, this has to correspond to a smaller C.
    B) The area (of overlap) decreases. Therefore, less charge contributes to the capacitance, while the voltage stays the same. Using Q=CV, this has to correspond to a smaller C.

    Both views are just approximations to the real setup, and you would need a numerical calculation to do it properly, but they give the right direction.
     
  4. Apr 2, 2013 #3
    But according to the superposition formula the electrical field strenght is supposed to get weaker when the distance increaces?

    Isn't it true that the capacitance will vary on the horizontal axis, since the plates are not aligned in a symmetrical way?
     
  5. Apr 2, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    For point-charges, not for "infinite" areas.
    Capacitance is a property of the whole system.
     
    Last edited: May 23, 2013
  6. May 23, 2013 #5
    I wrote also an equation that looks like this:

    [tex]\hat{E}=\frac{\rho}{\epsilon_0}[/tex]

    My teacher said it is a fundamental error here. What is it?

    I guess that it might be [tex]\epsilon_0[/tex] is the permittivity in free space. But since the free charges are on a metal piece then this should be a fundamental error since it is not free space. So a correct answer should be just [tex]\epsilon[/tex].

    Or is it because i forgot vector notation on the right side of the equation?
     
  7. May 23, 2013 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If ##\hat{E}## is a vector, then the equation cannot be right - the other side is a scalar.
    Apart from that, I don't see how the equation takes into account that you have two plates.
     
  8. May 24, 2013 #7
    Hi mfb,
    In the picture, the upper plate is only moved horizontally, and I think the distance d is constant.
    Can you explain more?
    Can I use the formula C = εS/d and because the overlap decreases and d is constant, then C decreases?
     
  9. May 24, 2013 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The vertical distance is the same, but there is an additional horizontal distance afterwards. In particular, the outer parts of the plates have a larger separation from the other plate.
    What is S? I don't think there is a nice general formula for shifted capacitor plates. Just using the area of overlap gives a value which is too low.
     
  10. May 24, 2013 #9
    Thanks,
    Do you mean that the non overlap between two plates also contribute to the capacitance?
    Here I consiered the capacitor as a two plate in parallel capacitor. Is this wrong?
    I ignored every part non-overlap between two plates and the remain is like a parallel plate capacitor.

    If this case is wrong then I think I have to use the formula C = Q/V
    charges are traped in the plate and therefore Q isn't changed but how can you know V increases?
     
  11. May 24, 2013 #10

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Sure, but not so much as the other parts.
    Then you underestimate the capacitance.

    C decreases, and you can show this with the arguments I gave in post 2. If you need a quantitative answer, simulate it or hope that someone else did that.
     
  12. May 24, 2013 #11
    I need to understand it first.
    I am confused about this part. I don't really know which path you refer to.
    If so, then V = ∫Edl and the voltage between two plate isn't the same at every two points between two plate?
     
  13. May 24, 2013 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Sketch (very qualitative):

    attachment.php?attachmentid=58995&stc=1&d=1369418480.jpg

    E is not constant any more, it depends on the position.
     

    Attached Files:

  14. May 24, 2013 #13
    Nice pic! I see it now, thanks mfb.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Capacitance value between two conducting plates.
Loading...