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Homework Help: [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


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    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
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