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Electric Field of Continuous Charge Distributions

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A disk with a radius of 0.6 m is given a uniform charge density of -7.2*10-9 C/m2. The disk is in the xz plane, and centered at the origin.
    A. What is the size and direction of the electric field at the point A, whose coordinates are (0m, 1.5m, 0m)?

    B. You now add an infinite line with a uniform charge density of +400 pC/m

    What is the position of the line if the electric field at point A now has a size of 100 N/C and points upward
    Relevant equations
    • ∫E(vector)*dA(vector)=qenclosed/( ε0)
    • Adisk=πr2
    • ASphere=4πr2
    • σ=Q/A
    • λ=Q/l
    • ε0=1/(4πke)


    3. The attempt at a solution
    Part A. So I first made a sphere around point A as my gaussian surface. I next solved for Q=σA and used Q as my enclosed charge. Next, I got rid of the dot product in ∫E(vector)*dA(vector) by making it negative. My reasoning was that the surface charge density was - so the E would be opposite the dA(vector). The E is constant at all points of the sphere so I was able to bring the E out of the closed integral. Next I did ∫dA and got the ASphere=4πr2. So, my equation turned into -E(4πr2)=(Adisk)(σ)/(ε0). With algebra I solved for E and got E= 32.54. This doesn't seem like a correct answer and there seems to be some uncertainty in my method which doesn't seem correct.

    Part B. I cant seem to understand how I am able to find the position of the line with Gauss's law. I can find the length, but I cant seem to understand how to start this part.
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  2. jcsd
  3. Sep 15, 2015 #2

    blue_leaf77

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    This is the source of your problem. By symmetry analysis and the fact that the coarge is distributed on a disk, the E field on any sphere cannot be constant. Think of the disk as being composed of concentric rings of varying radius, this way of course you also need to know the E field at the axis of a uniformly charged ring.
    In which Cartesian direction is this "upward" refering to?
     
  4. Sep 15, 2015 #3
    Thanks for replying. My assumption is that it is in the y direction upward.

    So, the E isn't constant around the sphere. Would it be constant if I changed the guassian surface to a disk itself?
     
  5. Sep 15, 2015 #4

    blue_leaf77

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    A surface of constant field magnitude does exist, but it's not a disk. The shape of such surface may be too complicated to be described by the existing mathematical functions. To put it short, using Gauss method here is unwise, try to consider what I suggested above. Do you know the field at the axis of a uniformly charged ring?
     
  6. Sep 15, 2015 #5
    I don't know the E field of a UC ring(or its not given). Can that be calculated with the information that is given? So, I am a little confused as to what method I should use to answer this problem. I was thinking of using the Fe=qE, but that requires a force to exist(which there probably is, but it is not given).
     
  7. Sep 15, 2015 #6

    blue_leaf77

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    The electric field on an axis of a uniformly charged ring situated on xz plane, centered at the origin is
    $$
    \mathbf{E} = \hat{\mathbf{y}} \frac{kyQ}{\left(y^2+r^2\right)^{3/2}}
    $$
    where ##r## the ring radius, ##Q## total charge of the ring, and ##y## distance of the observation point from the ring center.
    Now imagine building a disk from concentric rings of thickness ##dr## from zero radius up to the radius of the desired disk. The electric field at the axis due to the disk will then be the sum of fields due to individual rings constituting the disk. So,
    $$
    \mathbf{E}_{disk} = \int_{r=0}^{r=R} d\mathbf{E}_{ring}
    $$
    with ##R## the radius of the disk. The next thing to do is to express ##d\mathbf{E}_{ring}## using the first formula above in term of ##dr##.
     
  8. Sep 15, 2015 #7
    Thanks for the help. So the integral is used because we want to add up all the small rings energies?
     
  9. Sep 15, 2015 #8

    blue_leaf77

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    It's electric field, not energy.
     
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