Electrostatic force exist on the plate of a capacitor

[qqpost]

Hi everyone,

I am reading Rebeiz's RF MEMS book. It says as follow:' When a voltage is applied between a fixed-fixed or cantilever beam and the pull-down electrode, an electrostatic force is induced on the beam. This is the well known electrostatic force which exists on the plates of a capacitor under an applied voltage'.

I always take it as granted. But I don't know why. Can anyone give me a detail explain? How does the electrostatic force come out?

Thank you

 

[sv1437]

I think one plate/electrode is connected to ground and voltage is applied on the other plate, there by positive and negative charges develop on the plates, and the electrode which is not fixed, bends down due to force of attraction.

 

[qqpost]

From what you said, you mean it is the coulomb force between the positive and negative charge? Or the electric field E=U/d and the charges? I saw several equations like F=QE/2 and F=1/2*U^2*dC(g)/dg to calculate the force. But I have no idea what's the meaning of them. Could you help me?

 

[sv1437]

It is the coulomb force of attraction between the positive and negative charges.

Electric field = Voltage/distance (E=V/d)
For a capacitor , energy = 1/2* C*V^2. What is 'g' stand for in your equation ?

-sv

 

[qqpost]

g stands for the gap between the two plates of the capacitor, which in MEMS structure is the distance between the beam and dielectric layer

 

[Milad-D]

The energy stored between the two plates as you said is 1/2 c v^2 . As we know f=dw/dx, and c=eps*A/x which x is the gap between the two plates. Therefore, the force can be easily found.