Uku said:
Calculate the force dF that each surface element dS of charge dQ experiences
That is the part with the insane integral... If I put myself in perspective of the charge, looking inside the sphere, I have to scan trough all the dS'es contained in S=4\pi r^{2}, because they all affect me. That is one intense integral.
I guess you can find this argument in most classical electrodynamics textbook, but let me introduce it to you (again). No need for an insane integral.
See the attached picture. Here I consider a charge element dQ. Assume that
the conductor is in electrostatic equilibrium, then the total E-field outside is E_{out}=\sigma / \epsilon_0 and the one inside is 0.
Now thanks to the principle of superposition, we can split the total E-field into 2 parts: one due to dQ only E_{dQ}, the other due to the rest (dQ excluded) E_{other}.
For E_{dQ}, it's just like E-field due to an infinite uniformly charged plane. That is because of this assumption: we consider that the width of the charge layer is PHYSICALLY 0, i.e. even if how small dS is, dS is still a lot larger than that width. This assumption is the basic assumption in many continuous charge distribution model; otherwise, we would have to zoom it at the microscopic scale, and deal with a chain of separate electrons and atoms, which is crazy and unnecessarily complicated. Therefore, E_{dQ}=\sigma / 2\epsilon_0. Notice that E_{dQ} has different directions on different side of dQ.
For E_{other}, we don't have to calculate it now. But one thing we can be sure about is that it is CONTINUOUS at the position of dQ. This is due to an important characteristics of E-field: the E-field due to a distribution of charge is only discontinuous when passing through charges. Here the distribution of charge excludes dQ, and we only consider the E-field due to this distribution (i.e. ignore the existence of dQ), so passing through the position of dQ, there is no charges and thus E_{other} is continuous, which corresponds to the unchanged direction of E_{other} at the position of dQ.
So from the conclusions about the directions, we have:
OUTSIDE: E_{out} = E_{dQ} + E_{other}, so: E_{other} = E_{out} - E_{dQ} = \sigma / 2\epsilon_0
INSIDE: E_{in} = E_{dQ} - E_{other}, and we can see that the above result E_{other} = \sigma / 2\epsilon_0 is consistent with the fact that E_{in} = 0.
We know that the force on dQ is due to E-field of the rest: dF = E_{other}dQ. So now do you need a crazy integral?
granpa said:
I see your point now about the 'caution part'.
if the mass is spread out in a perfectly 2d layer of zero width
then the gravity goes from g at the surface to zero in the interior an infinitesimal distance away.
So you take one half.
(just imagine that the width is arbitrarily small but non-zero and imagine how the field changes over the width)
Theres a theorem somewhere about that but I can't think of the name of it.
http://en.wikipedia.org/wiki/Heaviside_step_function#Zero_argument
The Heaviside step function, using the half-maximum convention
I don't think this works for gravity (the "changing from g to zero" part), because the gravitational field (magnitude) is the same on both sides of the surface (I'm not sure if I get your point correctly). But this can be applied to the surface charge of a conductor in electrostatic equilibrium. Anyway the argument for taking 1/2 is applicable to both cases.
Applying this argument to my proof above, we don't even need to find E_{dQ}. We only need to know that
due to symmetry of the E-field, E_{dQ} has the same magnitude on both sides, and of course, opposite directions (this is also due to symmetry!). Therefore:
E_{out} = E_{other} + E_{dQ}
E_{in} = E_{other} - E_{dQ}
Summing the 2 equations side by side, we obtain:
E_{other} = \frac{E_{out}+E_{in}}{2}
From here, we can deduce the expected result of E_{other}.@Quinzio: No comment. I don't like dealing with crazily crazy integral. Good luck
