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Homework Help: Net force exerted between hemispheres of uniformly charged sphere

  1. Feb 24, 2005 #1
    Find the net force that the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere. Express your answer in terms of the radius R and the total charge Q
    [the "model" answer is [tex]\frac{1}{4 \pi \epsilon_0} \frac{3 Q^2}{16 R^2}[/tex]]

    my attempt:

    regard two hemispheres as two point charges located at their center of mass, [tex]\frac{3 R}{8}[/tex] from the center.

    so
    [tex]

    F = \frac{1}
    {{4\pi \varepsilon _0 }}\frac{{\left( {Q/2} \right)^2 }}
    {{\left( {2 \times \frac{3}
    {8}R} \right)^2 }} = \frac{1}
    {{4\pi \varepsilon _0 }}\frac{4}
    {9}\frac{{Q^2 }}
    {{R^2 }}

    [/tex]

    but I got it wrong....

    so, can anyone tell me how should I start?
     
  2. jcsd
  3. Feb 24, 2005 #2
    this is not valid... this only work for a solid or hollow sphere...

    Don't be lazy, you need integral in this problem... and a ungly one... show me some of your thought....
     
  4. Feb 27, 2005 #3
    [tex]

    \[
    \begin{gathered}
    {\text{OK, so I}}{\text{ try to start with potential, which is easier, then obtain the electric field by}} \hfill \\
    E = - \nabla V \hfill \\
    {\text{and finally }}F = qE.{\text{ So the potential at the point }}\left( {{\text{x}}_{\text{0}} ,y_0 ,z_0 } \right){\text{ due to the south hemisphere is}} \hfill \\
    V = \frac{\rho }
    {{4\pi \varepsilon _0 }}\int_0^{ - R} {\int_{ - \sqrt {R^2 - z^2 } }^{\sqrt {R^2 - z^2 } } {\int_{ - \sqrt {R^2 - z^2 - y^2 } }^{\sqrt {R^2 - z^2 - y^2 } } {\frac{{dxdydz}}
    {{\sqrt {\left( {x - x_0 } \right)^2 + \left( {y - y_0 } \right)^2 + \left( {z - z_0 } \right)^2 } }}} } } \hfill \\
    {\text{using cylindrical coordinates,}} \hfill \\
    V = \frac{\rho }
    {{4\pi \varepsilon _0 }}\int_0^{ - R} {\int_0^{2\pi } {\int_0^{\sqrt {R^2 - z^2 } } {\frac{1}
    {{\sqrt {\left( {r\cos \theta - x_0 } \right)^2 + \left( {r\cos \theta - y_0 } \right)^2 + \left( {z - z_0 } \right)^2 } }}rdrd\theta dz} } } \hfill \\
    \left( {r\cos \theta - x_0 } \right)^2 + \left( {r\cos \theta - y_0 } \right)^2 + \left( {z - z_0 } \right)^2 \hfill \\
    = r^2 + z^2 + r_0 ^2 + z_0 ^2 - 2r\left( {x_0 \cos \theta + y_0 \sin \theta } \right) - 2zz_0 \hfill \\
    {\text{put }}a = z^2 + r_0 ^2 + z_0 ^2 - 2zz_0 ,{\text{ }}b = x_0 \cos \theta + y_0 \sin \theta \hfill \\
    {\text{then}} \hfill \\
    \left( {r\cos \theta - x_0 } \right)^2 + \left( {r\cos \theta - y_0 } \right)^2 + \left( {z - z_0 } \right)^2 \hfill \\
    = r^2 - 2br + a \hfill \\
    = \left( {r - b} \right)^2 + a - b^2 \hfill \\
    \end{gathered}
    \]


    [/tex]




    [tex]
    \[
    \begin{gathered}
    {\text{so}} \hfill \\
    \[
    \begin{gathered}
    V = \frac{\rho }
    {{4\pi \varepsilon _0 }}\int_0^{ - R} {\int_0^{2\pi } {\int_0^{\sqrt {R^2 - z^2 } } {\frac{{rdrd\theta dz}}
    {{\sqrt {\left( {r\cos \theta - x_0 } \right)^2 + \left( {r\sin \theta - y_0 } \right)^2 + \left( {z - z_0 } \right)^2 } }}} } } \hfill \\
    = \frac{\rho }
    {{4\pi \varepsilon _0 }}\int_0^{ - R} {\int_0^{2\pi } {\int_0^{\sqrt {R^2 - z^2 } } {\frac{{rdr}}
    {{\sqrt {\left( {r - b} \right)^2 + a - b^2 } }}} } } d\theta dz \hfill \\
    = \frac{\rho }
    {{4\pi \varepsilon _0 }}\int_0^{ - R} {\int_0^{2\pi } {\left( { - b\ln \left( {\sqrt a - b} \right) + b\ln \left( { - b + \sqrt {R^2 - z^2 } + \sqrt {R^2 - z^2 + a - 2b\sqrt {R^2 - z^2 } } } \right) - \sqrt a + \sqrt {R^2 - z^2 + a - 2b\sqrt {R^2 - z^2 } } } \right)} dz} \hfill \\
    \end{gathered}
    \]

    \end{gathered}
    \]
    [/tex]

    It's too complicated ........
    any other suggestions?
     
  5. Feb 27, 2005 #4
    use maxwell stress tensor to approach this problem.... using potential field will lead you to a mess (sorry, forgot to warn you about that).....
    show me some of your work so that i can further help you
     
  6. Feb 27, 2005 #5
    I don't know tensor......
    is there no other way to do this problem?
     
  7. Feb 28, 2005 #6
    sorry, as far as I know, solving hard integral and maxwell stress tensor are the only way doing this problem, if you have a simpiler method, pls let me know
     
  8. Feb 28, 2005 #7
    but this question is an exercise in chapter 2 of "Introduction to Electrodynamics" ........
    how come I need tensor to solve the problem......
     
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