# Electic field / capacitance of parallel plates

Upon calculating the electric field for a plate of charge, you arrive at:

E = Q/Ae (Q the charge on the plate, A the area of the plate, e the permittivity of the dielectic)

But then, isn't the total electric field between the plates twice this because of the equal but opposite charge on the other plate?

I always see it written without a factor of 2.

which way is the E field going , make a Gauss pillbox around the capacitor .

Drakkith
Staff Emeritus
Upon calculating the electric field for a plate of charge, you arrive at:

E = Q/Ae (Q the charge on the plate, A the area of the plate, e the permittivity of the dielectic)

But then, isn't the total electric field between the plates twice this because of the equal but opposite charge on the other plate?

I always see it written without a factor of 2.

Is that equation only for the charge on 1 plate, or for both plates? Looks to me like it is only for the 1 plate based on your post.