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Electrostatic pressure

  1. Apr 5, 2007 #1
    I have a metal sphere with the net charge q. And i'm trying to calculate the force that southern hemisphere exerts to northern hemisphere... and I get 0.

    now, the electrostatic "pressure" is
    [tex]\mathbf f = \sigma \mathbf E = (q/4\pi R^2) (q/4\pi \epsilon_0 r^2) \mathbf {e_r}[/tex]

    due to the symmetry, the force will point at z direction, so integrating only the z component of "pressure" over the northern hemisphere should do it, right?

    [tex]\int f_z dA = \int_{-\pi/2}^{\pi/2} \int_0^{2\pi} (q/4\pi R^2) (q/4\pi \epsilon_0 R^2) Rcos(\theta) R^2 sin(\theta) d\phi d\theta = 0???[/tex]

    I found out that this's a question from Griffiths', and the answer manual says the [itex]\theta[/itex] integral should be within [itex][0,\pi/2][/itex], not [itex][-\pi/2, \pi/2][/itex].

    Why is that?
     
    Last edited: Apr 5, 2007
  2. jcsd
  3. Apr 5, 2007 #2
    It is the same reason that an integral over all space will have phi from 0 to 2 pi and theta from 0 to pi.

    It has to do with not double calculating a section of the sphere, not passing over it twice.
     
  4. Apr 5, 2007 #3
    You're only integrating over the top half of of the sphere, not the whole thing, so your theta integration should be truncated to only that side.

    What you actually found is that the net force on the entire sphere that it exerts on itself is zero, which is a nice statement of Newton's third law.
     
  5. Apr 6, 2007 #4

    Doc Al

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    Staff: Mentor

    Just to restate what Crosson and StatMechGuy already explained: Realize that your integral over [itex]\phi[/itex] goes from 0 to [itex]2 \pi[/itex], so counting negative values of [itex]\theta[/itex] counts those points twice. The point [itex](\theta, \phi) = (\theta, 0)[/itex] is the same point as [itex](\theta, \phi) = (-\theta, \pi)[/itex].
     
  6. Nov 6, 2010 #5
    In classical electrodymamics BY jackson this is explained.... clearly
     
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