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**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 100cm

^{2}. 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

ε

_{0}is equal to 8.854x10

^{-12}/m

**2.**I used C = (ε

_{0}ε

_{r (1 in this case)}A)/(d) and E = V/d and Q = CV

**3.**The area is 0.01m

^{2}, and using the first equation, along with the given capacitance, I found the distance to be 2.53x10

^{-5}m. 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 0.01m

^{2}. 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!