How does the area of the negatively-charged metal conductor of a capacitor affect its capacitance?

And…

How does the distance between the two conductors of a capacitor affect its capacitance?

Variables and Procedure

What is a Capacitor? and what is capacitance?

A capacitor is essentially two conductors separated by some medium, which is almost always an insulator. What it does is it holds an electric charge in a circuit. So, basically a battery, right? The only difference is that a capacitor doesn't create electrons and only stores them, allowing them to do several things a battery can't, like get rid of all of its charge immediately (camera flash, lasers), or balance out a voltage. Capacitance is a measure of its charge in Farads, which a 1 Farad-capacitor can hold 1 coulomb at 1 volt or 1 amp-second at 1 volt. Since this is very high, measurements are typically in microfarads.

The Results

**AP Physics Independent Project**

Interpretation

This graph shows that the area of the negatively-charged foil is powered in proportion to the capacitance. So, as the area of the negatively-charged foil gets bigger, the capacitance will grow at a faster rate. (C = a *(a^x) + b)

This graph shows that the distance between the foils is inversely powered in proportion to the capacitance. So, as the distance between the foils gets bigger, the capacitance gets smaller at a slower rate. (C = a *(1/a^x) + b)

Independent: Area of negatively-charged (first) metal conductor of the capacitor and distance between the two conductors

Dependent: Capacitance

Control: Temperature, distance/area depending on which experiment, material of foil, resistance, dielectric (same paper), time.

Get proper supplies including a meter stick, aluminum foil, scissors, a multimeter, two wires and a book or adequate paper.

Cut out one square of aluminum foil with a set size, as this will be the positive conductive foil of the capacitor.

Cut out several different sized squares of aluminum foil to be each of the negative conductive foils.

(It is recommended to fold the foils once on each side to make it more sturdy)

Prepare the capacitor by taking one of the negative foils and the positive foil and placing a set number of pieces of paper between a book, serving as the dielectric, in between them. Make sure that the distance between the two foils is the same in each trial. Also, take two wires and attach one of each onto one of the foils.

Press down on the book to get rid of air holes or inconsistencies.

Using a multimeter, plug the black cord in common and the red cord in the one with the sign denoted with ├ ┤or something similar to this (usually most right plugin point).

Placing the red on negative wire and the black on the positive wire. Turn multimeter to the sign previously shown and record the capacitance with the corresponding negative foil’s area.

Repeat steps 4-6 with each sized negative foil.

After that, repeat this experiment but make only one positive and one negative foil, made of same size

Record distance of several paper pieces (like 10) to find the distance of thickness of 1 piece of paper.

Then, change the number of paper pieces between each and measure capacitance.

Each time record capacitance.

Why this makes some sense...

The area vs capacitance graph should increase

because as the negatively charged foil

gets bigger, the more electrons it can hold at one time and be able to transfer to the positively charged foil. But, the positive plate can only accept so many, thus leading to a decimal power graph (like square root).

The distance vs capacitance graph should decrease because as the distance between the conductors increases, the attraction and force between the two foils will decrease, and the charge will not be stored and want to transfer to the positive foil as much.

Issues

The first issue involved the conductors, because aluminum foil is so hard to work with. Tears in the "plates" caused a random source of error. The fact that I couldn't place aluminum foil in between paper and attach wires to it and keep a constant pressure on it (to ensure equal distance) allowed for a the charge to take a different medium and reduce the capacitance.

The Real Equation plus Derivation

The real equation is C = ε(A/d), where ε is the permittivity of the dielectric, A is area of the plates, and d is the distance between the plates.