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Homework Help: Electric Field

  1. Feb 4, 2005 #1

    I have been assigned a problem that I can't solve.

    I have a rotating axis to which a thread is connected in one end. The thread is perpendicular to the rotating axis. The thread has a charge density [tex]\lambda[/tex] and a length [tex]L[/tex].

    First of all, I need a mean value of the charge density of the circular disc described by the rotating thread with respect to time. I have interpreted this as the charge density:
    [tex]\sigma(r) = \frac{\lambda \mathrm{d}r}{2\pi r\mathrm{d}r} = \frac{\lambda}{2\pi r}

    The charge density is supposed to be a function of [tex]r[/tex]; the distance to the center of the disc. However, in the density above, the charge density is infinite close to the center. I can't interpret this conceptually.

    Second, I am supposed to determine the electrical field a distance [tex]r_0[/tex] from the center of the disc along the rotational axis. Coulombs law yields:

    [tex]E = \frac{1}{4\pi\epsilon_0}\iint_\Omega \frac{\mathrm{d}q}{R^2} = \frac{1}{4\pi\epsilon_0}\iint_\Omega\frac{\sigma\mathrm{d}x\mathrm{d}y}{r_0^2 + x^2 + y^2} = [\mathrm{Polar\ coordinates}] = \frac{\lambda}{4\pi\epsilon_0}\int_0^L\frac{\mathrm{d}r}{r_0^2 + r^2} = \ldots = \frac{\lambda\theta}{4\pi\epsilon_0r_0}[/tex]

    This result is a bit strange, if you consider the extreme values. For instance:
    \lim_{\theta\rightarrow 0}E = 0 & \mathrm{Ok!} \\
    \lim_{\theta\rightarrow \frac{\pi}{2}}E = k & \mathrm{Ok?} \\
    \lim_{r_0\rightarrow 0}E = \infty & \mathrm{Not\ Ok??} \\
    \lim_{r_0\rightarrow\infty}E = 0 & \mathrm{Ok}

    From the third extreme value, I must conclude that the result is wrong, as it should be 0 in the disc (the forces cancel eachother)

    What is wrong? Is it the mathematics or the physics that fail?

    Please Help!

  2. jcsd
  3. Feb 4, 2005 #2
    1) First of all, it seems that the analogy of the rotating thread with a charged disc is not very appropiate (in my opinion).

    In any point of the rotating axis you will have an electric field with two vectorial components:
    The normal component will rotate syncronic with the thread whereas the axial component will be constant. Because of sinusoidal variation of the normal component, its average is zero. Then you have to evaluate only the axial component as the wire is at rest.

    2) When you evaluate the electric field, you add vectors, then I think you need a [tex]cos(\alpha)[/tex]. The expression for [tex]\vec{E}[/tex] which you use seems to be wrong even if the analogy with the charged disc is correct.
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