I was reading "point charge in the presence of a charged, insulated, conducting sphere" from Jackson's Classical Electrodynamics.

Let initially the sphere had on its surface $Q$. When $q$ is brought at $y$ from the center of the sphere, it must provide force to the charges residing on the surface owing to its electric field. However, he immediately jot down that only $q' = -\frac{a}{y}q$ will experience force from $q$ & that $q'$ would balance the electrostatic force of $q$; thus $(Q - q')$ will spread uniformly on the surface as this suffers no external force.

If we wish to consider the problem of an insulated conducting sphere with total charge $Q$ in the presence of a point charge $q$, we can build up the solution for the potential by linear superposition. In an operational sense, we can imagine with a grounded conducting sphere(with its charge $q'$ distributed over its surface). We then disconnect the ground wire & add to the sphere an amount of charge $(Q - q')$. ... .To find the potential, we merely note that the added charge $(Q - q')$ will distribute itself uniformly over the surface since the electrostatic forces due to point charge $q$ are already balanced by the charge $q'$.[...]

I'm confused how really $q'$ 'balanced' the electric field of $q$. I thought only a same magnitude of charge can balance the field of an equal but opposite charge exactly at the mid-point between them. So, I couldn't understand how/why $q'$ balanced the field of $q$. Was there no field of $q$ anymore, after $q'$ 'balanced' its field?? Can anyone explain me why & how $q'$ cancels or "balances" the field of $q$ even it is not of the same magnitude as that of $q$? Does all the field of $q$ vanish after being balanced by $q'$??

If I bring a charge near a system of $q$ & $q'$, it still gets force from the configuration even if the field of $q$ is balanced by $q'$ at the middle. So, why doesn't the charge $Q - q'$ experience force from $q$ despite its field got balanced by $q'$??


1 Answer 1


$q$ and $q'$ together produce a potential that is zero on the surface of the sphere. If you find the surface charge density determined by the radial change in that potential you get a charge distribution whose total over the whole surface is $q'.$

If you take that surface charge and $q$ then they together make a combined field that is exactly zero throughout the sphere so any excess charge ($Q-q'$) at the center just exerts forces on each other pushing each other away from the center and out to the surface of the sphere.

  • $\begingroup$ +1; So, I can conclude from your argument that in the region where the two charges make zero potential, any other charge wouldn't experience any force right? $\endgroup$
    – user36790
    Commented Aug 25, 2015 at 4:01
  • 1
    $\begingroup$ The two charges make zero potential just on the surface. When the $q'$ distributed through the surface, then the potential due to both the net $q'$ on the surface and the q outside the sphere together exert no net force inside due to themselves. But the $Q-q'$ at the center still exerts forces on each other. I think I'm just repeating myself. $\endgroup$
    – Timaeus
    Commented Aug 25, 2015 at 4:05
  • $\begingroup$ Would $Q - q'$ exert force on $q'$? The answer is no, as superposition principle tells you can't alter the distribution when you superimpose two cases. But I'm not getting why wouldn't there be force among them. $\endgroup$
    – user36790
    Commented Aug 25, 2015 at 5:06
  • $\begingroup$ Sir, I've made my above query as a separate question as: physics.stackexchange.com/questions/202415/…; thanks for help. $\endgroup$
    – user36790
    Commented Aug 25, 2015 at 12:34

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