I was reading about method of images for a point charge near a conducting sphere. There(Feynman Lectures) I found this:

What happens if we are interested in a sphere that is not at zero potential? That would be so only if its total charge happens accidentally to be q′. Of course if it is grounded, the charges induced on it would have to be just that. But what if it is insulated, and we have put no charge on it? Or if we know that the total charge Q has been put on it? Or just that it has a given potential not equal to zero? All these questions are easily answered. We can always add a point charge $q′′$ at the center of the sphere. The sphere still remains an equipotential by superposition; only the magnitude of the potential will be changed. If we have, for example, a conducting sphere which is initially uncharged and insulated from everything else, and we bring near to it the positive point charge q, the total charge of the sphere will remain zero. The solution is found by using an image charge q′ as before, but, in addition, adding a charge $q′′$ at the center of the sphere, choosing $q′′=−q′= \frac{a}{b}q$. The fields everywhere outside the sphere are given by the superposition of the fields of $q, q′, \quad \&\quad q′′$. The problem is solved. We can see now that there will be a force of attraction between the sphere and the point charge $q$. It is not zero even though there is no charge on the neutral sphere. Where does the attraction come from? When you bring a positive charge up to a conducting sphere, the positive charge attracts negative charges to the side closer to itself and leaves positive charges on the surface of the far side. The attraction by the negative charges exceeds the repulsion from the positive charges; there is a net attraction. We can find out how large the attraction is by computing the force on q in the field produced by $q′$ and $q′′$. The total force is the sum of the attractive force between q and a charge $q′=−(\frac{a/}{b})q$, at the distance $b−(\frac{a^2}{b})$, and the repulsive force between q and a charge $q′′=+(\frac{a}{b})q$ at the distance $b$.

So, he has asked for the sphere which is not grounded & has a certain potential. For that , he chose a charge $q''$ at the center; okay but what is the reason that he chose the magnitude of $q''$ to be equal to $-q'$?


If the sphere is initially uncharged, then the electric flux through its surface is zero, by Gauss' law. If we add just one point charge $q'$, then we will find a net flux through the surface, which is wrong. However, add a second charge $q^{''}=-q'$, then the net flux is given by the enclosed charge $q'+q^{''}=0$, and all is well.

  • $\begingroup$ They are image charges , aren't they? $\endgroup$ – user36790 Jul 9 '15 at 7:25
  • 1
    $\begingroup$ The whole point of image charges is that the field of the charges should be identical to the field of the system you're considering. If the image charge fields have a net flux through the surface of the sphere, then your choice of image charges doesn't produce the right field. $\endgroup$ – ragnar Jul 9 '15 at 7:29
  • $\begingroup$ I would generally recommend that you post this as a separate question so that other people with a similar question in the future can more easily find it. $\endgroup$ – ragnar Jul 9 '15 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy