So I'm trying to calculate some mechanics for an electrostatic attraction/repulsion system and I want to make sure that I'm using the correct numbers. I did a bunch of research on my own and I think I have it right, but I'm hoping that someone more experienced can help make sure that I am using the correct formulas.

As far as I understand coulombs law states that the force will be:

  • F = ( K * Q1 * Q2) / r^2


  • F = the attractive or repulsive force
  • K = 8.98 x 10^9
  • Q1 = the charge on the first plate
  • Q2 = the charge on the second plate
  • r = the distance separating the plates in meters

Is this correct? Is it different when it's plates and not points charges, or does it sum to the same thing either way?

  • $\begingroup$ kinda, but now I'm really confused as to why distance isnt a factor in the equation, I was under the impression that the force of electrostatic attraction fell of with the inverse square law. That equation would suggest that if I pulled plates appart with constant charge, that the force of attraction would increase proportionally with increasing distance. If I have 20 kv at 10 mm separation, pulling them to 100 mm separation at constant charge makes 200kv between the plates and would pull harder with that equation, doesnt make sense to me $\endgroup$
    – ADesilets
    Commented Nov 6, 2023 at 11:24
  • $\begingroup$ If the force is $\frac 12 QE$ and $E=\frac Vd$ and $Q=CV= \epsilon_0 A\frac Vd$, then keeping $Q$ constant means that as the separation of the plates increases so does the potential difference in proportion which in turn means that the electric field stays constant as does the force. $\endgroup$
    – Farcher
    Commented Nov 6, 2023 at 12:49

1 Answer 1


If the plates are much further apart than their linear dimensions (e.g. their diameters if they are circular) the force between them CAN be calculated approximately from Coulomb's law. This is because each plate subtends such a small solid angle as 'seen' by the other plate.

If the plate separation is of the same order of magnitude as the plates' linear dimensions you can't treat the plates as point charges, and you have quite a difficult problem. You can see why: each small charged patch on one plate attracts each small charged patch on the other plate, and the pairs of patches are different distances apart.

However if the plate separation is MUCH LESS than the plates' linear dimensions, you have another case where it's relatively easy to calculate the force, but the plates can't be treated as point charges. Instead it helps to appeal to symmetry and to apply Gauss's law (which, for static charges, can be deduced from Coulomb's law, but which is, in fact, more general). What emerges is that the field due to each plate individually is uniform and normal to the plate within a distance from it that is small compared with the plate's linear dimensions. So the force on the other plate, if close, is trivial to calculate.

  • $\begingroup$ Could you show me an example, like if I have 10cm sqare plates separated by 1 cm and I the charge remains constant but the separation distance has a travel of 10 cm and I want to be able to calculate the attraction between plates or the repulsion $\endgroup$
    – ADesilets
    Commented Nov 6, 2023 at 11:38
  • $\begingroup$ like I'm getting to the point of frustration where I'm almost at the point of building a physical model and measuring it $\endgroup$
    – ADesilets
    Commented Nov 6, 2023 at 12:07
  • $\begingroup$ ADesilets "I have 10cm square plates separated by 1 cm" If they carry equal (and opposite) charges you can calculate the force between them to a good approximation using Gauss's law (see third paragraph of my answer). But what do you mean by "the separation distance has a travel of 10 cm"? I hope you don't mean that you increase the plate separation to 11 cm, because then you are well into the hard-to-handle region I mention in my second paragraph $\endgroup$ Commented Nov 6, 2023 at 16:59
  • $\begingroup$ That's exactly my predicament.... I want to learn the math so I'm trying to calculate the results before I run my experiements... I got the impression from reading about Coulombs practical experimentation that his law may work provided I use spheres like he did (openstax.org/books/physics/pages/18-2-coulombs-law) is that link good info or am I being mislead? $\endgroup$
    – ADesilets
    Commented Nov 6, 2023 at 17:14
  • $\begingroup$ Uniformly charged spheres will indeed attract or repel according to an inverse square law. By "uniformly charged" I mean either with charge uniformly spread over the volume or over the surface. This is difficult to achieve... The obvious strategy is to use conducting spheres (e.g. table tennis balls coated with graphite applied using a pencil). You could give them equal and opposite charges by induction using a rubbed plastic rods with the two spheres temporarily in contact. Unfortunately, unless their separation is much greater than their diameter, $\endgroup$ Commented Nov 6, 2023 at 17:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.