1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

[Calc 3] Gauss' Law for Elecricity over a hemisphere

  1. Dec 7, 2008 #1

    Thank you very much for takign the time to help me out. This is an applied calculus 3 (multivariable calc) level problem. I have an answer, but it is nonsensical as the units are not in accordance with units of charge, C.

    1. The problem statement, all variables and given/known data
    Instructions: Use Gauss' Law for electricity and the relationship [tex]q=\int\int\int_{Q} \rho dV[/tex].

    Question: For [tex]E=\left\langle 2xz^{2},2yx^2,2zy^2\right\rangle[/tex], find the total charge in the hemisphere [tex]z=\sqrt{R^{2}-x^{2}-y^{2}[/tex].

    2. Relevant equations
    [tex]\nabla \bullet E=\rho/\epsilon_{o}[/tex]

    3. The attempt at a solution
    Finding the divergence of E:
    [tex]\nabla \bullet E = \frac{\partial 2xz^{2}}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}[/tex]
    Subsituting [tex]z=\sqrt{R^{2}-x^{2}-y^{2}[/tex] in the partial of x as z is a function of x:
    [tex]= \frac{\partial 2x(R^{2}-x^{2}-y^{2})}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}[/tex]
    [tex] \nabla \bullet E = 2(R^{2}-2x^{2}) [/tex]

    Now, this means that [tex]\rho = 2 \epsilon_o (R^{2}-2x^{2})[/tex]

    Using the relation [tex]q=\int\int\int_{Q} \rho dV[/tex]:
    and setting up the region of integration:
    [tex]q = 2 \epsilon_o \int^{R}_{-R} \int^{\sqrt{R^{2}-x^{2}}}_{-\sqrt{R^{2}-x^{2}}} \int^{\sqrt{R^{2}-x^{2}-y^{2}}}_{0}(R^{2}-2^{2}) dz dy dx[/tex]

    Seeing spherical symmetry, a conversion from rectangular to spherical coordinates will be convenient. *NOTE: My book and teacher use phi to indicate the angle from the z-axis and theta as the angle from the x-axis; like this: http://mathworld.wolfram.com/SphericalCoordinates.html * :
    [tex]R^{2} - x^{2} = \rho^{2} - (\rho sin\phi cos\theta)^{2} [/tex]

    Switching limits of integration and substituting [tex]\rho dV[/tex]

    [tex]q = 2 \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} \int^{R}_{0} \rho^{2} sin\phi (\rho^{2} - (\rho sin\phi cos\theta)^{2}) d\rho d\phi d\theta [/tex]
    Distributing and integrating [tex]d\rhp [/tex]
    [tex]q = 2 \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} [\rho^{5}/5sin\phi - 2\rho^{5}/5sin^{3}\phi cos^{2}\theta]|^{R}_{0} d\phi d\theta [/tex]

    After integrating wrt [tex]d\rho[/tex] and moving R^5 outside of the iterated integrals, the iterated integral remains as (Notice the weird R^5 units!):
    [tex]q = 2/5 R^{5} \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} [sin\phi - 2sin^{3}\phi cos^{2}\theta] d\phi d\theta [/tex]

    Once all the trigonometric integrations are done (which is just as long as the work up to this point), my finals answer is
    [tex]\frac{4}{15} \pi R^{5} \epsilon_o[/tex]

    The units of this answer are [tex][R^{5}][\epsilon_o] = m^{5} * \frac{C^{2}}{Nm^{2}}\neq C[/tex]

    Im sure there is an error I made as this is a law. However, my teacher was unable to find my error after visiting him in his office. I have been thinking about this problem for a few days now and ...well... Im still without a sensical answer. Any help?
    Last edited: Dec 7, 2008
  2. jcsd
  3. Dec 8, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    You should take note here that for E to represent an actual electric field, it must have units of N/C. The only way this can happen is if you assume (sort of hand-wavingly) that there is some constant [tex]k=1 \frac{\text{N}}{\text{m}^3\text{C}}[/tex] multiplying this expression. If you were to assume this, then when you found your answer to be proportional to [tex]\epsilon_0 R^5[/tex], you would have that it is actually proportional to [tex]k\epsilon_0 R^5[/tex] which would give you the correct units of charge. However, since this is a math course and not a physics course I doubt they expect you to correct the units of E in this manner.

    There is a mistake on your part though:

    When you do this substitution, you are effectively assuming that [tex]z=\sqrt{R^{2}-x^{2}-y^{2}[/tex] everywhere in the hemisphere; but that is only true on the surface of the hemisphere. At a general point in the hemisphere, z is independent of x and y, so [tex]\frac{\partial }{\partial x} (2xz^2)=2z^2[/tex] and the divergence of E becomes [tex]2z^2+2x^2+2y^2=2\rho^2[/tex].
    Last edited: Dec 8, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: [Calc 3] Gauss' Law for Elecricity over a hemisphere