10-13-14 potential of continuous charge distribution

This shows that V=k int(dq/r)=k int(dq/(x^2+a^2)^1/2)
replacing the r with an angular radius. as well the integration is for q and so the denominator can be pulled out of the integral and it become kq/(x^2+a^2)^1/2
BUT
dq=lambdadx
this way you can integrate along a rod instead of a disk
YAY!!!!1!

The potential from a continuous charge distribution can be calculated several ways.  Each method should yield approximately the same result.  First, we can use an integral method in which the potential dV from each element of charge dq is integrated mathematically to give a total potential at the location of interest.  Second, we can approximate the value of the potential V by summing up several finite elements of charge Dq by using a computer spreadsheet or hand calculations.  Finally, we can use Gauss’ law to find the electric field along with the defining equation for potential difference to set up the appropriate line integral.

            Again, let’s consider a relatively simple charge distribution.  In this case we will look at a ring with charge uniformly distributed on it.  We will calculate the potential on the axis passing through the center of the ring as shown in the diagram below.  (Later on you could find the potential difference from a disk or a sheet of charge by considering a collection of nested rings.)

There are several ways that you could evaluate this integral. If you have a program like Wolfram Alpha or Maple, you could get the symbolic solution directly. For example using wolfram Alpha you get the following
One way to see if our answer is in the correct ball park is to imagine that we replace the line charge with a point charge at the center of the rod. What is the equivalent magnitude of the point charge, given that it has to be equal to the entire charge of the rod

This is a huge mess.

10-9-14 Power and Potential

This is a drawing of a circuit. These ancient pieces of art reflect early understandings of parallel and series circuits. It includes how to draw a switch in the lower corner as well as a bulb, a resistor, and a power supply. Truly the people who drew these were wise beyond their years.

This circuit is running power through a coil and then into a current sensor using logger pro to collect data. the coil can then be used to heat up water in a cup and the transfer of heat between the coil and the water will be an added amount of resistance with is relative to the temperature of the water.

Because of the water, the slope of the graph is not ohmic. Therefor the constant is no that of resistance and the equation ust be looked at differently.

This equation is more specific for the relationship between the change in temperature and the resistance.
Alternate forms of Joule's Law for Ohmic Materials: V = I R

P=VI=V^2/R=I^2R

• Like gravity the electric force is a conservative force. Therefore it should be possible to talk about an electrical potential energy in the same way we talk about a gravitational potential energy. This is not surprising because of the similarity between symbolic structure of the gravitational force and the electrical force. 

This is the conversion of potential energy due to a difference in location of particle charges.

This is the connection between work and voltage. It takes positive external work to move a positive charge into a reion of higher voltage. In this case, you are doing external work against the electrc force field. Force fields are real. Likewise, it takes positive external work to move a negative charge into a region of lower voltage.
Work is path independent. the work to go from point A to point B is the same no matter which path is taken to go from A to B.

10-7-14 Current and Resistance

This is a simple application which is done by people with doctorates in Electrical Engineering from Harvard cannot do. One wire, one battery, and one light bulb. recognizing how to complete the circuit by attaching signal and ground

This box detects electric fields.

This picture reflects in laymen terms, what current is. It shows a closed loop and the position of electrons in the battery and the movement of energy and charge throughout the wires.

This is all that is required to make the light bulb shine bright like a diamond. A diamond with a light on it of course. 

This is the derivation of power. P=IV. The voltage that is carried in a current equates to the amount of joules per second, or the power of the process.

This picture is disturbing. The excercise is to show that the current is the same throughout. I have no idea how these inequalities could work out. This dumbfounded and in all honesty hurts my brain to look at.

These are a collection of ammeters. The are implemented into a circuit and measure the current. Current is measured in Amperes, which is a measurement Coulombs per second, or the change in charge over the change in time. Another equation for currant is that I=pqva. where v is the drift velocity, measured in meters/second. p is the number density, or the number of charges per unit volume (#/m^3). q is the magnitude of the charge, for example the magnitude of the charge on an electron is 1.602x10^-19. and finally a is cross sectional area perpendicular to the flow, measured in m^2. this is often known as a gauge for the wire.

This circuit uses an ammeter to measure the current, as well as a switch to open and close the circuit. It shows that without a complete loop there is no current and no voltage gets to any of the hardware.

This board is looked at a wire. it is measure charge relative to the volume of the wire and the number of charge carriers in the wire. It uses the conversion for current and implements it into the charge  and then calculates the drift velocity.

This graph shows the relationship between current and voltage. This relationship can be put in the form y=slope x as well it crosses through the origin, which is good.

In this exercise the slope of the line reflects well in OHM's law. The V=IR. This means that the slope of the line is the resistance, this is the resistance in the wire. 

• When a potential difference V is applied across the ends of a wire a steady current I quickly begins to flow. In different materials the amount of current that flows will be different even if the voltage is the same.
• The scattering of charge carriers off impurities, imperfections, and vibrations of the atomic lattice creates the resistance to the flow of charges in the wire.

These are three different positions and measuring the resistance in the wires. All of them behave ohmicly and show the relationship V=IR.

This is Georg Simon Ohm, he discovered Ohm's Law. He is an all around badass.

This is Ohm's Law. 

9_30_14 Gauss' law

HEY PROFESSOR MASON

9_25_2014 Electric Flux

This equation shows the similarity between charge and projectile motion.

This is a problem looking at the torque on two particles in a field. 

This display is showing the field between to charges.

This is the basic formula for flux.

These are the conceptual rules that apply to flux.

9_23_2014 Gauss Law

These are general standards that are found when looking at E fields.

These problems pertain to a particle in between to bars in a charged field. They are problems one concepts from activphysics, provided by pearson.

There are two particles equidistant from a particle, which is a small number above the line between them. Both have opposing charges, so one pushes it up while one pulls it down, therefor it has no direction in the j-hat direction. Because they opposite charges, but they both posses the same magnitude of charge, the force is just doubled in the direction of the negative charge.

This is a simple problem in which multiple points are looked at for the field effect of the charges in those locations.

The above problem was solved using a formula in which you can choose any point and then adjust its relative location to a charge and then sum the values to calculate the Field at that location.

This bar has a uniform charge through out, what is the effect of the field cause by the bar on the point .05 meters above the center of the bar? What's wrong? you don't know? Don't worry, I have the data tabulated on the next slide so you can see the parts and the formula is printed on the board above.

This is used as though the charge is at one end of the bar, instead of directly over the bar. and then it is recalculated with at the point 0.05 m above the middle of the poll.

This is the effects of the field using smaller dx. and then after words (after A21) looking at the effect on random points to find the magnitude of the E along 1 dimensions.

I am the master of this game, there is no one better than me.

After completing this, I was emailed by IEEE about a research position. They would research me and try to recreate me.