Find dialectric constant in a capacitator

[romakarol]

Homework Statement

A parallel plate capacitor has a capacitance of 1.5µF with air between the plates. The capacitor is connected a 12V battery and charged. The battery is then removed. When a dielectric is placed between the plates, a potential difference of 5.0V is measured across the plates. What is the dielectric constant of the material?

Homework Equations

not sure on this, possible culprits are:
q=CV
where as far as I understand q is magnitutde of charge on each plate
c is capacitance
and v is voltage

k=E0/E
This is the only formula in our notes that doesn't involve plate area and distance between plates, which isn't given in the question. What E0 and E are isn't explained though which is causing me no end of confusion

Is it fair to say V=12 in this question and C=1.5(10^-6)?

3. The attempt at an answer

I tried googling the question and funilly enough it came up here before. The OP was equally clueless and the advice given was:
a dielectric will increase the capacitance but charge on the the plates will be same. Can you write an equation with this information and find the new capacitance? Comparing the two capacitances, you can calculate the dielectric constant of the new capacitor.

It's late at night so can someone please just tell me what the E0/E formula means and what formulae are relevant to the question? I asked around but the question is "sorcery" to everyone I know. Is the formula I attached below relevant? It is dependent on both capacitance and dialectric constant which makes it seem relevant but it invovles variables not given in the question (distance and area).

 

[cnh1995]

The formula attached is correct. It tells you that the capacitance will increase k times when you insert a dielectric of dielectric constant k. But final charge on the capacitor doesn't change after insertion of dielectric. Q=CV will be useful here.

 

[romakarol]

The formula attached is correct. It tells you that the capacitance will increase k times when you insert a dielectric of dielectric constant k. But final charge on the capacitor doesn't change after insertion of dielectric. Q=CV will be useful here.

How can the attached formula be used without knowing either k (dialectric constant) or A (area of the plate)? Also in an example the 3o is 8.85X10^-12, is this a constant?

 

[cnh1995]

How can the attached formula be used without knowing either k (dialectric constant) or A (area of the palte)? Also in an example the 3o is 8.85X10^-12, is this a constant?

That formula can't be used directly. It only tells you that capacitance increases k times.

Also in an example the 3o is 8.85X10^-12, is this a constant?

Yes it's a constant. But all you need is Q=CV, Q being same in both the cases.

 

[romakarol]

That formula can't be used directly. It only tells you that capacitance increases k times.

Yes it's a constant. But all you need is Q=CV, Q being same in both the cases.

Cheers that's something I may be able to work with I'll get cracking.

 

[romakarol]

That formula can't be used directly. It only tells you that capacitance increases k times.

Yes it's a constant. But all you need is Q=CV, Q being same in both the cases.

So you mentioned in the other post that there are two formulas, and q is the same in both. What I didnt realize originally is that the 5V potential difference in the question is simply voltage (at least i think it is) so as I gather there are 2 equations:
q=CV
q=(12)(1.5X10^-6)
q=1.8X10^-5C
and
q=CV
q=(5)(1.5X10^-6)
q=7.5X10^-6C

But since you said q is the same in both, I'm guessing k is figured out by setting the second equation equal to the value of the first one:

(5)(1.5X10^-6) =/= 1.8X10^5C
Presumably k slots in here somewhere to make it equal? I don't know the relationship k has to C however, if this is right can you tell me where k goes in the equation?

 

[romakarol]

Can anyone help me finish this off? I want my sleep.

 

[cnh1995]

q=CV
q=(12)(1.5X10^-6)
q=1.8X10^-5C
and
q=CV
q=(5)(1.5X10^-6)
q=7.5X10^-6C

When voltage is 5V, capacitance is no longer 1.5uF.
So,
12*1.5=5*C_new.
Find C_new and take ratio of C_new with C_old to get k.

 

[romakarol]

When voltage is 5V, capacitance is no longer 1.5uF.
So,
12*1.5=5*C_new.
Find C_new and take ratio of C_new with C_old to get k.

Thanks I think I have it:
But why are you using just 1.5 as the old capacitance, isn't it 1.5x10^-6 as given in the question. Is the constant the 1.5 is being multiplied by being ignored for some reason?

Taking it as 1.5 however:
Cnew=[(12*1.5)/5]
Cnew=3.6

k=3.6 : 1.5
k=1 : 2.4

 

[cnh1995]

But why are you using just 1.5 as the old capacitance, isn't it 1.5x10^-6 as given in the question.

It is in farads. I wrote directly in microfarads. But it doesn't matter because ultimately the units are going to cancel out.

 

[cnh1995]

k=1 : 2.4

It should be 2.4:1...k=2.4

 

[romakarol]

It is in farads. I wrote directly in microfarads. But it doesn't matter because ultimately the units are going ti cancel out.

Right thanks for all the help, only took till 2:30AM to get the right answer :D. Totally worth it.

 

[cnh1995]

Right thanks for all the help, only took till 2:30AM to get the right answer :D. Totally worth it.

You're welcome..have a good night!