How Does Charge Affect Boundary Conditions in an Electric Field?

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neworder1
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A conducting ball of radius R and total charge Q is placed in a homogenous electric field E. Find the potential V everywhere.

Some attention needs to be paid to boundary conditions. I know how to calculate this when there is no charge, i.e. Q=0. Then I put V=0 on the surface and V=Ez at infinity, and using these two conditions I can obtain the coefficients in the general solution. My problem is - why these boundary conditions don't work when there is nonzero charge? I don't how to turn the total charge into some kind of boundary condition for V. My reasoning is - I can choose V freely on the surface, since V is determined only up to a constant, and at infinity there is only homogenous field E, so the condition at infinity is the same as before. So the calculations are the same, and so is the result. Where do I take the charge Q into account, then?
 
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I can't follow you completely, but since the ball is conducting, the charge Q will be on the surface of the ball only, right?
 
neworder1 said:
A conducting ball of radius R and total charge Q is placed in a homogenous electric field E. Find the potential V everywhere.

Some attention needs to be paid to boundary conditions. I know how to calculate this when there is no charge, i.e. Q=0. Then I put V=0 on the surface and V=Ez at infinity, and using these two conditions I can obtain the coefficients in the general solution. My problem is - why these boundary conditions don't work when there is nonzero charge? I don't how to turn the total charge into some kind of boundary condition for V. My reasoning is - I can choose V freely on the surface, since V is determined only up to a constant, and at infinity there is only homogenous field E, so the condition at infinity is the same as before. So the calculations are the same, and so is the result. Where do I take the charge Q into account, then?
For a charge Q, first solve the Q=0 problem as you say.
Then just add the potential of a uniformly charged sphere (Q/r).