Question: A metal sphere of radius R carries a total charge Q. What is the force of repulsion between the "northen" hemisphere and the "southern" hemisphere?(adsbygoogle = window.adsbygoogle || []).push({});

The force for a small area 'da' on the outer surface should be

[tex] df = \sigma da E [/tex]

What E to use? My guess is that it should be averaged over both sides of the surface.

So,

[tex]E = \frac{1}{2} (\frac{Q}{4 \pi \epsilon_0 R^2} + 0)[/tex]

[tex] \vec{E} = \frac{Q}{8\pi \epsilon_0 R^2} \vec{e_r} [/tex]

and

[tex] \sigma = \frac{Q}{4 \pi R^2} [/tex]

And the answer should be

[tex] \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \frac{Q^2 \sin\theta}{32 \pi^2 \epsilon_0 R^2}(\sin\theta \cos\phi \vec{e_x} + \sin\theta \sin\phi \vec{e_y} + \cos\theta \vec{e_z}) d\theta d\phi [/tex]

Is my approach correct?

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# Homework Help: Question: A metal sphere of radius R carries a total charge Q

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