4/16/15 - Electric Potential Cont.

Today we start off with an example. Given a ring of uniform charge Q, we were asked to find the electric potential at points around the ring. We found that the radius was the only thing changing and it was the square root of the distances, or the hypotenuse of triangles.

Method 1: We split the charges up into segments and summed up their effect on the point.

Method 2: We substituted in the radius and found the electric potential as a function of the angle between the point and the bottom of the ring.

Method 3: We too the integral in the X direction of the electric field on that point and found a formula for electric potential.

Using the methods that we just derived formulas for, we attempt to find the electric potential at a given point above a uniformly charged rod.

Method 1: We broke the rod into 16 segments and used superposition to find the total electric potential at the point.

Method 2: We used an integral to find the electric potential using a differential radius.

Method 3: We integrate the electric field with the differential radius and found the electric potential.

In the end, we used 3 methods to find the same answer and, thus, reinforcing our understanding of electric potential.

Here we have different electric fields and we find different lines that have the same potential along the field. The parallel lines and point charge are simple enough, but when there are two point charges, the field lines bend and more interesting surfaces are created.

A web simulation of the above that Professor Mason showed us.

A set of point charges is set up next to each other.

Here we see their electric field.

And here we see their electric potential alongside the electric field.

Here is a negative point charge and the electric field associated with it.

A set of 2D lines with equal electric potential.

This is a piece of conductive paper, we hook up the voltage supply to two points on the paper and measure the electric potential relative the the negative point. We find a linear relationship between the two points and as shown below, we have a table that records all the values taken during our experiment with a multi-meter.

4/14/15 - Electric Potential

This class session starts off similar to last weeks where we are given batteries, wires, and light bulbs. We found that putting them in different configurations yielded different results in the brightness of the bulbs. If the bulbs are in series, then the bulbs are dimmer than if they were in parallel configuration.

We take our well known induction heater that we know to have an output of about 300 Watts out of a wall socket and hook it up to a voltage supply. We then heat a small bowl of water and record its temperature change in LoggerPro.

We set a linear fit to the graph and, using the slope, we found the power of the induction heater.

Second run with the voltage doubled, we can clearly see that the power is not just doubled from the voltage being doubled.

Since we have Power = Voltage x Current and we also have Voltage = Current x Resistance, we found that though the voltage double, the resistance stays about the same; then we know that the current is also doubled and therefore, the power should be almost quadruped. The graph doesn't quite should a factor of four, but it explain how it would increase given only a change of the voltage.

Professor Mason testing the resistance of skin on students, we found Robert to not be very resistant.

Now given our resistance rating, we found how much current would actually go through us if we were shocked by a standard 120V wall socket using the formula V = IR.

Now if there was an electric field and we wanted to find the work to get from one point to another, we can see that the work done is path independent, and the same amount of work will be done regardless of the path taken.

Now we rated the work done by each path the particle takes. Depending on the cosine of the angle the path makes with the electric field, it determines the amount of work done by the movement.

Given our general formula for work, Force x Distance, we can substitute in out electric field and find an integral that we can modify to find voltage.

Here we have the general form for electric potential from an infinite distance to a distance r. The diagram depicts a charge with a lines perpendicular to the electric field.

Now using this formula and superposition, we can find the net electric potential at any given point such as the example above. There we have two point charges and two observation locations, we find the potential at each point.

We were also assigned a VPython project that's due in a week, Professor Mason kindly corrected our definition of plagiarism for us.

Here's what Mario actually wrote.

4/9/15 - Introduction to Circuts

A simple question, can you make this light bulb turn on with a battery, a wire, and nothing else? Well, we sort of got it. The class starts off by allowing the students attempt this task. After successfully lighting it up, we concluded a few things on the whiteboard.

More pictures of lighting the bulb up.

Two circuits that will make the bulb light up and two that don't. The two that work have the the bottom of the bulb touching the battery and the wire connecting the side of the bulb.

Two two that don't work doesn't finish the connection back to the battery and the second doesn't allow the charge to pass through the filament.

This is an electroscope. It has a conducting rod on top that can accept charge and will allow it to pass downward to two metal plates. Professor Mason touches the rod with a charged rod and gives the rod a charge. The charge flows down and, since there are two plates, allows each plate to repel one another as shown in the picture above.

Given these observations, we predict what would happen if the top of the battery was applied to the rod. Our prediction was incorrect, when the battery was applied on both sides, nothing had happened.

We also explain why the bulb has to be connected the way it was in order to work properly. The charge has to have a steady rate of flow in order the work to be done.

In this picture, we are given the analogy of water to better understand current. We find two aspects that would increase this water flow in a given amount of time. The two we came up with for a waterfall were the height of the waterfall and the amount of water flowing through.

