An interesting problem in electrostatics.

1. Mar 26, 2009

the-genius

There is a cunducting hollow sphere. We know if we charge it all the charges will stay at the outer surface. But if we drill a hole of radius r(say) then what fraction of the charge will remain inside and outside surface

I think it is logical to think that less charge will stay in the inner surface than the outer surface and that more charge will stay inside if the size of hole increases.

I attempted to find the solution using Gauss law, by assuming the condition as that of a complete charged sphere with a oppositly charged sphericall circular plane overlayed on it to create hole of charge at that place. But since the electric field won't be uniform, I failed.

2. Mar 26, 2009

rl.bhat

In a conducting hollow sphere all the charges will stay at the outer surface, because due to repulsive force they move as far away as possible. Therefore only in charged conducting sheet, charge will be distributed uniformly on both sides. In any curved surface they will be only on the outer surface.

3. Mar 26, 2009

the-genius

well, rl.bhat, I don't think if you very slightly curve a charged plane sheet of charge, the charge on one surface will now totally move to the other. Do you have a proof that it does.?

4. Mar 26, 2009

rl.bhat

Even in a slightly curved surface, the largest distance will be between the two point on the outer surface only. You can draw the diagram and verify.

5. Mar 26, 2009

the-genius

Usually it is proved that no charge resides on the inner surface of the spherical shell by drawing a spherical gausian surface that lies withing the cunductor and arguing that there can be no electric field inside a cunductor (otherwise the cunductor won't be in electrostatic equilibrim as the electrons inside the cunductors will move, creating currents) so no electric field passes through the gausian surface, hence by applying
Interation(E.dA)=q / epsilon
the charge inside the gaussian surface is proved to be 0.
But now if we drill a hole in the spherical shell, Electric field needn't be 0 at the gausian surface at the region of hole. Hence, Interation(E.dA) needn't be 0, so is the charge q.

6. Mar 26, 2009

the-genius

Is there any reason that all the charges should crowed to one surface for a slightly curved plane cunucting sheet?

7. Mar 26, 2009

Count Iblis

Yes, there will be a charge in the inner surface in this case. If you use Gauss' law and make some reasonable estimates assuming that the hole is small and that the hollow sphere is a thin shell, so that the inner surface is colose to the poiuter surface, then you find:

Q_in = Q_t A_h/[2A_t - A_h]

Here Q_in is the charge on the inner surface, Q_t the total charge, A_h the area of the hole and A_t the outer surface area.

Note that the hole has to be present also in the inner surface.

8. Mar 26, 2009

the-genius

How did you derive the result, Count Iblis? Why should the hole be small? Pease explain.

9. Mar 26, 2009

Count Iblis

It is bedtime for me now, so I'll be brief. Just think about what happens if you integrate E dot dS over a surface that is inside the sphere. Because of the hole you can't say that E is zero everywhere. If you deform the surface you ae integrating on by letting it move outside the sphere when it is at the hole, then you know that near the hole all that is changed is that the surface charge density is zero instead of sigma (note that we don't know yet what the value of sigma is).

Now, you can use the superposition principle by writing the surface charge density as sigma everywhere and that at the hole you have an extra surface charge density of minus sigma. Then just above the hole you have the same electric field as just outside any other point on the sphere plus the electric field due to a surface charge of minus sigma.

This then means that the electric field just above the hole would be

E = sigma/epsilon_0 - sigma/(2epsilon_0 ) = sigma/(2 epsilon_0)

Just inside the hole, the electric field of the fictitional minus sigma charge points in the opposite direction, so for positive sigma it points in the normal direction. This has to be added to the electric field of a sphere woith surface charge sigma without a hole. But inside the spere that electric field would be zero, so we have:

E = 0 + sigma/(2epsilon_0 ) = sigma/(2 epsilon_0)

This is the same as above. Of course, this had to be the case as moving into the hole shouldn't lead to a discontinuity in the field.

If you now evaluate the integral of E dot dS inside the sphere, you get a contribution from the part where the hole is of:

A_h sigma/(2 epsilon_0)

where A_h is the area of the hole.

This then has to be the total charge divided by epsilon_0 that is enclosed by the surface, which must be the interior surface.