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.


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 $q=-q=0$.

  • $\begingroup$ If we literally consider the case where a grounded metallic sphere of radius $a$ is placed in an electric field generated by a single huge charge $Q$ located at a huge distance $r$, there will be an induced charge of $−aQ/r$ on the sphere. Intuition tells us that it is a wrong way to represent the problem of a grounded metallic sphere in a uniform electric field, but arguing why it is so looks subtle to me. I guess it has to do with what we really mean by grounding. $\endgroup$ – higgsss Aug 25 '14 at 22:26
  • $\begingroup$ it is wrong way to represent the problem because it doesn’t have the symmetries of original problems. $\endgroup$ – seyed Aug 26 '14 at 14:53
  • $\begingroup$ literally because the boundary conditions isn't the same. actually the original problem's boundary condition is ill-defined. for example if the field is in the $\hat x$ direction, the $x=constant$ planes would be the equipotential surfaces. so the $(0,0,0)$ and $(0,0,\infty )$ has the same potential. but in the $Q$ charge model the potential deference between the origin and the $(0,0,\infty )$ point would tend to $\infty$ after taking the limit. $\endgroup$ – seyed Aug 26 '14 at 15:04
  • $\begingroup$ I guess we can consider a Neumann boundary condition at a sufficiently large spherical surface for the original problem. I see that the notion of being grounded, i.e., having the same potential as the spatial infinity, is ill-defined in this case. $\endgroup$ – higgsss Aug 26 '14 at 16:10
  • $\begingroup$ On the other hand, in the model with the charge Q, grounding makes the metallic sphere have the same potential as the spatial infinity, which is much farther away than the charge Q. By this, the role of Q is not only to generate a constant electric field near the metallic sphere but also to affect the boundary condition on it. This is certainly not what we have intended. I agree with you that that the trouble here is the boundary condition. It seems that the usage of the notion "grounding" is somewhat misguided in the original problem statement. $\endgroup$ – higgsss Aug 26 '14 at 16:30

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 ($R\rightarrow\infty$).

  • $\begingroup$ Kolahal, but what's the mathematical point of view of needing two charges? The BC is that we have a uniform field far away from the sphere, and one single charge can fulfill the purpose. By the way, the sphere is grounded, not neutral, although it will remain so in the solution using two charges. $\endgroup$ – velut luna May 20 '14 at 5:58
  • $\begingroup$ Even if the sphere is grounded, in a uniform electric field it will remain neutral. In fact, in this example there is not difference between grounded/non-grounded (if you have access to Greiner's electrodynamics, you can refer to that book for this example). The electric field is not itself the boundary condition. Boundary condition relates to the potential on the boundary of the sphere - that the potential is uniform across the surface. And in that case, the symmetry should be maintained. $\endgroup$ – kolahalb May 20 '14 at 6:11
  • $\begingroup$ If uniform field is not a BC, what is the BC at infinity? $\endgroup$ – velut luna May 20 '14 at 6:18
  • $\begingroup$ There is no boundary at infinity...so why are you worried about boundary condition there? $\endgroup$ – kolahalb May 20 '14 at 7:50
  • $\begingroup$ How can you obtain a unique answer without specifying a BC at infinity?? $\endgroup$ – velut luna May 20 '14 at 8:17

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