This is analogous to the Voltage and Current of the power flow.

Further down, we are given that the power is Voltage x Current which has units of Joules / Coulomb and Coulombs / Second - canceling out to Joules / Second.

Here, we hook up our original setup to a amp-meter. Ours was a little faulty, but we found the average amperage going through our circuit according to the meter.

The measured values are with the meter hooked up before the bulb and afterwards. We find that there is almost no change in the current so there is no current lost after flowing though the bulb.

Here we define the four things that we should know in order to find the charge, and their common symbols used.

Given the formula for current, we're asked to find the drift velocity of the electrons passing through the wire. By rearranging the formula to use the givens, we find that the drift velocity is relatively small compared to speeds we normally see in mechanical physics.

Here is a induction coil from a toaster, it's hooked up to a voltage supply and a watt-meter.

The graph shows the relation as voltage was changed during the experiment, Through the graphs we can see that voltage and current have a generally linear relationship.

Our predicted graph before the experiment, we thought it would have a linear slope with an inital vlaue.

Since our itinal prediction was correct, we found the slope to be voltage over current and a differential version as well.

We then set resistance equal to this slope and found the units of Ohms, being voltage over current.

Here we try different coils of resistance made of different materials and length. We connect them to a multi-meter set for ohms.

After a various amount of tests with diffident lengths and materials, we find a general formula for resistance. We found that the resistance is proportional to the length and inversely related to the area. Given these, we set the resistance equal to the \rho density of the material, length of the segment and all over the area of a cross-section.

3/31/15 - Flux cont.

Here we start off the class by using a computer simulation to visualize the flux created by some point charges. We then attempted to copy what we saw onto the whiteboard and, given our knowledge of flux at the time, we can indicate the direction of each line.

The red are arbitrary surfaces that we made to enclose some charge, given the direction of the flux lines, we summed up the net flux in and out of each surface.

Therefore, we conclude a statement about the flux relating to the net charge enclosed.

Here we have a metal cylinder hooked up to a Van der Gaff generator, since the attachment is a conductor, all the charge is diverted to the metal cylinder. However, since the cylinder is a conductor itself the charge enclosed in the cylinder is zero. We will see the results of this hypothesis from the aluminum foil that is attached at the sides of the cylinder.

As by the picture, there is an electric field on the outside of the cylinder, but not on the inside. The foil on the inside is stationary while the foil on the outside is experiencing a force due to the electric field.

As clearly indicated by our in-class prediction, our group was wrong about what would happen to the foil when the charge started flowing. We had forgotten that there would be no electric field inside the cylinder.

With our original statement that flux is related to the charge enclosed, we derive the general formula for flux. Since the electric field is normal to all points on a surface, we find that the cosine factor simply becomes 1. We can then write a generic integral and equate it Gauss' law for flux.

Given charges that are allowed to flow onto a conducting ring, the charges will naturally tend to repel each other and move the furthest away from each other. Maximum distance is achieved at the very edge of the ring and equal-distant from each other.

A conceptual question where we are given the condition of a lightning storm and options where the safest location to be in is. We concluded that simply staying in the car would be the optimal choice due to the insulating factor of the car. The charge that flows onto the car will be dissipated around the body of the car and not into the space inside.

Other options were far less..conventional.

On the left-hand side, we find the factor of change of circular and spherical objects when we modify the radius of said object. We find that it's proportional to the power of the radius on the equation.

We then relate these findings to the charge enclosed in a sphere. Given charge density and a factor of change in the radius, we can find a proportional relationship between the charges.

In this picture, we derive Gauss' law through the general flux formula. We use the standard equation for the electric field and find the surface area of a sphere. The surface area cancels out some terms and we are left with Gauss' law.

From out previous calculations, we wrote our proportionality scalar in terms of variables. We then substituted back into the general formula for an electric field and found a formula to find the electric field given a change in radius and a known charge.

Along with the electric field, we also found a scalar for the charge enclosed in a cylinder based on the volume of the cylinder and a change of radius in the cylinder.

We also found a simplified formula for the sum of surface area of a cylinder on the bottom.

If we have a uniform electric field, we can solve for this field by equating the general flux formula to Gauss' law. Since the electric field is constant, we can factor it out and we're left with the integral of the surface area. We apply this to the cylinder problem we had previously and found an equation to find the electric field at any radius given the charge density.

Applying our proportionality scalar equation and the electric field equation we had just derived, we found a formula to find the electric field at any radius, even inside the cylinder.

Finally, to conclude the day, we wrote a general equation with a constant for gravitation. We then plugged in the numbers, and found our gravitational constant at the Earth's surface.