vPython Activity

Here is the vPython activity that we were asked to do at home. The only main difference between this and the provided code was that this one required a loop. The condition was that the loop could only run when the loop is less than 2pi. I incremented it at pi/18 so that it would display a total of 36 electric charges. 

4/16 Electric Potential

We derived equations that would calculate the electric potential at different points. We calculated first when the point is a distance x from the center of the ring. We found that we could use the radius and the distance x to calculate the distance from the point to the ring at all parts of the ring. We then calculated the electric potential at a point x away from the top of the ring. This was more difficult because we did not have a constant distance away from the ring at all points on the ring. We had to use trig to come up with a way to sub in for the distance from the point to the ring.

We calculated for the electric potential a different way by using the integral of E * dA. We calculated for the electric potential of when the point is a distance x from the center of the ring and found that we ended up with the same answer as the one we did when he used V = kq/r to derive the answer.

We did the same problem except we used excel to calculate the numerical values of the electric potential at each segment of the ring. Since there were 20 segments, we came up with 20 different values for the electric potential at each segment of the ring. 

We are given a a problem of a point that is above a rod (not above its center) and were asked to find the electric potential of the point on any part of the rod. The distance from the point to the rod was not uniform and ended up with a radical in the denominator that could not be integrated without using integration techniques.

We did an experiment with a multimeter to find the difference in electric potential from a negative charge to another point on the conductive paper. We used the voltage meter to see that the electric potential between the negative and positive was 15V. We then grabbed the two ends of the wire and put them on the conductive paper to see what the electric difference was when the two points were at different distances.

We recorded the electric potential at the different points on the table to the right and answered the questions.

4/14 Electric Potential Energy and Work

We designed a closed circuit that resulted in the two light bulbs to be the brightest. We found that putting the batteries next to each other and then connecting one light bulb to the other via direct contact was the best way to achieve this.

This is how we designed a closed circuit to yield extremely dim light bulbs. To do this, we put the batteries parallel to each other so that there wouldn't be as strong of an attraction in the wire and we made use of the lengths of the wires so that it would give us a dimmer light.

We drew the pictures of the ways you can design the closed circuit to have the dimmest light bulbs and the brightest light bulbs. We also learned how to properly draw the loop below the sketch with the proper notations. The different notations are on the top right showing the correct way to draw a battery, light bulb, and resistor.

We did an experiment where we heated a cup of water with a water heater that was connected to a voltage meter. This way we could see how much voltage the heater was using to get the water to a certain temperature.

This shows the temperature vs time graph of the experiment. 

We repeated the experiment except this time we doubled the voltage of the heater. The blue line is the new graph and the black line is from the previous run. We see that the increase in voltage lead to an increase in the rate at which temperature increased.

We were asked to determine the relationship of voltage and temperature given that the experiment showed that more voltage results in increased rate of temperature increase. We also said that we needed 3 things to calculate this relationship: mass of the water, power, and time.

We calculated the amount of work done on the scenario depicted on the top left. The scenario was how much work is needed to move the car from a -> d -> c, a -> b -> c, and a -> b on a 30 degree incline. We calculated that the work done for all 3 scenarios is the same.

We derived a formula to calculate the work needed to move between any two points on an electric field. We symbolically represented this as D. Since point A is parallel to the electric field, it did the least amount of work to move across the electric field. Point B is perpendicular to the electric field so it did the most work to move along its dotted line. Point C goes diagonally through the electric field so it does the 2nd most work out of all the points.

We drew what equal potential lines would look like on a positive point charge. At every intersection, the two lines make a perpendicular angle. We calculate on the left what the change in voltage would be from an infinitely large distance to r. The integral works out so that infinity ends up below a fraction which makes our answer kq/r. 

We drew how we thought the provided python code would look like when we ran it. Our predictions were correct.

We were asked to add another charge and electric potential of the charge to the provided code. The new charge is the one in yellow and the green dot closest to the yellow circle is the electric potential.

We calculated the net electric potential and resulted in zero because the charges are opposite.

We were asked to explain the difference between plagiarism and collaboration. Since the school did not have a moral code, Mason took it upon himself to teach us the difference between plagiarism and collaboration.  

4/9 Current, Amp, and Resistance

We are given the task of producing light using batteries, a wire, and a light bulb. We determined that the way to light the bulb was to have the wire be at 1 positive and 1 negative end of the batteries. This way, the bulb can receive a current of charge that can be used to power the bulb. It does not matter if the wire goes from positive to negative or vice versa.

We are then asked to strengthen the light emitting from the light bulb using a second battery. The loop must go from positive to negative for there to be current so we stack the batteries so that both batteries can increase the amp in the current.

This is an electroscope we use to tell if an object has charge. If an object has charge, the two plates inside the box will move.

We came up with 2 ways to successfully light the bulb and 2 ways to incorrectly light the bulb using the battery, wire, and bulb.

We were asked what would happen if the positive side of the battery were to touch the electroscope and we predicted that nothing would happen. Our predictions were correct because in order for there to be charge, the circuit must be a closed loop.

We used a milliamp to observe if there was energy lost when the circuit lights a light bulb and when it doesn't. We observed that our milliamp returned the same number so there was no energy lost during the closed loop.

We were asked to come up with 4 things that are required to find the charge. We then used a formula consisting of all 4 variables and used it to calculate drift velocity given the cross-sectional area, number density, and the amps.

We performed an experiment in class to find the relationship between potential energy and current. We determined that their are proportional because their graphs are both linear.

We predicted the graphs we would yield from the experiment and were correct on thinking that the graphs were going to behave linearly.

We made a table to see the relationship between the length of a wire, the material used, and the diameter of the wire. We determined that all 3 variables contribute to the resistance of the wire.

3/31 Gauss' Law

We drew the flux of 3 charges: 1 negative charge and 2 positive charges. We calculated the flux but did not divide out the epsilon not because it is a constant.

We related charge to flux by saying charge is equal to flux times a constant k and flux is equal to the closed integral of Ecostheta with respect to area. After drawing the relationship, we observed the behavior of their graph to see what kind of value we should yield for k.

This is a metal cylinder with plastic tape hanging off the inside and outside of the cylinder. This picture is the cylinder without charge. 

This is the what happens when a current is run through the metal material and it can be observed that the plastic on the outside of the cylinder is repelled and the plastic inside the cylinder remains the same.

This is our predictions for the lab and we were correct in predicting that nothing will happen to the plastic on the inside of the cylinder when there is a current run through the metal.

We found the relationship of how area, volume, and circumference change when radius changes. From this, we can derive a relationship between the change in volume (dV) and change in radius (dr).

We calculated the flux of an electric field going through a plane. Since the angle at which the electric field is perpendicular to the plane, we can say cosine will be 1 and are left with EdA which we then substitute using the dV formula found in the previous picture.

We use the gravitational force formula to draw comparisons to the Gauss' Law equation.

We are asked the question "Where is the safest place to be during a lightning storm". We chose the option to stay in your car because the rubber tires would act as insulators preventing the passengers inside the vehicle from being shocked.

We calculated for the surface area of a cylinder in order to derive the formula for flux in the next picture.

Using the calculated dA from the previous picture, we were able to derive a formula that calculates the electric field inside a cylinder.

Mason demonstrates how the electric field created by the microwave effects various metal objects when exposed to it.