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Homework Help: Electric field between two plates

  1. Jul 4, 2010 #1
    1. The problem statement, all variables and given/known data
    We have two vertical plates with the surface area of A. The charge of the left one is -(Q-q), while the charge of the one on the right is +Q. The left plate is held stationary. The question is, what is the force pulling the plate on the right side? The plates are made of metal (conductor)


    2. Relevant equations
    Gauss's equation (E=Q/e0)
    E=F/Q
    Laws of superposition
    +the fact that the charge in a conductor, when in an electric field, is positioned on the surface of the conductor.


    3. The attempt at a solution
    I actually know the solution, but I don't understand it. In the attachment I tried to draw down the two solutions I came up with, the first one (I) being the one that's supposed to be correct, and the second one (II) being the one I don't see why it isn't correct.

    The first method is simple: we find what electric field the left plate makes, and using F=EQ, we simply multiply that by the charge of the right plate, Q.
    The second method is more complicated. I calculated the eletric field made by both plates, added them together (purple arrows, summa E); and I also checked how the charges are positioned on the right plate. Then I also used F=EQ to see what is the summa force acting on the right side plate.

    If anyone can tell me why the second one is wrong, please do; it's been driving me crazy for a few days now. The second solution is exactly two times as much as the first one. Sorry if the images are hard to get, I'll explain anything in more detail if asked for.
     

    Attached Files:

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  2. jcsd
  3. Jul 4, 2010 #2
    By the way, the problem is located here (page 8, 3/B): http://www.oh.gov.hu/letolt/okev/doc/oktv_2009_2010/fiz2_javut2f_oktv0910.pdf

    It's in Hungarian, but you might understand the equations. Anything below the very top of page 9 isn't concerned here. The thing I don't understand is, if you look at the long line starting with F3 = ... at the bottom of page 8, E(2<-3) and E(3->) are both divided by two (those are the two electric fields that appear in my second attempt). That's exactly what's wrong with my second solution, I see no reason where I need to divide it by two... I'll try to translate that problem with the solution, once I figure out how to make readable equations... until then dictionary.com may be sufficient.

    -Tusike
     
  4. Jul 4, 2010 #3
    How do you calculate the force on a charge point placed in an electric field [tex]\vec{E}[/tex]? Do you take into account the electric field created by the charge when calculating the force? :smile:
     
  5. Jul 4, 2010 #4
    Normally, I don't. But I was going over the solution given on that site, and I noticed that's how they calculated. And the only difference I noticed in their calculations, is that in the end, instead of F=EQ, for some reason they used F=(E/2)Q, and that gave the correct solution. What I'm interested in is that those this /2 thing work every time, or those the number to divide by, (2), only apply in this specific situation? And what I'm more interested in, is how did it even get there?

    The beginning of the solution I translated, applied to the two-plate image I posted (P2 is the left plate, P3 is the right plate):

    SOLUTION.
    The electric field between plates (2) and (3) can be calculated using Gauss's equation and superposition:

    E2<-3 = (Q-q/2) / (e0A)

    where the arrow in the index shows which way the electric field is pointing. On the outer surface of plate (3), the charge is +q/2, and in the inside there is +(Q-q/2). The electric field to the right of plate (3):

    E3-> = (q/2) / (e0A)

    The force acting upon plate (3):

    F3<- = (Q-q/2)(E2<-3 / 2) - (q/2)(E3-> / 2)

    END OF TRANSLATION.

    Now, if you place the two formulas for E2<-3 and E3-> into that equation, you get the correct solution. This equation is supposed to be simply the format F=EQ+E'Q'; what I don't understand, is why does E=E2<-3 and E'=E3-> get divided by 2???
     
  6. Jul 4, 2010 #5
    The only explanation I can find is somewhat silly. Here it is:
    Consider a charged plate placed in a uniform electric field [tex]\vec{E}[/tex] which points to the right. The charge is split into 2 parts, one to the left [tex]q_l[/tex] and the other to the right [tex]q_r[/tex]. Since the E-field inside the plate is zero, we have:
    [tex]E + \frac{q_l}{2e_oA} - \frac{q_r}{2e_oA} = 0[/tex]
    The force on the plate:
    [tex]F = E(q_l + q_r)[/tex]
    From the 2 equations, we get:
    [tex]F = \frac{1}{2}(\frac{q_r}{e_oA}q_r - \frac{q_l}{e_oA}q_l)[/tex]
    This looks similar to your last equation: F3<- = (Q-q/2)(E2<-3 / 2) - (q/2)(E3-> / 2), right? But this way is unnecessarily complicated.
     
    Last edited: Jul 4, 2010
  7. Jul 5, 2010 #6
    Yeah OK, so it was probably just a mix up from the fact that the E lines are halved, since one half leaves from one side of the surface, while the other half from the other side.
     
  8. Jul 5, 2010 #7
    I don't get what you meant.
     
  9. Jul 5, 2010 #8
    In the problem, after calculating what charges were on each plate and what the E fields were in between each plate, i got that E2<-3 = (1/e0)*Ql, where Ql is the charge on the left side of plate 3 (Ql=Q-q/2, but it doesn't really matter). This is exactly the same amount of charge, only opposite to what is on the right side of plate 2. So, we can say that the E field between the two plates are cause by these two charges. However, as you said, when calculating the force by the E-field on a charge, we mustn't take into account the charge's own E-field (what Ql itself is making). That's why I think it has to be divided by 2.

    (I'm still thinking about why E3-> has to be divided by 2 as well...)
     
  10. Jul 5, 2010 #9
    "when calculating the force by the E-field on a charge, we mustn't take into account the charge's own E-field" so yeah I did include the physical explanation as well. But yeah you did a way better job of fully summarizing everything, it's much more logical the way you put it. So I guess problem's solved:) Thanks!
     
  11. Jul 5, 2010 #10
    Ahh, I accidentally deleted my post when editing it :frown: But I guess you read it.

    Anyway, I think there is no reasonable physical reason for the solution they gave, and if there is, why choose the harder and more vague way?
     
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