• Support PF! Buy your school textbooks, materials and every day products Here!

Net force exerted between hemispheres of uniformly charged sphere

  • Thread starter Kelvin
  • Start date
52
0
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?
 
609
0
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....
 
52
0
[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?
 
609
0
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
 
52
0
I don't know tensor......
is there no other way to do this problem?
 
609
0
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
 
52
0
but this question is an exercise in chapter 2 of "Introduction to Electrodynamics" ........
how come I need tensor to solve the problem......
 

Related Threads for: Net force exerted between hemispheres of uniformly charged sphere

  • Last Post
Replies
4
Views
486
Replies
1
Views
8K
  • Last Post
Replies
8
Views
12K
Replies
2
Views
7K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
8K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
3
Views
1K
Top