Capacitance, Charge and Voltage (P.D.)

[MemoNick]

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 = (ε₀ε_{r (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!

 

[Enigman]

...
2. I used C = (ε₀ε_{r (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?

 

[gneill]

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 = (ε₀ε_{r (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?

 

[MemoNick]

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]

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!

Glad to help!