Aakash PHYSICS JEE (Main & Advanced) Study Package - 5 & 6 (Class XII)

Chapter - Electric Charges and Field

Assignment (page 12)

SECTION - A; Q.no - 1

The force of repulsion between two point charges is F, when they are d distance apart. If the point charges are replaced by conducting spheres each of radius r and the charge remains same. The separation between the center of sphere is d, then force of repulsion between them is

(1) Equal to F

(2) Less than F

(3) Greater than F

(4) Cannot be said

Answer - (2) Less than F

What I expected the answer to be was (1). The electric field due to a conducting sphere of charge Q is equivalent to the same due to a point charge at the center of the sphere as total charge on it appears as concentrated at the center for the points outside the charged sphere. So, the force should have remained the same.

Any suggestions?

  • $\begingroup$ Hi Paras, welcome to Physics.SE. I've added the "homework"tag per our homework policy (meta.physics.stackexchange.com/questions/714/…). $\endgroup$ May 17, 2013 at 19:54
  • $\begingroup$ If this ques. was in MAINS , ans:1 . If this comes in ADVANCED , ans:2 $\endgroup$
    – ABC
    May 18, 2013 at 14:36
  • $\begingroup$ @ 007 I don't think type of exam would change the answer. O.o $\endgroup$
    – Baahubali
    May 18, 2013 at 16:17

2 Answers 2


Your answer would have been correct if, for example, the spheres were non-conducting and if the charges were distributed uniformly over their surfaces.

However, since the spheres are conducting, the surface charge distribution on each sphere will be altered because of the repulsion from the charges on the other sphere. In particular, the charges on each sphere will be pushed away by the charges on the other sphere. This will cause the charges on opposite spheres to be further away from each other, and the force of repulsion to be less than in the case of a uniform surface charge distribution.

Addendum 1. Why would the force be the same if the spheres were non-conducting with uniform surface charge densities? Well, without going into mathematical detail, note that using Gauss's law on each sphere would show that the electric field outside of that sphere would be the field of a point charge with the same total charge. Therefore, each sphere would simply "see" the other sphere as a point charge, and the result follows.

Addendum 2. Let's be a bit more precise in showing that when the spheres are conducting, the force will be lower than in the point charge case. By addendum 1 above, we know that the force would be the same of the spheres had uniform surface charge distribution with total charge equal to that of each point charge. Therefore, it suffices to show that the the surface charge distributions in the conducting case would lead to a smaller force.

To show this, divide each sphere into a large number of charge elements, and suppose that the charge distributions on the surfaces of the conducting spheres begin uniform. How exactly do the charge elements rearrange themselves on each sphere because of the influence of the other sphere? Well, note that if the other sphere were not there, then the charges would stay put and the distributions would remain uniform. Since the other sphere is there, the charges feel a force that pushes them away from the other sphere. In particular, picking any pair of charge elements from opposite spheres, we see that the distance between such a pair will be larger in the new surface charge distribution than in the uniform case. In particular, if we perform a sum over all pairs of surface charge elements on opposite spheres of the coulomb force between them, then every term in the sum will be smaller than in the uniform case, so the total force will also be smaller.

  • $\begingroup$ Oh! Got it! The case I was talking about is for a test charge and it's assumed that a test charge is so small that its own electric field doesn't affect the source charge. But in this case, the charges on either sphere are large enough to produce their own electric field. Well, I think so. Is my approach correct? And would you please elaborate how would've been my answer correct if the spheres were non-conducting and if the charges were distributed uniformly over their surfaces. Won't the same answer apply on this case too? BTW, thanks for the help mate. :) $\endgroup$
    – Baahubali
    May 17, 2013 at 20:37
  • $\begingroup$ I don't remember about the 3D case, but it is interesting that in 2D this situation is exactly soluble. I mean, there is a simple formula for the force, not a mind-blowing series. $\endgroup$ May 17, 2013 at 20:38
  • $\begingroup$ @PeterKravchuk That's pretty cool. I've never actually seen the solution; is it simple enough that I shouldn't be scared to attempt to generate it on my own? $\endgroup$ May 17, 2013 at 21:33
  • $\begingroup$ @ParasLehana See the addendum. $\endgroup$ May 17, 2013 at 21:38
  • $\begingroup$ @joshphysics, the idea is simple. At least, I was able to figure it out on my own.) $\endgroup$ May 17, 2013 at 21:48

Your answer might be (1) i.e. equal to F. Physically no point charge exists because charge is not independent matter, it is just property of matter. Charge can not exists without mass. Charge on the spherical body is termed as point charge. We should not be confused point charge as point has no dimension.

If spheres of same charge equal to point charge are placed at same distance apart, then force of intraction will be same as F. Charges are uniformly distributed on outer surface of sphere and distance from this charge measured from its center i.e. from the point so we consider it as point charge.

  • $\begingroup$ No, my friend. I think joshphysics got it right. This way, we can't even apply the Coulomb's Law now (as it is only valid for point charges). Practically, point charge may not exist but I don't think this is the solution to the problem. Charges on the sphere may be uniformly distributed but at time = 0. After a fraction of second, they'd try to rearrange themselves such that each of the charges on one of the spheres would experience the least force due to the charges on the other sphere. This would lead the increase in the equivalent distance between the distributions and thereby, decrease F. $\endgroup$
    – Baahubali
    May 18, 2013 at 16:41

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