How Much Energy Can a Custom Car Capacitor Store?

AI Thread Summary
To design an energy storage system for a car using a custom capacitor, the geometry is defined as 150 cm wide, 2500 cm long, and 30 cm tall, with a sandwich structure of alternating conductive and insulating layers. The insulating material is paper, with a minimum thickness of 50 micrometers, and the goal is to maximize energy storage while considering the electrical breakdown factor. The formula for capacitance is given as C = e0 e A / d, where total capacitance is nC for n parallel capacitors. The total charge and energy stored can be calculated using Q = n C V and U = n C V^2 / 2, respectively. Clarification is needed on the breakdown factor, as it is distinct from the dielectric constant, and the wiring configuration determines whether capacitors are in parallel or series, though energy density remains consistent.
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!
 
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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.
 
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?
 
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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