A parallel plate capacitor of area 2 cm^2 is filled with three dielectrics with constants k1=3.5, k2=1.7 and k3=8.5. The dielectrics are placed as shown in the figure, each occupying half the total area, while the volume of the first double of the other two. If the separation distance between the plates is 2 mm.
a. What is the capacitance before including the dielectrics?
b. What is the capacitance after including the dielectrics?
c. If we connect the capacitor to a battery of 12 V, What is the energy stored?
Image of the exercise: http://i41.tinypic.com/jhg7l1.jpg
Part a. C=(E0)*(Area)/(distance)
Part b. C=(Keq)*(E0)*(Area)/(distance)
Part c. V = 1/2 * E0 * E^2
The Attempt at a Solution
Part a. I solved this part of the exercise by just plugging in the value given by the exercise.
Part b. I am assuming I have to use the formula I denoted as Part b. I looked at the exercise and assumed it was a circuit... When I looked at it that way, I calculated the Equivalent Capacitance. I had two capacitance in Series then that went to be in parallel with the other capacitance. I calculated the Keq like that. Then since we applied dielectrics, I am assuming the Area would half of what it used to be. This is the part where I am stuck... trying to understand clearly how to solve part b.
Part c. I think I need the capacitance of part b to solve part c using the formula presented above?
Thanks for all the help that can be provided!