- #1
erst
- 21
- 0
So for just vacuum in between one could say F=q*E = (σA)*(σ/2ε0) = A*σ^2 / (2ε0)
Alternatively, F = dU/dx, where x is the separation. U = 1/2 CV^2 = 1/2 QV = x*A*σ^2 / (2ε0). To get V, we just add up pieces of E-field*distance.
But now let's say part of x is a dielectric of thickness t and ε = εr*ε0, so vacuum is of thickness x-t.
Unfortunately, I get the exact same answer with the F = dU/dx approach since U = (A*σ^2 / (2ε0)) * (x - t + t/εr).
And with F=q*E, I'm not even sure what to do. If the dielectric was "suspended", the E-field would change right back to the vacuum value after passing through the dielectric so it's like it's not even there. Or if the dielectric is right at the plate's boundary, it would feel the reduced E-field.
I'm getting something wrong conceptually. I feel like there should be a difference in the force between the plates if there's a dielectric in there...
Alternatively, F = dU/dx, where x is the separation. U = 1/2 CV^2 = 1/2 QV = x*A*σ^2 / (2ε0). To get V, we just add up pieces of E-field*distance.
But now let's say part of x is a dielectric of thickness t and ε = εr*ε0, so vacuum is of thickness x-t.
Unfortunately, I get the exact same answer with the F = dU/dx approach since U = (A*σ^2 / (2ε0)) * (x - t + t/εr).
And with F=q*E, I'm not even sure what to do. If the dielectric was "suspended", the E-field would change right back to the vacuum value after passing through the dielectric so it's like it's not even there. Or if the dielectric is right at the plate's boundary, it would feel the reduced E-field.
I'm getting something wrong conceptually. I feel like there should be a difference in the force between the plates if there's a dielectric in there...