A parallel plate capacitor of plate area A and separation distance d contains a slab of dielectric of thickness d/2 (see Figure 8) and dielectric constant [kappa]. The potential difference between the plates is [Delta]V.
a) In terms of the given quantities, find the electric field in the empty region of space between the plates.
b) Find the electric field inside the dielectric.
c) Find the density of bound charges on the surface of the dielectric.
a) Suppose the electric field in the capacitor without the dielectric is equal to E0. The electric field in the dielectric, Ed, is related to the free electric field via the dielectric constant [kappa]:
The potential difference between the plates can be obtained by integrating the electric field between the plates:
The electric field in the empty region is thus equal to
b) The electric field in the dielectric can be found by combining eq.(41) and (43):
c) The free charge density [sigma]free is equal to
The bound charge density is related to the free charge density via the following relation
Combining eq.(45) and eq.(46) we obtain
Information provided by: http://teacher.nsrl.rochester.edu