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Physics Midterm Question 3 Help

  1. May 10, 2009 #1
    Ok, well I asked this question somewhere else and he said the same answer I wrote on my exam. But I got docked heavily for that. Ok well here's the problem:

    A negative charge - Q is distributed uniformly over a half circle with radius 'R'. P is at the center of the half circle.

    A) Show that the electric field at point 'p' is given by E = -(2kQ/πR2)j

    Magnetic Flux = Integral of E dot da = Q enclosed/Epsilon knott
    = E is uniform therefore it pops out of the integral and you get E integral of da = Qenclosed/Epsilon knott. Intergration of da is A. So it forms into E dot A = Qenclosed/Epsilon knott.

    E = -Q/Epsilon knott x 2/Pi r^2
    E= -2/Epsilon knott x Q/Pi r^2

    What's Wrong with my process because I ONLY got 3/15 on this.
     
  2. jcsd
  3. May 10, 2009 #2

    diazona

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    Is it half of a filled circle or half of a circular ring?

    [tex]\iint \vec{E}\cdot\mathrm{d}\vec{a}[/tex] is electric flux, not magnetic flux

    Why do you say [tex]\vec{E}[/tex] is uniform?

    Also, for what it's worth, it's [tex]\epsilon_0[/tex], "epsilon-zero" or "epsilon-naught" by some (not "epsilon-knott")
     
  4. May 10, 2009 #3

    Doc Al

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    Are you trying to use Gauss's law? Over what surface? Why do you think the electric field is uniform? (That's electric flux--not magnetic flux--of course.)

    The way to solve this is to by direct integration of the electric field due each element of charge around the half circle.
     
  5. May 10, 2009 #4
    It's half of a circular ring. Sorry I meant Electric Flux. Magnetic Flux = Integral of B dot Da = mew-naught x Ienclosed. So you are saying the E Field is not uniform? Is it because its not an enclosed circle?
     
  6. May 10, 2009 #5

    Doc Al

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    To apply Gauss's law you need a closed surface and enough symmetry to make the electric field uniform over that surface. Gauss's law won't help you solve this problem.
     
  7. May 10, 2009 #6

    Redbelly98

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    E is not uniform because it is not a flat plate.

    Re-read the last sentence in Doc Al's post #3.
     
  8. May 10, 2009 #7
    So Magnetic Flux = integral of E? Im confused about what formula to use on that question.
     
  9. May 10, 2009 #8

    Doc Al

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    Forget about electric flux and Gauss's law.
    Think of the half circle as a bunch of small charge segments. What's the field from a point charge? Add them up! (Integrate.)
     
  10. May 10, 2009 #9
    Integration of Kq/r^2?
     
  11. May 10, 2009 #10

    diazona

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    Yep...but you have to remember that the direction of the electric field coming from each little segment is different. So you can't just multiply Kq/r^2 by the length of the half-circle; you need to account for the different directions.
     
  12. May 10, 2009 #11

    Doc Al

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    Yes. Express the element of charge, dq, in terms of charge density and the element of length along the half circle. Note that the field is a vector; make use of symmetry to simplify the integral.
     
  13. May 10, 2009 #12
    So lambda= Q/Pi R since the total Length is half the circumference of the circle is 1/2 (2pi R). dq = lambda dl
    dq = Q/Pi R dl

    E = integration of k dQ/R^2
    E = k integration of Q/Pi R/ R^2

    Is this right step so far?
     
  14. May 10, 2009 #13

    diazona

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    If you mean
    [tex]\mathrm{d}q = \frac{Q}{\pi R}\mathrm{d}l[/tex]
    then yes, that is right. And if you mean
    [tex]E = \int \frac{k\mathrm{d}q}{R^2} = k\int \frac{Q}{\pi R}\frac{1}{R^2}\mathrm{d}l[/tex]
    that's almost right - but remember, you do need to account for the direction of the electric field.
     
  15. May 11, 2009 #14
    [tex]\frac{k}{\pi}\int \frac{-Q}{R^3}\mathrm{d}l[/tex] =

    [tex]\frac{-k}{\pi R^2}\int {Q}\mathrm{d}l[/tex]

    Where do I go from here?

    Because its supposed to be

    E = [tex]\frac{-2kQ}{\pi R^2} [/tex]
     
  16. May 11, 2009 #15

    Doc Al

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    You need to take into account the direction of the electric field from each point. Hints: Consider horizontal and vertical components. Express the field in terms of an angle.

    Rewrite your integral with that in mind.
     
  17. May 11, 2009 #16

    diazona

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    Let me clarify my previous post:
    [tex]E = \int \frac{k\mathrm{d}q}{R^2} = k\int \frac{Q}{\pi R}\frac{1}{R^2}\mathrm{d}l[/tex]
    is wrong. It's very close to being right, but it's not actually correct.

    As I said (and apparently, as Doc Al said while I was writing this), you need to account for the direction of the electric field. Now, the electric field from each little segment of charge points directly away from that segment. Hopefully you can see that at the center of the (half-)circle, the electric fields from all those little segments are pointing in different directions. For example, the field from the left end of the loop points to the right, and the field from the right end of the loop points to the left - those two will cancel out. The field from the segment directly above the center of the half-circle points straight down, but because it's a half-circle, there's no segment below the center to cancel that one out. And so on.

    Mathematically, the way you take that into account is by inserting a unit vector into the integral:
    [tex]E = k\int \frac{Q}{\pi R}\frac{(-\hat{r})}{R^2}\mathrm{d}l[/tex]
    where [tex]-\hat{r}[/tex], for each small segment of the loop, is a vector of length 1 which points directly away from that segment. You can't pull [tex]\hat{r}[/tex] out of the integral because it's different for each segment - but you can express it in terms of other vectors, [tex]\hat{x}[/tex] and [tex]\hat{y}[/tex] (the unit vectors in the x and y directions), which are the same for all segments. Can you figure out how you'd do that? (It'll probably help to draw a picture)
     
    Last edited: May 11, 2009
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