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Finding potential difference across capacitor plates?

  • Thread starter cheap_noob
  • Start date
1. Homework Statement
Two capacitors are identical, except that one is empty and the other is filled with a dielectric (k=3.6). The empty capacitor is connected to a -11V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacity?


2. Homework Equations
C=q/V C=kε°0)A/d E=q/(ε°)A
Estored=1/2qv = 1/2 cv^2


3. The Attempt at a Solution
I don't really know what to do. I'm guessing that there's a lack of information but however the part about the empty capacitor, that would mean Estored = 0 right? But even though I plug in something with zero and solve for C, C would be zero which would mean 0=q/V or the other equation and if I solve for a variable it would be just zero.

I'm guessing my logic is probably wrong but I'm really lost. I basically have only 2 known variables from the question.
Any help would be awesome!
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

ehild

Homework Helper
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Empty capacitor means that the space between the plates is not filled with dielectric.

ehild
 
So does that mean I let C=kε°A/d equal to each other so kε°A/d=kε°A/d?
And then from that plug in 3.6 for one of the k, and 1 for the other?
And from that, the k's will cancel, the A's will cancel b/c identical, and I'm not sure about the d's. Well is that the right method or am I still doing something wrong?
 

gneill

Mentor
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2,612
So does that mean I let C=kε°A/d equal to each other so kε°A/d=kε°A/d?
And then from that plug in 3.6 for one of the k, and 1 for the other?
And from that, the k's will cancel, the A's will cancel b/c identical, and I'm not sure about the d's. Well is that the right method or am I still doing something wrong?
How would the k's cancel if they are not the same?
 
How would the k's cancel if they are not the same?
Oops sorry. Well I meant that there will be a ratio when you divide one by the other; when transferring k over to solve for the unknown variable, there will be an actual number
Anyways I'm still lost
 

gneill

Mentor
20,486
2,612
Oops sorry. Well I meant that there will be a ratio when you divide one by the other; when transferring k over to solve for the unknown variable, there will be an actual number
Anyways I'm still lost
Start by working out how the values of the two capacitors are related (write an expression).
 

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