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Hard Electrostat Prob

  1. Feb 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A hemisphere of radius a centered at the origin carriers a total charge Q uniformly distributed over its curved surface. Find the e-field at a point h on the z-axis.

    2. Relevant equations
    [tex]d\vec{E} = \frac{1}{4\pi\epsilon} \frac{dq}{R^2}\hat{R}[/tex]

    3. The attempt at a solution
    Can someone please tell me if this answer is correct (my professor and I both had a real hard time figuring this one out):

    [tex]\vec{E}(0,0,h) = \frac{Q}{4\pi\epsilon}\Big(\frac{a}{h^2\sqrt{h^2+a^2}} + \frac{1}{h^2}\Big)\hat{z}[/tex]
    Last edited: Feb 18, 2007
  2. jcsd
  3. Feb 18, 2007 #2


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    I don't think that's right.

    One way of doing this, is to first calculate [itex]V(z)[/tex], and then get [itex]\vec{E}[/itex] from that.
    Last edited: Feb 18, 2007
  4. Feb 18, 2007 #3

    Tha Gauss law should be the simplest way.
    The equation you mention for calculating the electric field is a consequence of the Gauss law.
    You can use you "relevant equation" too. But in this case, try to simplify the calculations by using the symmetries of your problem.

  5. Feb 18, 2007 #4


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    The Gauss law won't be useful to obtain the field in this question.

    Since the object is a hemisphere, you won't be able to construct a gaussian surface which will give you the Electric field. There's no symmetry.
  6. Feb 18, 2007 #5
    Here is another answer I got using the principle of superposition:

    [tex]\vec{E}(0,0,h) = \frac{Q}{4\pi\epsilon h^2}\Big(\frac{1}{\sqrt{a^2+h^2}} - \frac{1}{\sqrt{a^2-2ha+h^2}} + \frac{h}{a\sqrt{a^2-2ha+h^2}}\Big) \hat{z}[/tex]

    Is this closer to the right answer?
    Last edited: Feb 18, 2007
  7. Feb 18, 2007 #6


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    As siddharth mentioned it is best to obtain the potential at a point first. Do you know the general equation for obtaining the potential of a charge distribution?
  8. Feb 18, 2007 #7
    Using the potential method, I got the following equation for the electric field:

    [tex]\vec{E}(0,0,h) = \frac{Q}{4\pi\epsilon}\Big(\frac{h^3-a^3}{a^2 h^2\sqrt{h^2+a^2}}\Big)\hat{z}[/tex]

    How is this?
    Last edited: Feb 18, 2007
  9. Feb 18, 2007 #8

    Sorry, I read you question too fast and did not notice the "hemi"-sphere.
    I calculated the potential for the hemisphere.
    For a charge density of 1, and droping the permitivity, this gives:

    V = (r^2 + z^2)^(1/2)/(2*z) - (r^2 - 2*r*z + z^2)^(1/2)/(2*z)

    and the electric field is a longer expression:

    E = -(r - z)/(2*((r - z)^2)^(1/2)*z) - ((-r + z)^2)^(1/2)/(2*z^2)
    - 1/(2*(r^2 + z^2)^(1/2)) + (r^2 + z^2)^(1/2)/(2*z^2)

    In the above expressions, you have to be careful if taking the square roots.
    I did the same exercice for the full-sphere, and I got also similar square roots. Taking the correct sign is essential to get the right solution inside or outside the sphere. For the hemisphere this should be also taken care for.

    Last edited: Feb 18, 2007
  10. Feb 18, 2007 #9
    BTW, would you get a different answer if you assume that h < a?
  11. Feb 18, 2007 #10
    Here is another solution I got by redoing the surface integral for the E-field very carefully. This time I am pretty confident on my answer. I have checked and double checked everything.

    [tex]\vec{E}(0,0,h) = \frac{Q}{4\pi\epsilon h^2}\Big(\frac{h-a}{|h-a|} + \frac{a}{\sqrt{a^2+h^2}} \Big)\hat{z}[/tex]

    Anyone else agrees?
    Last edited: Feb 18, 2007
  12. Feb 18, 2007 #11
    Actually, that is the right answer (but only for h > a). I got the same answer both by doing the surface integral of the E-field and also by calculating the potential and then finding the E-field.
  13. Feb 19, 2007 #12


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    I don't think so.

    Can you post the actual steps instead of just the final answer? It'll be easier to spot the errors that way, if any.

    The form of this solution seems correct.
  14. Feb 19, 2007 #13

    The potential is given by the following integral (Mathematica notations):

    Integrate[(2 Pi r Sin[teta]) / (4 Pi Sqrt[r^2Sin[teta]^2 + (r cos[teta]-z)^2]),{teta,0,Pi/2}]​
    The numerator (2 Pi r Sin[teta])*d(teta) is the elementary charge from a circle on the sphere.
    (for a charge density of 1, and dropping the eps factor)

    Going to the Mathematica integrator site you can check the following result:
    (but it is not very difficult with paper and pencil)

    Integrate[(2 Pi r Sin[x]) / (4 Pi Sqrt[r^2 Sin[x]^2 + (r Cos[x]-z)^2]),x] = ...​

    From this you can get the potential for a sphere or for an hemisphere:

    VSphere = ... ​

    VHemisphere = ... ​

    You get the end result by introducing the correct charge density and eps.

    Last edited: Feb 19, 2007
  15. Feb 19, 2007 #14


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    Michel, my question was directed to Swapnil, and I apologize for not stating that clearly.

    Your method is indeed correct. But, could you modify the post so as to not give away the answer? I'd prefer it if Swapnil could figure it out on his/her own.
  16. Feb 19, 2007 #15
    Sorry siddharth,

    I saw many answers and I did not realize this was the homework section.
    (I see the PF through an RSS feed and didn't look further than the question asked)
    I hope anyway that I motivated Swapnil and others.
    I really think that using Mathematica and other software can boost students,
    even if paper and pencil is absolutely needed to get all the details.

  17. Feb 19, 2007 #16
    I think that michel and I are on the same track(and so are you siddharth since you agree with michel's answer). My answer is just simplified further than michel's. Notice that

    [tex]\sqrt{r^2 - 2rz + z^2} = \sqrt{(r - z)^2} = |r - z|. [/tex]

    If you realize this fact and then take the derivative of the above expression for the potential with respect to z, you will get an answer which has the same form as the answer I originally mentioned. I would have posted all the steps myself but with my typing speed, it'll take me hours.
    Last edited: Feb 19, 2007
  18. Feb 20, 2007 #17
    I tried the question under the assumption that you're only looking for the z component of the field and here's my answer:

    <equation removed>

    I just plain integrated it dividing it into small half rings if infinitesmal width.
    Last edited: Feb 20, 2007
  19. Feb 20, 2007 #18
    This doesn't make sense. How can the e-field be proportional to [tex]\frac{1}{h}\ln\Big(\frac{1}{h}}\Big)[/tex]? It should more or less follow an inverse square realtionship.
  20. Feb 20, 2007 #19
    Oops sorry I made a 'slight' mistake in the calculation. I've some doubts:

    In what direction is the hemisphere oriented? Besides the mistake I made in my previous answer I also assumed that the hemisphere had its circular face lying on the XZ plane. That is the say that its axis of symmetry was assumed to be coincident with the y axis. Is this the case? Or is the hemisphere having its axis as Z-axis?
  21. Feb 21, 2007 #20
    The axis of symmetry is the z-axis in my OP.
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