Grounded conducting sphere inside uniform electric field The problem of a grounded conducting sphere inside a uniform $E$ field in the $z$ direction can be solved by imagining the field be produced by a pair of charges $Q$ and $-Q$ put on the $z$ axis, taking the limit that the charges are very very far away, $Q\rightarrow\infty$, and using the image method. Under this limit, the field produce by $Q$ and $-Q$ will be uniform in the vicinity of the sphere. My question is, why we need two charges? A single charge $Q$ can also produce a uniform $E$ field under the same limit.
 A: you don't need two charges, you can use only one charge and the answer will be the same, you only should note to add a second image charge in the centre of the sphere to make the sphere neutral.
in both cases, when you take the limit, you get same electric dipole in the centre of sphere.
but because the symmetry of two charges, calculations are somehow easier.  
with any relevant boundary condition, the sphere should remain neutral. to see this suppose it gains charge $q$. if you change $E \to -E$ then the charge of sphere should change to $-q$ (you can see this by dimensional analysis, since the $q$ should be proportional to $E$). but the new setup are the same as the first one, because letting $E\to -E$ is the same as rotating the setup by 180 degrees. it certainly shouldn't change the sphere's charge. so $q=-q=0$.
A: If we forget about the image charges for a moment, there should be a symmetry between the front part and rear part of the sphere. Because, $e^-$s drawn from 
the back of the sphere to the frontal part will expose equal amount of +ve ions in the back. A single charge will note restore this symmetry. Also, there will be unbalanced amount of charge on the sphere of magnitude $-\frac{aQ}{R}$(which is not supposed to be, as to start with the sphere was neutral). Here, $a$ is the radius and $R$ is the distance where the charge $Q$ has been kept ($R\rightarrow\infty$).
