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A metal sphere of radius $R$ carries a total charge $Q$. What is the force of repulsion between the northern hemisphere and southern hemisphere?

In this problem you consider the force per unit area on the surface of the northern hemisphere and then add it up basically: You get the average electric field as$$ \frac{1}{2}(E_{\text{outside}} + E_{\text{inside}}) ~=~ {E_{\text{average}}} ~=~ \frac{1}{2}\frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2}\hat{r} \,,$$because of the symmetry the horizontal components cancel for the force per unit area you get $f_{z} = {\sigma \left(E_{\text{average}}\right)}_{z}$ where $\sigma = \frac{Q}{4 \pi R^2}$. Then the total force on the northern hemisphere is$$ F_{z} ~=~ \int f_{z} \, \mathrm{d}a ~=~ \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \left(\frac{Q}{4 \pi R^2}\right)\frac{1}{2}\left(\frac{Q}{4 \pi \epsilon_0 R^2}\right) \cos{\theta} \, R^2 \sin{\theta} \, \mathrm{d} \theta \, \mathrm{d} \phi ~=~ \frac{Q^2}{32 \pi R^2 \epsilon_0}. $$

Question: Are the forces of repulsion on the northern hemisphere a result of only the charges from the southern hemisphere while there is symmetric cancellation from charges in the northern hemisphere, or is the force on each unit area in the northern hemisphere due to the whole sphere?

The way the question is worded seems to suggest that the answer is the first case, i.e. that the force on the northern hemisphere is due completely to the southern hemisphere.

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  • $\begingroup$ Possible duplicate : Find the net force the southern hemisphere of a uniformly charged sphere exerts on the northern hemisphere: physics.stackexchange.com/q/23071 $\endgroup$ – sammy gerbil Jul 29 '16 at 13:46
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I recognize that problem from Griffiths's Introduction to Electrodynamics. The net force on the northern hemisphere is due entirely to the southern hemisphere but that's not to say that there are no forces that parts of the northern hemisphere exert on other parts of the northern hemisphere, there are such forces. If you take away the southern hemisphere, the northern hemisphere isn't going to propel itself anywhere---all of its forces cancel each other out as you should know from introductory physics.

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