Net force exerted between hemispheres of uniformly charged sphere

AI Thread Summary
The discussion focuses on calculating the net force exerted between the hemispheres of a uniformly charged sphere. The correct approach involves integrating the electric potential and using the electric field to determine the force, rather than treating the hemispheres as point charges. Participants emphasize the complexity of the integrals involved and suggest using the Maxwell stress tensor for a more straightforward solution. The model answer provided is \(\frac{1}{4 \pi \epsilon_0} \frac{3 Q^2}{16 R^2}\), highlighting the need for a rigorous mathematical approach. Overall, the conversation underscores the challenges of solving this problem and the necessity of advanced techniques in electrodynamics.
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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?
 
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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...
 
<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?
 
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
 
I don't know tensor...
is there no other way to do this problem?
 
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
 
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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