Electrostatic force on a parallel plate

[serverxeon]

I'm tasked to find the force one parallel plate exert on the other plate in a capacitor.

somehow the force isn't just F=qE, where E=Q/2εA

What am I missing here?

 

[Simon Bridge]

What am I missing here?

Erm - direction?

You can check by using another method ... i.e. Coulombs law for a point charge and add them up, or start with the electric field due to a sheet of charge by itself.

 

[serverxeon]

i mean, my solution is lacking so coefficient.
based on my (confused and vague) understanding, the electric field is actually 1/2 E or something?

 

[Simon Bridge]

Is it? Why not calculate the field?

 

[Nickie]

The electrostatic force between two parallel plates in a capacitor is not simply given by the equation F=qE. This equation only applies to a point charge in an electric field, and does not take into account the geometry of the plates. In order to calculate the force between the two plates, you must consider the electric field between them and the area of the plates. This is why the equation for the electric field between parallel plates is E=Q/2εA, where Q is the charge on each plate, ε is the permittivity of the medium between the plates, and A is the area of the plates. To find the force, you would need to multiply this electric field by the charge on one of the plates, and then by the area of the other plate. This will give you the total force exerted by one plate on the other. Additionally, other factors such as the distance between the plates and any dielectric material between them may also affect the force. It is important to consider all relevant factors when calculating the electrostatic force between parallel plates.