- #1
MemoNick
- 12
- 0
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!
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!