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Why do the Horizontal components cancel in E-Field

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Plane X=0 has a uniform charge density, [tex]\rho[/tex]s while plane X=a has -[tex]\rho[/tex]s. Determine the electric field intensity in region X<0.


    2. Relevant equations

    E=[tex]\frac{Q}{4\Pi\epsilon0R2}[/tex]aR


    3. The attempt at a solution

    I already know the answer to this question. What I'm trying to do is figure out a small part of the math behind it. Let's start by looking at one plate instead of two. How can I prove mathematically that all of the horizontal components cancel out leaving behind the component pointing away from the plate. I understand why but I need to understand the math behind it. This makes working problems so much easy when I can understand the math behind it easily. So here goes what I tried to do.

    First I need to define a point to examine my e-field at. r=(x,y,z) next i need to define the different parts of the equation above. Best way to do this is to look at very very small peices and then adding them up.

    dE = [tex]\frac{dQ}{4\Pi\epsilon0R2}[/tex]
    dQ = [tex]\rho[/tex]sdS
    dS = dY dZ
    R = ?????

    I want to put that R=r-r' and that r' would the location of the very very small plate which would be (0,y',z') which would make R=(x,(y-y'),(z-z')) The y and z would stay constant because they are the point that we are evaluating E at. This is were I get lost. I know technically that y and z can be ignored because they will add to zero but, how can I prove this. If I keep going with my attempt the next step would be to find the magnitude of R and the unit vectors of R. I'm going to leave these out for now since they are long to type. So now I would put everything together and integrate. This is where everything gets messy. I know that if I put in a value for y say 4 and then integrate (4-y') from 0 to 8 the answer is 0. How can I write this out when even though y is constant because I don't know what it is I can't really prove that it ends up being 0.
     
  2. jcsd
  3. Sep 18, 2010 #2
    I worked on the problem a little bit more and I believe that I've worked out the correct solution. First when you write everything out it takes up a lot of paper and it is somewhat complicated. So the first thing is to go ahead and integrate the y and z components since they are the horizontal components that should go away by the end of the solution. If you do this they do indeed go to zero. I'll write out the y component so that it's easier to understand what I'm writing.

    ay=[tex]\frac{\rhos}{r\Pi\epsilon0}[/tex][tex]\int\int\frac{(y-y')}{(x2+(y-y')2+(z-z')2)\frac{3}{2}}}[/tex]

    This ends up equaling zero because the integral from infinity to negative infinity of (y') is 0. So this leaves the X component as the only one that has to be deal with. The X component is


    ax=[tex]\frac{\rhos}{r\Pi\epsilon0}[/tex][tex]\int\int\frac{(x)}{(x2+(y-y')2+(z-z')2)\frac{3}{2}}}[/tex]

    First thing to do to evaluate this integral is to realize that X is constant and can be relabeled H to be a little less confusing. The next thing if you draw everything out you realize that converting to polar coordinates would be a whole heck of a lot easier. This changes the equation to look like

    ax=[tex]\frac{\rhosH}{r\Pi\epsilon0}[/tex][tex]\int\int\frac{(1)}{(H2+\rho2)\frac{3}{2}}}[/tex][tex]\rho[/tex] d[tex]\rho[/tex] d[tex]\Phi[/tex]

    The limits of integration would be 0 to infinity and 0 to 2pi.

    Integrating this will leave

    [tex]\frac{2\pi}{H}[/tex]

    Which makes the final equation of

    [tex]\frac{\rhos}{2\epsilon0}[/tex]

    Then going back to the original problem and noticing that H isn't involved in the final equation means that the e-field in the region X<0 is equal to 0. My text book does point the final equation out but, I don't like using these equation with out understanding how they were derived. Hopefully I did all of the work correctly.
     
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