These days I encoutered the famous grounded sphere near a charge problem, and I saw a pretty straight forward solution(for the image charge induced on the sphere). I am not sure if this solution is OK... so here it is:

Consider a charge $q$ at a distance $r$ from the center of a grounded, conducting sphere of radius $R$. Find the charge induced on the sphere. ($r>R$)

Solution(seen by me)

Because the sphere is grounded it has the potential(inside and on the sphere): $$V=0$$ The charge q' induced on the surfuace of the sphere(and the charge q) ,regardless of how it is distributed , will give in the center of the sphere the total potential: $$V=\frac{q'}{4\pi\epsilon R} + \frac{q}{4\pi\epsilon r} = 0$$ So $$q'=-q \cdot \frac{R}{r} $$

Do you find this ok? If yes , please explain.

(rigurous solution: http://www.youtube.com/watch?v=KoQ3KP2oSMo)

  • $\begingroup$ Your approach is absolutely correct. If you can read Russian I would suggest a very good article "Electrostatics for smart ones" (Zilberman, Kvant Feb 2016, kvant.mccme.ru/pdf/2016/2016-02.pdf). Something tells me it will be very interesting to you. $\endgroup$
    – lesnik
    Feb 8, 2017 at 7:03

1 Answer 1


This doesn't give you the distribution of the charge on the surface, but it's an elegant way of getting the total induced charge on the sphere.

A couple of things I notice:

  1. The larger $r$, the smaller the induced charge
  2. The sign is correct (the charge must be the opposite of $q$)

So yes, I "find this OK". Plus - it matches the expression in your rigorous derivation.


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