Induced charge in a mobius strip I know that when a positive charge is placed at the center of a circular ring, the "inside" of the ring will have a negative induced charge and the "outside" will have a positive induced charge.
Consider the following mobius strip

If a positive charge is placed at the center, what will be the induced charge in this case as both "inside " and "outside" are basically , well, the same "side" ?
EDIT
I thought making a new question thread would be unnecessary so here's the edit - 
Okay, now I understand about the charge distribution , but another thing keeps bugging me-
If the outer side of the annulus was grounded, the charge from outer surface would get drained but the inner surface won't as it is bound by the  positive charge at the center. What will happen when the "outer" surface of the mobius strip  is grounded? 
 A: So in the first case, when talk about a plain circular ring, I assume you mean an annular ring, with a well defined inner radius and a different well defined out radius. With a positive charge at the center of the annular ring, positive charges will be repelled outward and negative charges attracted inward. Incidentally, not all the positive charge will go to the outside edge, since as more charge builds up there, it becomes energetically less favourable to move another positive charge there. In a 3-D conductor with 2-D surfaces, everything goes to one of the surfaces, but in a 2-D (or lower) conductor (with 1-D surfaces), there's not enough space on the surface, so you can have charge in the 'bulk' of the conductor, away from the edges, as long as the field everywhere in the conductor is zero. (References: [1] R. Friedberg, Am. J. Phys. 61, 1084 (1993), [2] D.J. Griffiths and Y. Li, Am. J. Phy. 64, 706, (1996))
With a mobius strip like the one in your illustration, Even if the concepts of inside and outside are not well defined, you can still talk about distance from the central point charge, and about electric field. Assuming that your mobius strip is made of a conductor, the charge will redistribute themselves until the electric field everywhere inside the material of the conductor is zero. For portions like the right hand portion of the strip in your picture, that means you will have a well defined separation of charge, because the local inner edge is definitely closer than the local outer edge. But on the left hand portion of the strip in your picture, both inner and outer edges are roughly the same distance from the center, so there will not be any separation of charge (there's no point to it, energetically). Being a 2-D conductor, not all the charge will have to be at the edge. So my rough impression is that you will have two 'circles/loops' of charge, one positive and one negative, which are well separated on the right, but converge and merge on the left.
A: Just answering the second question (since the first was well covered already):
Grounding works the same that it always does - it means there is no longer a constraint that the net charge on the ring is zero. Charges will distribute to minimize energy - if there is a negative charge in the center of the ring, negative charges will try to get as far away as possible. The amount of charge collecting on the "inner" surface (shortest distance to the charge in the middle) will try to cancel out that charge, but since there is no perfect spherical symmetry the charge distribution will not be exactly uniform. However I think that the net charge on ring + point will be zero as before.
A: I want to respond to your edited question.
Firstly, if your annular ring is a conductor, you don't ground a side, you ground the whole thing (because it is an equipotential surface), and then yes an amount of charge equal to the charge in the center flows away.
For any conductor, there is a charge distribution $\rho_1$ associated with it being charged but with no external charge.  And there is a charge distribution $\rho_2$ associated with it being neutral but with an external charge.  And finally, when there is an external charge and the conductor is grounded, you take the $\rho_2$ associated with the external charge and take the $\rho_1$ associated with placing an amount of charge equal (and opposite) to the external charge.  The sum $\rho_1+\rho_2$ is the charge distribution of the grounded conductor with an external charge. 
As for the Möbius strip there is no such thing as an "outer surface" on a Möbius strip, there is a just a surface. But luckily for us you don't ground surfaces, you ground conductors.  If you ground the whole Möbius strip, an amount of charge equal to the charge in the center will flow away. Like all conductors the charge distribution is the sum of the charge for a neutral isolated conductor plus as if you'd charged the Möbius strip in isolation due to the amount of charge equal and opposite to the external charge.
