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

1. Jan 31, 2006

siddharth

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?

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

$$df = \sigma da E$$

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

$$E = \frac{1}{2} (\frac{Q}{4 \pi \epsilon_0 R^2} + 0)$$
$$\vec{E} = \frac{Q}{8\pi \epsilon_0 R^2} \vec{e_r}$$

and

$$\sigma = \frac{Q}{4 \pi R^2}$$

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

Is my approach correct?

Last edited: Jan 31, 2006
2. Jan 31, 2006

Meir Achuz

Yes, but you should follow Tesla's advice and recognize from the beginning that the x and y components of F vanish due to the axial symmetry.

3. Jan 31, 2006

siddharth

Thanks Meir Achuz. That would have been my next step. Stating that F_x and F_y are zero by symmetry.