Chapters

Page 1
This text is meant to accompany class discussions. It is not everything there is to know about the basics of torque and the two conditions of equilibrium. It is meant as a prep for class. More detailed notes and examples are given in the class notes, presentations, and demonstrations (click here.)

Capacitance
Take two metal flat metal plates of equal size. Position them such that they are parallel to each other. Connect one plate to the negative side of a battery and the opposite plate to the positive side of a battery. An equal amount of charge will move off of one plate and onto the the other plate. As this happens an electric field will build between the plates. This will not continue forever. once the potential difference across the plates equals the potential difference of the battery, the charges will stop moving. Given enough time the plate that is connected to the positive side of the battery will positively charged and the side connected to the negative side will be negatively charged. The potential difference will match the potential difference of the battery.

Because these two plates have a potential difference, the charged plates are energy storage devices. This is called a parallel plate capacitor.

Capacitance is defined as the ratio of the charge on the plates, measured in Coulombs, to the potential difference, measured in Volts. Capacitance is measured in farads -abbreviated with a capital "F" because it is named in honor of Michael Faraday.

C is the capacitance [ Farads: F ]
Q is the charge on one the plates [ Coulomb: C ]
V is the potential difference across the plates [ Volts: V ]

Each plate has the same charge on it just with an opposite sign. When physicists talk about the charge on this simple capacitor, they are referring to the charge on one plate. This is the "Q" in the formula for capacitance.

Sometimes it is convenient to think of the equation above by isolating the charge on the plates.

This shows that the charge, Q, is directly proportional to the potential difference, ΔV.

 Parallel Plate Capacitor

Furthermore, for a parallel plate capacitor, the formula for determining the capacitance.

εo describes the ability to store energy in the presence of an electric field. Sometimes a polarizable material such as glass, mylar, paper, etcetera. These materials are called a dielectric and they increase the capacitance of a capacitor. The dielectric constant, K, measures how the permitivity constant is affected. A higher dielectric constant means a higher capacitance and therefore a higher amount of energy can be stored.

Every dielectric has a limit to how much energized charge is can hold back. The breakdown voltage is the amount of voltage needed to turn the normally insulating dielectric into a conductor. In a capacitor if the dielectric breakdown voltage is reached, the capacitor usually shorts out and becomes nothing more than wire. Some capacitors can explode and spray their electrolytes out. These electrolytes are dangerous and can harm your skin or blind you if they get in your eye.

 Material Dielectric Constant Breakdown Voltage (V/m) Vacuum 1 - Pyrex Glass 5.6 14 x 106 Mylar 3.1 430 x 106 Air 1 3 x 106 Teflon 2.1 60 x 106 Paper 3.7 16 x 106 Polyethylene 2.25 50 x 106

 Example 1
 Question Solution When accounting for the amount of humidity in the air, the breakdown voltage of air is between 20,000 and 70,000 V/inch. On a clear, dry, day in the winter time, it is not unusual for the air's breakdown voltage to be 50,000 V/inch. If you rub the fur of a pet and then as you touch a door knob get an spark of between your hand and the door knob that is 1/2 an inch long, then what is the potential difference between the two ends of the arc of electricity? When accounting for the amount of humidity in the air, the breakdown voltage of air is between 20,000 and 70,000 V/inch. On a clear, dry, day in the winter time, it is not unusual for the air's breakdown voltage to be 50,000 V/inch. If you rub the fur of a pet and then as you touch a door knob get an spark of between your hand and the door knob that is 1/2 an inch long, then what is the potential difference between the two ends of the arc of electricity? This question is solved using the breakdown voltage and dimensional analysis. Givens Vbreakdown =50,000V/in d = 0.5 inches

by Tony Wayne ...(If you are a teacher, please feel free to use these resources in your teaching.)