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Finding Electric Field of A Sheet Infinite In Magnitude

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A large, flat, horizontal sheet of charge has a charge per unit area of 3.15 µC/m2. Find the electric field just above the middle of the sheet.


    2. Relevant equations
    [itex]dq = \sigma dA[/itex]

    [itex]\oint \vec{E} \cdot d\vec{A} = \frac{q_enc}{\epsilon_0}[/itex]

    [itex]\epsilon_0 = \frac{1}{4 \pi k_e}[/itex]



    3. The attempt at a solution

    I figured that if I could generate a Gaussian surface--specifically, a cube--,then I could solve this problem. Since two sides of the cube won't experience electric field lines propagating through, we won't have to consider those two sides; but the side of the cube above the sheet, and the side of the cube below the sheet, will experience electric flux. Furthermore, since the electric field lines are either emanating upwards or downwards, the amount of electric field lines will be the same through each of the two sides will be the same.

    Hence, [itex]\vec{E}\oint d\vec{A} = \frac{q_enc}{\epsilon_0}[/itex].

    In order for me to substitute in[itex]dq[/itex] for [itex]q[/itex], they need to be equal. I know they are equal, but I am having trouble justifying why the two are equal.

    By doing this substitution, and solving for the electric field [itex]\vec{E}= \frac{\sigma dA}{\oint d\vec{A} \cdot \epsilon_0}[/itex].

    Again, I know that that [itex]dA[/itex] and [itex]\oint d\vec{A}[/itex] cancel out, but I just can't qualify it.

    Finally, since we are speaking about a point immediately above the center of the sheet, if we were to place a test particle at this position, it would only endure the electric field lines emanating from the top face of the sheet. Therefore, the electric field line at this point would be [itex]\frac{||\vec{E}||}{2}[/itex]

    By using this method, I was able to solve the problem; but, as you can see, there are portions of this method I can't really justify. If someone could help me with using correct reasoning and terminology, I'd really appreciate it.
     
  2. jcsd
  3. Feb 11, 2013 #2

    mfb

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    dA is a differential, the integral is not, they cannot cancel.
    I think the electric field should be part of the integral.

    The electric field is perpendicular to the surface and the cube is small, so you can assume that the electric field is constant in this area, and get E*A as result.
     
  4. Feb 11, 2013 #3
    How come this method worked?
     
  5. Feb 11, 2013 #4

    mfb

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    Is there any reason to expect a problem? ;)
     
  6. Feb 11, 2013 #5
    I don't understand. Also, [itex]dq = q[/itex] is still needed, but why is it true?
     
  7. Feb 11, 2013 #6

    mfb

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    This is not true. You have to integrate q over the same area.
     
  8. Feb 12, 2013 #7
    All right, I am quite confused at the moment. Could someone please show me how to start a solution for this problem?
     
  9. Feb 12, 2013 #8

    mfb

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    Take a small cube (better: the limit to size 0), symmetric around the center of the sheet with the obvious orientation. Based on symmetry, just the two sides above/below contribute to the flux of electric field in/out. The field is perpendicular to the surfaces and constant everywhere.

    Gauß => ##2EA=\frac{Q}{\epsilon_0}## where the 2 comes from the two sides and Q is the charge of an area A of the disk. Therefore, E=... (independent of A)
     
  10. Feb 15, 2013 #9
    I think I solved it, by discovering why precisely q and dq are equal. In a sense, they are equal. Since we know the charge per unit area, it is safe to assume that the charge is evenly distributed throughout this sheet that is infinite in magnitude. By summing up all of the infinitesimal areas, each containing the same charge, we get [itex]dq = \sigma dA_1 \rightarrow \int dq = \sigma \int dA_1 \rightarrow q = \sigma \cdot A_1[/itex] This particular q, is the q that is enclosed within the cube, thereby allowing me to substitute it into my Gauss law equation: [itex]E= \frac{\sigma \cdot A_1}{\oint d \vec{A_2}}[/itex], where A_1 is the area of the flat sheet, and A_2 is the surface area of the Gaussian cube. My question is, how would A_1 and A_2 cancel out? Would they completely vanish, or would only some of the terms from each one cancel out? One side of the Gaussian cube that is parallel to the sheet has the same area as the sheet; however, we have two sides of the Gaussian cube that are parallel to the sheet. So, would the relationship between A_1 and A_2 be: [itex]A_1 = \frac{A_2}{2}[/itex]
     
  11. Feb 15, 2013 #10

    mfb

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    Right. That is the same "2" I have in post 8.
     
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