# Capacitance, Charge and Voltage (P.D.)

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!

.......
2. I used C = (ε0εr (1 in this case)A)/(d) and E = V/d and Q = CV

....... Then I used the last equation to find the charge to be 4.2x10-8.

...... 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.

...... for a new capacitance of 1.22x10-8. So far, so good.

...... the change in the potential difference between the two plates. How do I find that? • 1 person
gneill
Mentor
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.
Okay.
Using the second equation I found the magnitude to be 474308N.
Incomplete units.
Then I used the last equation to find the charge to be 4.2x10-8.
Units?
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.
Units?
So far, so good.
Except for your units. You WILL lose marks if your units are not correct or not given.
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?
What capacitor formula do you have that relates capacitance, charge, and potential difference? What do you know about the charge on the "new" capacitor?

• 1 person
So the charge doesn't change between the two parts? Hence I can use Q/C = V? If yes, then the voltage is 3.44V.

Also, sorry about the units. I had been going at this question for an hour, and I was extremely frustrated. Thanks a lot guys! I really appreciate your help!

gneill
Mentor
So the charge doesn't change between the two parts? Hence I can use Q/C = V? If yes, then the voltage is 3.44V.
Yup. The charge can't change if the capacitor is not connected to an external circuit.
Also, sorry about the units. I had been going at this question for an hour, and I was extremely frustrated. Thanks a lot guys! I really appreciate your help!