Net force exerted between hemispheres of uniformly charged sphere

[Kelvin]

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 \frac{1}{4 \pi \epsilon_0} \frac{3 Q^2}{16 R^2}]

my attempt:

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

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

but I got it wrong...

so, can anyone tell me how should I start?

 

[vincentchan]

regard two hemispheres as two point charges located at their center of mass, 3R/8 from the center.

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...

 

[Kelvin]

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




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

It's too complicated ...
any other suggestions?

 

[vincentchan]

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

 

[Kelvin]

I don't know tensor...
is there no other way to do this problem?

 

[vincentchan]

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

 

[Kelvin]

but this question is an exercise in chapter 2 of "Introduction to Electrodynamics" ...
how come I need tensor to solve the problem...