How Does a Spherical Shell Affect Forces Between Charges?

kihr
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Homework Statement



A thin metallic spherical shell contains a charge Q on it. A point charge q is placed at the centre of the shell and another charge q1 is placed outside it. All the three charges are positive. Find (a) the direction of the force on the charge at the centre, and (b) the direction of the force on the charge at the centre only due to the shell.



Homework Equations



Gauss's Law

The Attempt at a Solution



(a) The electric field inside the shell due to the external charge q1 is zero as the flux crossing the sphere due to this charge is zero.
The electric field inside the shell due to the charge on the shell is zero.
The force on the charge q would be the product of q and the electric field in which this charge is located (which is zero in this case). hence the force on the central charge is zero.

(b) By the logic applied in the case of (a), the force should be zero. However, the answer is "towards right". I need some help to understand this, as it is likely that the logic applied to solve (a) may not be entirely correct. Thanks.
 
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kihr said:
(a) The electric field inside the shell due to the external charge q1 is zero as the flux crossing the sphere due to this charge is zero.
It's correct that the net flux crossing your sphere due to a charge outside the sphere is zero. But that does not mean that the electric field due to that charge is zero.
kihr said:
The electric field inside the shell due to the charge on the shell is zero.
Why? (Or why not?)
kihr said:
The force on the charge q would be the product of q and the electric field in which this charge is located (which is zero in this case). hence the force on the central charge is zero.
You seem to have arrived at the right answer, but without the right justification.

kihr said:
(b) By the logic applied in the case of (a), the force should be zero. However, the answer is "towards right". I need some help to understand this, as it is likely that the logic applied to solve (a) may not be entirely correct. Thanks.
Yep, once you figure out part (a), this should make more sense.
 
I would request for more elaboration on the concept of the net flux due to an external charge being zero, and also why there would be an electric field inside the sphere due to the charge q1.
I took the electric field inside the sphere due to the charge on the sphere to be zero as I assumed that this was to be evaluated considering that the central charge was not there. This is because the force on the central charge would be the product of the charge and the external electric field acting on it. This external electric field would arise due to the charge on the sphere. Please help me to understand the error with this logic.

One more thing. I forgot to indicate in the problem that the charge q1 is located to the right of the spherical shell.
 
You've worked out from Gauss' law that the net force on the charge at the centre should be zero. Now you can say that the net force on this charge is actually due to a contribution from the shell and the point charge to the right... hope this helps?
 
Yes, but how to find out the directions of the field inside the shell due to the charge on the shell and also due to q1. I am still confused!
 
There is no net field inside the sphere and so the field inside the sphere due to only the shell is just the opposite to the field due to q1 (the charge of the shell is irrelevant, this equal and opposite field would be created whether the sphere had charge or not).

The question, however, asks only to find the force on the centre charge due to the shell only; this is going to be equal and opposite to the coulomb force the centre charge would experience from q1 if the shell wasn't there.
 
Yes. Could you please explain as to how to calculate the Coulomb force on q by q1 when the shell intervenes between the two? Do we have a situation of induction of a charge -q1 on the outer surface of the sphere, and a charge of +q1 on its inner surface due to induction? If this becomes clear the rest of the explanation falls in place. Many thanks.
 
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