# Another Electrostatics Problem

1. Mar 6, 2005

### AKG

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

My book gives a formula for the electrostatic pressure:

$$P = \frac{\sigma ^2}{2\epsilon _0}$$

pushing the surface outwards (and hence pushing the northern hemisphere away from the southern one). I can find $\sigma$:

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

so I get:

$$P = \frac{Q^2}{32\pi ^2 \epsilon _0 R^4}$$

The force acting on one hemisphere will be the pressure times the area of that hemisphere, which is just $2\pi \R^2$, so I get:

$$F = \frac{Q^2}{16\pi \epsilon _0 R^2}$$

Is this right? The whole problem seems weird to begin with, I'm not sure if the numbers I'm using actually give me the quantity I'm looking for, and not just some other quantity related to the system. So is this right? Thanks.

2. Mar 7, 2005

### StatusX

That looks ok. I'm not sure what your confused about.

3. Mar 7, 2005

### AKG

Thanks. I was confused because the book doesn't talk about spherical objects (although it says that the formula for pressure applies to any conductor), but things like flat surfaces. In a flat surface, you would still see this force, but it wouldn't make sense to think of the force in terms of a repulsion from some other conducting surface, since there is none. I wasn't sure why in this case the force was being called the force of repulsion. The force acting on one hemisphere didn't seem to be the force of repulsion due to the other hemisphere, but rather just the same type of force that exists for "open" surfaces when there isn't even any other hemisphere to repel it.