# Attractive or repulsive force between two statically charges plates [duplicate]

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

where:

• 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?

• 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 Commented Nov 6, 2023 at 11:24
• 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. Commented Nov 6, 2023 at 12:49