# Force of repulsion between two hemispheres of a metal conductor

1. Aug 26, 2014

### ELB27

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
The force per unit area on the surface of a conductor:
$$\vec{f} = \frac{1}{2\epsilon_0}\sigma^2\hat{n}$$
where $\hat{n}$ is the unit vector perpendicular to the surface.
Electrostatic pressure (if I understand it right - just the magnitude of the force per unit area expressed in terms of $E$ just outside the surface):
$$P = \frac{\epsilon_0}{2}E^2$$

3. The attempt at a solution
Since it's a sphere the charge will be evenly distributed, thus the surface charge density is: $$\sigma = \frac{Q}{4\pi R^2}$$ Substituting into the equation for the force per unit area:
$$\vec{f} = \frac{Q^2}{32\epsilon_0\pi^2R^4}\hat{r}$$ Now to find the force only on one hemisphere I multiply by the surface area of it: $$\vec{F} = \vec{f}2\pi R^2 = \frac{Q^2}{16\epsilon_0\pi R^2}\hat{r}$$ This approach sounds logical to me, but after searching through google to check my answer (including this forum - it was asked 3 times) I find 2 different approaches. One is the same as mine but another one asserts that the above result for $\vec{F}$ needs to be integrated over the hemisphere and take the $z$ component only. (see this and this ; My approach is suggested here)
While the integration approach seems to dominate (2 out of 3 results in this forum) I don't see why the need for it. My only thought is that if I take the hemisphere as a whole, the x and y components cancel, but I don't understand why my approach doesn't account for that.
I would appreciate if someone can elaborate on this.

Edit: I think I know what's the problem. The question is asking for the force acting on the hemisphere as a whole. In this case the net force will be in the z direction and I need to integrate the z component of the result above. However, I found the repulsion force at every point on the hemisphere. Any thoughts on this?

Last edited: Aug 26, 2014
2. Aug 26, 2014

### Simon Bridge

A $\hat n$ turned into a $\hat r$ ... how did you do that?

I agree - the question wants the net force between the hemispheres.
The repulsion at every point on the hemisphere is like air pressure on a parachute - each surface element is pushed away from the center, that's why the charge distribution is uniform. Summed over a single hemisphere, the (cylindrical) radial components cancel, leaving the axial components.

Have you considered a brute force approach - starting with a charge element on the north hemisphere and working out the force on it due to each charge element on the southern hemisphere, then adding all those forces up ... i.e. integrating over the surface of the southern hemisphere. Then you ca integrate over the surface of the northern hemisphere to get the net force on the north due to the south?

3. Aug 26, 2014

### ELB27

Thank you for your reply. $\hat{n}$ is the unit normal vector to the surface charge. In the case of a sphere centered at the origin, $\hat{r}$ is playing this role for every point on the sphere. As for the problem itself, to find the net z-force I think that all that's required now is to take the z-component of my result:
$$\vec{f}_z = \frac{Q^2}{32\epsilon_0\pi^2 R^4}\cos\theta\hat{z}$$
...and integrate over northern hemisphere:
$$\vec{F} = \hat{z}\int_0^{2\pi} \int_0^{\pi/2} \frac{Q^2}{32\epsilon_0\pi^2 R^4}\cos\theta\sin\theta d\theta d\phi = \frac{Q^2}{32\epsilon_0\pi^2 R^4}\frac{1}{2}2\pi \hat{z} = \frac{Q^2}{32\epsilon_0\pi R^2}\hat{z}$$

This is also the result I get from the brute-force method and what the other people I quoted seem to get. Looks like that's it (still, that's a weird problem in my opinion).
Thanks again!

4. Aug 26, 2014

### Delta²

Can i ask how one derives the equation $\vec{f}=\frac{\sigma^2}{2\epsilon_0}n$

5. Aug 26, 2014

### ELB27

There is a derivation of this equation quoted from Griffiths' 3rd edition of "Introduction to electrodynamics" here.

6. Aug 27, 2014

### Simon Bridge

So long as you are sure you've accounted for how each surface element contributes a different amount to the z-component. I haven't checked myself - it's your work after all.

7. Aug 27, 2014

### ELB27

The $\cos\theta$ accounts for that as far as I understand - the force that I found in the OP is constant and is the force that acts on every point on the hemisphere in the $\hat{r}$ direction. To find the contribution of each point in the Z direction I multiply by $\cos\theta$ and integrate. Sounds logical to me - the force in the r direction is constant but the Z component is not because of different angle with the vertical.

8. Aug 27, 2014

### Delta²

I ve looked at that link, i am not sure but i think some of the force you calculated in the z direction, is due to some parts of the northern hemisphere affecting other parts of the same hemisphere. Because in the exrpession for the force per unit area, if i udnerstand correctly, it is the force that is exerted on an infinetsimal area, by the rest surface area of the conductor, not only by the southern hemisphere.

9. Aug 27, 2014

### Delta²

Ah ok i see now my mistake, due to Newton's 3rd law, the total force a hemishpere exerts to itself is always zero.