# Electric field on the surface of charged conducting sphere?

1. Apr 10, 2015

### mohamed el teir

just above the surface it's (kq/r^2) where r is the radius of the sphere and just below the surface it's zero, so is the electric field zero also exactly on the surface ? (as the q enclosed then will be zero since the flux is coming from the surface and not actually penetrating it)
and concerning the point charge, is the electric field also zero at the exact position of the point charge ?

2. Apr 10, 2015

### physichu

I think the answer to this question is hiding in another question:
How do you measure the the electric field on the surface?
How do you measure the electric field at the point of the chrage?

about the point charge i think that:
1. the formula (kq/r^2) is not corect when you get microscopicly close to the electron.
2. the notion - " the exact position of the point charge" is undefine.

3. Apr 11, 2015

### DaPi

Mohamed, you must realise that you are learning physics in an idealised world - sometimes called a model. In the real world conductors aren't spherical (they're bumpy due to the atoms making them up) and point charges don't exist. These are mathematical fictions that avoid having to deal with the tremendous complexity of real life (TM).
You can easily see one problem of a point charge by taking your charged sphere, calculating the field at the centre and at 2r from the centre, and letting r go to zero. You end up with two different values (one infinite, one zero) for the field at the same point r=0. It took several centuries for the charge of the electron to be understood - you'll need to have faith and be patient when the models reach their limits and can't provide an answer.

4. Apr 13, 2015

### vanhees71

The point is that a charged conducting surface in the stationary state carries a surface charge, and thus the elecric field's radial component makes a jump across the surface equalling the surface-charge density (charge per area) (modulo artificial factors if SI units are used). Inside the electric field is 0, outside along the surface it's $\sigma/\epsilon_0=Q/(4 \pi \epsilon_0 R^2)$, where $R$ is the radius of the sphere, and $\epsilon_0$ is the artificial conversion factor due to the use of SI units.