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Electric Potential Hemisphere Problem

  1. Aug 9, 2006 #1
    There is a hemisphere of radius R and surface charge density [itex]\sigma[/itex]. Find the electric potential and the magnitude of the electric field at the center of the hemisphere. ​
    I started by saying [tex]V = \int dV = \int \frac{\sigma dA}{4\pi \epsilon_0 R}[/tex]. This, at least, I am confident is correct.

    Then, I changed [itex]dA[/itex] into [itex]Rd\theta[/itex] and [itex]Rd\alpha[/itex], where the first one goes across the hemisphere's surface, and the second one goes around the hemisphere's circular edge.

    Continuing:
    [tex]
    V = \frac{\sigma R}{4\pi \epsilon_0}\int_{0}^{\pi/2}d\theta \int_{0}^{2\pi}d\alpha[/tex]
    Which, when evaluated, gives me the wrong answer.

    Oblivious to my error at the time, I continued to the second part.

    [tex]E = \int dE = \int \frac{\sigma dA}{4\pi \epsilon_0 R^2}\hat{r}[/tex].
    I said that [tex]\hat{r} = \cos{\theta}\hat{i} + \sin{\theta}\hat{j}[/tex]. Using similar logic as I did for the potential part of the problem, I continued:

    [tex]
    E = \frac{\sigma R}{4\pi \epsilon_0}\int_{0}^{\pi/2}\cos{\theta}\hat{i} + \sin{\theta}\hat{j} d\theta \int_{0}^{2\pi}d\alpha[/tex]

    Which, when evaluated, also gives me the wrong answer. The correct answers for the parts are, respectively, [tex]V = \frac{\sigma R}{2 \epsilon_0}[/tex] and [tex]E = \frac{\sigma}{4\epsilon_0}[/tex].

    I have no idea what I'm doing incorrectly. I've never done a three-dimensional integration before, but this is the way that I thought it should be done. What have I done wrong?
     
  2. jcsd
  3. Aug 9, 2006 #2

    nrqed

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    I am not sure what you mean by this but from the rest of your calculation (which I did not check), you used [itex] dA = R^2 d\theta d\alpha [/itex]. But the correct relation is [itex] dA = R^2 sin \theta d\theta d\alpha [/itex]. You can see that your equation was not right since the total integral of da over a sphere should give [itex] 4 \pi R^2[/itex]. Your expression would give [itex] 2 \pi^2 R^2 [/itex].
     
  4. Aug 10, 2006 #3
    Yes - I don't understand why it's supposed to be [itex]dA = R^2 \sin{\theta} d\theta d\alpha[/itex]. Why is there a sine? I realize that [itex] dA = R^2 d\theta d\alpha [/itex] doesn't give the right answer, but conceptually I don't understand why there needs to be a [itex]\sin{\theta}[/itex].

    I still don't get the second part. The solution puts in a cosine as well as a sine. If I understand why there's a sine in the equation for potential, I'll understand why it's here. However, I don't understand why there should be a cosine. The solution says that [tex]dE_z = dE \cos{\theta}[/tex], but I thought it would be [tex]dE_z = dE \sin{\theta}[/tex].
     
  5. Aug 11, 2006 #4

    mezarashi

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    The reason the sin is there is because of the way the integral is done. You are integrating over two variables, the azimuthal and the zenith coordinates. The sin is introduced when integrating the azimuthal variable. The radius is actually Rsin(zenith angle). You then add up these strings into a hemisphere through the second integration.

    [​IMG]
     
  6. Aug 11, 2006 #5
    Ok, I think I get it now. The only way I'll see if I get it is by doing more problems, but your explanation makes me understand what's going on. Thanks!
     
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