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Capacitors with dielectrics

  • Thread starter star2003
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star2003

I am supposed to design an energy storage system for a car, so I want to calculate how much energy can be stored in a capacitor whose geometry matches the car. As a baseline, we have an area 150 cm wide by 2500 cm long by 30 cm tall. The capacitor is a sandwich geometry , one layer of conductor, one layer of insulator, one layer of conductor etc. There are n+1 sheets of conductor and n sheets of insulator. The thickness of the insulating sheets is equal to the thickness of the conucting sheets, d. Paper is the insulating material. It has a minimum thickness d, 50 micrometers which is twice the thickness of paper. The conducting sheets alternate in voltage between +V and -V. I am trying to figure out how much energy can be stored in this capacitor if a factor of 2 for the electrical breakdown of the insulating sheets and adjusting the thickness d to maximize the energy stored. I also need to find the maximum voltage. I am not even sure where to start. Please help!
 
40
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Your multi-plate capacitor is equivalent to n 2-plate capacitors connected in parallel.

The capacity of each 2-plate capacitor is given by

C = e0 e A / d

where e0 is a constant (epsilon-0), e is the dielectric constant of the paper (assuming that it fills all the volume between the conducting plates), A is the area of the plate, d is the distance between 2 conducting plates.

The total capacity is nC (because the capacitors are connected in parallel).

The total charge stored in the system is:

Q = n C V

The energy stored is:

U = Q V / 2 = n C V^2 / 2

I don't understand the piece of information you gave about the breakdown. Please explain it better.
 
140
1
I believe the breakdown factor and dielectric constant are one in the same.

Why aren't the n parallel plate capacitors considered to be connected series?
 

NateTG

Science Advisor
Homework Helper
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Originally posted by discoverer02
Why aren't the n parallel plate capacitors considered to be connected series?
It depends on how they're wired. It also really doesn't matter. The energy density is the same.
 

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