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I have a metal sphere with the net charge q. And I'm trying to calculate the force that southern hemisphere exerts to northern hemisphere... and I get 0.

now, the electrostatic "pressure" is

[tex]\mathbf f = \sigma \mathbf E = (q/4\pi R^2) (q/4\pi \epsilon_0 r^2) \mathbf {e_r}[/tex]

due to the symmetry, the force will point at z direction, so integrating only the z component of "pressure" over the northern hemisphere should do it, right?

[tex]\int f_z dA = \int_{-\pi/2}^{\pi/2} \int_0^{2\pi} (q/4\pi R^2) (q/4\pi \epsilon_0 R^2) Rcos(\theta) R^2 sin(\theta) d\phi d\theta = 0?[/tex]

I found out that this's a question from Griffiths', and the answer manual says the [itex]\theta[/itex] integral should be within [itex][0,\pi/2][/itex], not [itex][-\pi/2, \pi/2][/itex].

Why is that?

now, the electrostatic "pressure" is

[tex]\mathbf f = \sigma \mathbf E = (q/4\pi R^2) (q/4\pi \epsilon_0 r^2) \mathbf {e_r}[/tex]

due to the symmetry, the force will point at z direction, so integrating only the z component of "pressure" over the northern hemisphere should do it, right?

[tex]\int f_z dA = \int_{-\pi/2}^{\pi/2} \int_0^{2\pi} (q/4\pi R^2) (q/4\pi \epsilon_0 R^2) Rcos(\theta) R^2 sin(\theta) d\phi d\theta = 0?[/tex]

I found out that this's a question from Griffiths', and the answer manual says the [itex]\theta[/itex] integral should be within [itex][0,\pi/2][/itex], not [itex][-\pi/2, \pi/2][/itex].

Why is that?

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