Calculate Electrostatic Force Between Two Plates

[ak_47_boy]

Hello, first post here. I hope this is in the right section.
I have spent an hour or so surfing the net and could not figure this out (i am bad at physics).

How do you figure out the electrostatic force attracting two plates? The plates are 2sq/m, have a 0.006m spacing, and have 10000 volts of potential difference.

 

[Shooting Star]

Welcome. Since you said you were “bad” at Physics, I’m trying to give you a detailed solution.

When a voltage V is applied across the plates, suppose the plates receive charges +Q and -Q. The surface charge density on the plates are +s and –s, where s=Q/A. If the plates were infinite in extent, then each would produce an electric field of magnitude
s/(2e) = Q/(2Ae) --(1), where I’m writing e for epsilon_naught. Let d be the separation between the plates.

The sum of the fields of both the plates E = Q/(Ae) and V=E*d
=> Q = AeV/d --(2).

So, force on one plate due to field of other = Q*field = Q*Q/(2Ae) (from 1)
= eAV^2/(2d^2) (from 2).

Note that the electric field of each plate has been calculated for the case when each plate in infinite in extent. In practice, some correction factor is introduced.

 

[ogb p]

Sure, I'd be happy to help you with this calculation. The electrostatic force between two parallel plates can be calculated using Coulomb's Law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the plates have a potential difference of 10000 volts, which means they have a charge of 10000 volts x 2 square meters = 20000 Coulombs (since the units for voltage are equivalent to joules per coulomb). The distance between the plates is 0.006 meters, so the force can be calculated as:

F = (k * q1 * q2) / d^2

Where k is the Coulomb's constant (8.99 x 10^9 N*m^2/C^2), q1 and q2 are the charges on the plates (20000 Coulombs), and d is the distance between the plates (0.006 meters). Plugging in these values gives us:

F = (8.99 x 10^9 N*m^2/C^2) * (20000 C) * (20000 C) / (0.006 m)^2

= 8.99 x 10^9 N*m^2/C^2 * 400000000 C^2 / 0.000036 m^2

= 8.99 x 10^9 N * 400000000 / 0.000036 m^2

= 8.99 x 10^9 N * 11111111111.11 / m^2

= 9.99 x 10^20 N

Therefore, the electrostatic force between the two plates is approximately 9.99 x 10^20 Newtons. I hope this helps and let me know if you have any further questions or need clarification.