How to Make a Capacitator Using Aluminum Foil and Wax Paper?

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Homework Help Overview

The discussion revolves around constructing a capacitor using aluminum foil and wax paper, focusing on estimating dimensions and capacitance. The original poster seeks guidance on how to approach the problem and calculate the excess charge when connected to a voltage source.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of aluminum foil as conductive plates and wax paper as an insulator, questioning how to measure dimensions and calculate capacitance. There is uncertainty about the thickness of the wax paper and the appropriate formula for capacitance.

Discussion Status

Participants are actively engaging with the problem, offering equations and discussing the properties of materials involved. Some guidance has been provided regarding the capacitance formula and the importance of using correct units. There is ongoing exploration of how to maximize capacitance through design choices.

Contextual Notes

There are concerns about the dimensions of the wax paper and the need for accurate measurements to ensure proper calculations. Participants are also clarifying the distinction between area and volume in their calculations.

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Homework Statement


You have aluminum foil and wax paper and you want to make a capacitator.
a- describe how you would make a capacitator and and then estimate all appropriate dimensions in order to estimate the capacitance of your design, in Farads.
b- Say you connect your capacitator to a 9volt battery, so that the potential difference between each side is 9 volts. (V) Find the amount of excess charge on each side of the capacitator.



Homework Equations



C= lQl/ delta V

The Attempt at a Solution


a- I don't even know how to approach this problem. I would assume that the aluminum can be used as the positive and the wax paper can be used as the negative. Maybe we can twist off strips of the aluminum foil and use them as conducting wires? We would need to measure the distance between the [+] and [-] sources. I am totally lost
b- I don't know
 
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Welcome to the PF, aquabum. The relevant equation that you listed is correct for part (b) of the problem statement. But first you need to calculate the capacitance that you can get by sandwiching an insulating layer (the wax paper) between two plates (using two pieces of the aluminum foil).

There is an equation that is commonly used to calculate the capacitance C in terms of the dielectric properties of the insulator and the plate area A and the separation distance between the plates d. Are you familiar with that equation?


EDIT -- BTW, I moved your post from Advanced Physics Homework to Intro Physics Homework. The Advanced Phyisics section is more for complex upper division homework problems.
 
capacitance = epsilon naut x area / distance?
 
Correct

C = \frac{\epsilon A}{d}

So now you need to figure out as much as you can about the epsilon and d (thickness) of wax paper, and decide what dimensions you want to make the Al foil plates. Then calculate the resulting capacitance that you can expect, being careful to use mks units and get your answer in Farads (it will be in more like uF, but that's still in "Farads").

Quiz Question -- What happens if you stack several layers on top of each other? Al - Wax - Al - Wax - Al ? How can you connect the various Al layers to maximize the resulting capacitance?
 
the Al layers need to be touching each other?
 
The Al foil dimensions will be 2x2x2, therefore the area of each one will be 8cm^3.

The distance or thickness of the wax paper will be 1cm.

now, do I take area and divide it by distance?
 
Just a note. Area is measured in squared units. Volume is measured in cubic units.
 
ranger said:
Just a note. Area is measured in squared units. Volume is measured in cubic units.

right, sorry.

So The Al foil dimensions will be 2x2, therefore the area of each one will be 4cm^2.

The distance or thickness of the wax paper will be 1cm.

now, capacitance = 4cm^2 / 1cm = 4 epsilon naut? Now I have to figure out what epsilon naut is. How could I do that?
 
The thickness of the wax paper is definitely not 1cm. You want to use the thinnest insulator possible in order to maximize the capacitance, right?

And on the epsilon, you do not use \epsilon_0. Why not? You could probably estimate the value of epsilon for wax paper from just looking up what a typical plastic's value is.
 
  • #10
ah i think i got it. I have to run to class now. Thanks for the help!
 

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