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Homework Help: Electric field of semi circular piece of wire

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a semicircular piece of wire of length L and total charge of -Q where Q>0 (radius [itex]R=L/\pi[/itex]). What is the magnitude and direction of the electric field at the centre of the semi circle ?(if the wire was a full circle the centre would be the centre of the full circle).

    2. Relevant equations

    [tex]dE=k_{e}\frac{dq}{r^2}[/tex]

    3. The attempt at a solution
    I know what the direction of the electric field would be but not sure about the magnitude.
    Say for example if the semi circle passes through the points (0,R), (-R,0) and (-R,0) on the x-y axis then the direction would be to the left. The bit I am unsure about what the magnitude of dq is?
    I correctly guessed that it was
    [tex]dq=2\lambda dx = 2\lambda R sin\theta d\theta[/tex]
    where theta goes from 0 to Pi/2
    but there is no way I can convince myself that is right. Can someone please explain WHY dq is that?
     
  2. jcsd
  3. Jul 12, 2010 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    I'm not quite sure I'm following you. Is the segment of wire a half circle or a quarter of a circle? If it's a half circle, θ needs to cover a total distance of π, not π/2. Also, I'm not sure where you are getting your factor of "2" from in your dq equation.

    Perhaps I can help by getting your started. The differential electric field equation is

    [tex] d \vec E = k_e \frac{dq}{r^2} \hat{r}, [/tex]

    where [itex] \hat r [/itex] is the directional unit vector (your coursework might use some other notation for unit vectors such as [itex] \hat a_r [/itex]). Much of the "trickiness" of this problem comes from dealing with this unit vector, not the dq part.

    So let's break this equation down.

    If you wish to use the linear charge density λ, you can define it as λ = -Q/L, which is λ = -Q/πR.

    The differential length of wire is Rdθ.

    So the differential charge is dq = λRdθ.

    Now the tricky part, which I'll let you finish. We need to deal with that pesky unit vector [itex] \hat r [/itex]. You can express this unit vector in terms of the standard Cartesian unit vectors [itex] \hat x [/itex] and [itex] \hat y [/itex] (Your coursework might use some other notation for these unit vectors such as [itex] \hat a_x [/itex] and [itex] \hat a_y [/itex] or perhaps [itex] \hat \imath [/itex] and [itex] \hat \jmath [/itex] ). Note that:

    [tex] \hat r = \cos \phi \hat x + \sin \phi \hat y. [/tex]

    where ø is the angle from the positive x-axis.

    Now put everything together :wink:

    [Edit: By the way, the "tricky" part is to make sure you are careful in how you interpret the direction of this [itex] \hat r [/itex] unit vector. Remember, its direction is from the charge (dq) to the test point. In this problem, the test point is at the origin, not the charge. That's probably backwards from what you might be used to.]
     
    Last edited: Jul 13, 2010
  4. Jul 13, 2010 #3
    The wire is a half circle.

    The factor of 2 pops up because of the symmetry, using the orientation I gave earlier, the net electric field in the y direction is zero and breaking up the semi circle into 2 quarter circles results in those 2 quarter circles each contributing an equal amount to the electric field hence only the 0->Pi/2 and the factor of 2. I did that because I already know the direction of the field and only interested in calculating the magnitude.

    I will have to think about your method of the unit vector as it seems quite difficult to me.

    Thanks for the reply
     
  5. Jul 14, 2010 #4
    ok here is what i managed to think of

    [tex]\mathbf{E}=\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}d\mathbf{E} =\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} k_{e}\frac{\lambda }{R}\left(
    \begin{array}{cc}
    \cos\theta \\
    \sin\theta
    \end{array}
    \right)d\theta=-k_{e}\frac{Q }{LR}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(
    \begin{array}{cc}
    \cos\theta \\
    \sin\theta
    \end{array}
    \right)d\theta
    [/tex]
    [tex]\mathbf{E} = -k_{e}\frac{Q}{LR}\left(
    \begin{array}{cc}
    2 \\
    0
    \end{array}
    \right)
    [/tex]
     
  6. Jul 14, 2010 #5

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    Wow, I've never seen that notation before for problems like this. :smile: But it works!

    Actually, when I said "vector" I simply meant that it has a magnitude and direction. It can be represented in Cartesian coordinates, polar coordinates, or whatever. I didn't necessarily mean linear algebra notation for vectors. But either way, your notation and solution seemed to work.

    So your last part of your answer was

    [tex]
    \mathbf{E} = -2k_e\frac{Q}{LR} \hat x + 0 \hat y, [/tex]

    which is the same answer that I got.

    But noting that L/π = R, you could change it to be

    [tex]
    \mathbf{E} = -2 \pi k_e\frac{Q}{ L^2} \hat x [/tex]

    Meaning the magnitude of E is 2πkeQ/L2, and its direction is toward the charged semicircle. :cool:

    By the way, dealing the unit vectors (directional vectors) is often seen as one of the toughest aspects of EM field theory. There is another way to solve this problem that doesn't hinge so greatly on these pesky unit vectors. It involves electrical potential. But if you haven't learned about potential yet, I won't spoil it for you. :tongue2:
     
  7. Jul 16, 2010 #6
    OK thanks for that.

    Glad I am not the only one having a bit of trouble with the unit vector.
     
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