1. One question I'm having a lot of trouble solving is this: The plates of a parallel plate air capacitor have an area of 100cm2. The capacitor has a capacitance of 3.5nF and is connected to a 12V battery. It is fully charged. Calculate the: a) separation of the plates b) magnitude of electric field intensity between the plates c) the charge stored on the capacitor plates ε0 is equal to 8.854x10-12/m 2. I used C = (ε0εr (1 in this case)A)/(d) and E = V/d and Q = CV 3. The area is 0.01m2, and using the first equation, along with the given capacitance, I found the distance to be 2.53x10-5m. Using the second equation I found the magnitude to be 474308N. Then I used the last equation to find the charge to be 4.2x10-8. Now, the real difficulty... the circuit is broken, and an insulating material filling the whole distance between the two plates is inserted. It has a dielectric constant of 3.5. This material covers the same area of the plates, plus a border of 2cm. This border doesn't affect the area, since the first formula deals with the common area, in this case 0.01m2. I was asked for the new capacitance, for which I multiplied the one I was given (3.5nF) by the dielectric constant - for a new capacitance of 1.22x10-8. So far, so good. However, my stumbling block was when I was asked to find the change in the potential difference between the two plates. How do I find that? Thanks in advance, any help is extremely appreciated!