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Homework Help: E-field at the center of a semicircle

  1. Jan 16, 2008 #1
    [SOLVED] E-field at the center of a semicircle

    1. The problem statement, all variables and given/known data

    Calculate E at center of semicircle:

    A uniformly charged insulating rod of length L = 12 cm is bent into the shape of a semicircle as shown. The rod has a total charge of Q = -6 nC. Find the magnitude of the electric field at the point P -- the center of the semicircle. (Answer in units of N/C.)

    2. Relevant equations

    E = K((Q)/(r^2))

    3. The attempt at a solution

    To find the distance/radius from the point P to any point on the semicircle would be...
    12cm*2 = circumference
    2*pi*r = 24cm
    r = 3.8197cm or .038197m

    When I use the equation under the "relevant equations" heading, I'm getting a really huge (E) of about 36,970 N. This can't be right and I'm not sure of where to go from here. Any help would be greatly appreciated. Thanks!

    Attached Files:

  2. jcsd
  3. Jan 16, 2008 #2
    try to use gauss law... div(E)=rho.
    i suggest you spherical coordinates and divergence theorem..

  4. Jan 16, 2008 #3
    we haven't even started studying gauss law yet...
  5. Jan 16, 2008 #4
    What you did would be correct if the electric field from each infinitely small piece of semicircle was in the same direction.

    You need to integrate the field caused by each infinietly small peice of semicircle.
  6. Jan 16, 2008 #5
    When I integrate the field of the semicircle, I come up with a field of magnitude -147,858 N/C.......?
  7. Jan 16, 2008 #6
    The electric field, [tex]d\textbf{E}[/tex], due to a small piece of charge, [tex]dQ[/tex], is

    [tex]d\textbf{E}=\frac{kdQ}{r^2}(\cos\theta \textbf{i} + \sin\theta \textbf{j})[/tex].​

    Since we obviously can't take the integral with respect to charge, let's define the linear charge density of the rod as

    [tex]\lambda=\frac{Q}{L}[/tex]. ​

    Now, being a density, we can also say, for a small length of the rod, [tex]dL[/tex], that

    [tex]\lambda dL = dQ[/tex]. ​

    Also, for a small angle, [tex]d\theta[/tex],

    [tex]dL=rd\theta[/tex], so [tex]r\lambda d\theta = dQ[/tex]. ​

    Plugging this in above, we have

    [tex]d\textbf{E}=\frac{kr\lambda d\theta}{r^2}(\cos\theta \textbf{i} + \sin\theta \textbf{j}),[/tex]​

    So, simplifying and including the integral sign,

    [tex]\textbf{E}= \frac{k\lambda}{r} {\int^{\frac{3\pi}{2}}_{\frac{\pi}{2}} (\cos\theta\textbf{i} + \sin\theta\textbf{j})d\theta,[/tex]​

    which I will leave to you to solve.
    Last edited: Jan 16, 2008
  8. Jan 17, 2008 #7
    Thanks for your help, foxjwill!
  9. Jan 17, 2008 #8
    You're very welcom. :)
  10. Jan 19, 2008 #9
    I have the answer to an example problem of this type, but each time I calculate this my answer turns out to be different.

    Since the j component is symmetrical, they cancel each other out, so I came up with
    [tex]\textbf{E}= \frac{k\lambda}{r} {\int^{\frac{3\pi}{2}}_{\frac{\pi}{2}} (\cos\theta\textbf{i})d\theta,[/tex]

    Solving for the integral, I get

    [tex]\texbf{E}= \frac{k\lambda}{r} \left[\sin\theta\right]^{\frac{3\pi}{2}}_{\frac{\pi}{2}}[/tex]
    [tex]\texbf{E}= \frac{k\lambda}{r} \left[\sin\frac{3\pi}{2}-\sin\frac{\pi}{2}\right][/tex]
    -1 -1 is -2, so
    [tex]\texbf{E}= \frac{k\lambda}{r} \left(-2\right)[/tex]

    I'm sure my radius is fine, so where else am I going wrong?
    Last edited: Jan 19, 2008
  11. Jan 20, 2008 #10
    I'm not entirely sure what you're asking. Were you given values?
  12. Jan 20, 2008 #11
    You know, this kind of response isn't very helpful. Can you really solve this problem this way? Gauss's law is convenient for problems with lots of symmetry so that the surface integral becomes trivial, but for a problem like this, I don't see how it would be easy to do, especially compared to simply integrating Coulomb's law, as above.
  13. Jan 27, 2008 #12
    why is this true? [tex]dL=rd\theta[/tex] Maybe I am missing something here because this doesn't make sense to me...
  14. Jan 27, 2008 #13
    It's the differential line element along the circumference of a circle. Arc length on a circular arc is angle (in radians) times radius; here r is radius and [tex]d\theta[/tex] is the (differential) angle.
  15. Jan 27, 2008 #14
    I think that sometimes it helps people to give them just the hint and not all the calculus done by somebody else... Perhaps for me is very important to rember that using only 4 equations (Maxwell) and some calculus we can solve every C.E.M. problem.
    I thought that just saying Gauss+div theorem he could recognize on wich subset of (0,6.28) integrate...
    i see more foundamental to rember where equations come from, and try every time to "derive" all the connections and then solve the problem... maybe not only analitically.

    in any case i know that in this world is more important to give the right answer as soon as possible instead of thinking why??

    As an example: The lenght of every line on a plane is [tex]L=\int dl=\int\sqrt{dx^2+dy^2}[/tex]
    and only in this contest you have [tex]\int Rd\vartheta[/tex].
    this after differentiating the trasformation of coordinates:
    [tex]x=Rcos\vartheta; y=Rsin\vartheta[/tex]

    regards marco
  16. Jan 27, 2008 #15
    Marco_84: I agree, and I didn't mean to suggest that you were incorrect to suggest the use of Gauss's Law. My objection had to do with where you might lead the OP with this, since it is quite common in introductory courses to label certain problems as "Gauss's Law" problems when they involve symmetries that allow the reduction of the problem to a trivial one. Examples are those with spherical symmetry, linear symmetry, or planar symmetry, which allow one to make arguments for the integral over a suitably defined surface to be either zero-valued or equal to the field magnitude times the area (i.e. the field is either parallel to or normal to the surface). In such cases one need not even take any explicit integrals, since they all reduce trivially.

    This problem doesn't quite fall into that category, IMO, so I wouldn't want the OP to struggle to define a surface with those characteristics; it would not be easy, and it's not necessary.

    I guess we could agree to call the solution of this problem "explicit use of Gauss's Law" or something else to signify that one really has to do the integral. (?)
  17. Jan 27, 2008 #16
    Why do you go from pi/2 to 3pi/2?
  18. Jan 27, 2008 #17
    It doesn't matter what limits you pick, as long as the difference is pi, because it's a semicircle. Whether you go from zero to pi or pi/2 to 3pi/2 doesn't matter.
  19. Jan 27, 2008 #18
    Perfect you get it.... people reduce the differential equations to simpliest (already solved) finite quantities after a "smart" knowledge of the simmetries involved... but then we have to know how to simplify the integral knowing the meaning of what a flux is and so on..
    This is what i would like to suggest people when ask.

    best regards
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