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

1. Feb 24, 2005

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

but I got it wrong....

so, can anyone tell me how should I start?

2. Feb 24, 2005

### vincentchan

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....

3. Feb 27, 2005

### Kelvin

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

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

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

4. Feb 27, 2005

### 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).....

5. Feb 27, 2005

### Kelvin

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

6. Feb 28, 2005

### 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

7. Feb 28, 2005

### Kelvin

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