Flux

Above we see what would happen if an electron went in between two plates, one being positive and one being negative. We see that the electron would slowly go toward the positive plate and go away from the negative plate. On the right we see a dipole with two charges on it, and arrows illustrating an electric field. The dipole would want to straighten out since the field would cause the positive charge to go right and the negative charge to go left.

Here are a few equations we derived for torque. We found that the sum of torque is the cross product between the dipole moment vector and the electric field vector. On the bottom right we have some equations for work.

The above picture illustrates what we thought the program provided by Professor Mason would do ( in red ) and the correct path those vectors should take (in green). There is a positive charge (red small circle) and a negative charge (green small circle).

The picture above is the program Professor Mason provided modified to show how the field should look in 3 dimensions. The original program, prior to adding the Z axis, would draw the orange arrows.

The picture above shows the flow of the electric field between a positive and negative charge. Professor Mason added the blue negative charge to our illustration, and on the bottom right we show what affect that would have. The two negative charges would repulse one another.

First we found the units for flux in the top right corner then we determined the normal vectors for the objects drawn.

In the picture above we calculate flux on a box. In 3 dimensions it was hard to illustrate so we opened the box up. 4 of the sides of the box were parallel with the electric field so flux would be zero. We also see that the two sides affected by the field, colored in green, would cancel each other out, meaning that the total flux of the box would be zero.

Electric Fields

The above picture is a list that describes electric fields.

This picture shows instructions on what to do in VPython in order to recreate that picture with vectors representing the electric field.

The above picture shows my modification to the code provided by Professor Mason. This program illustrates the 3 dimensional electric field (the arrows) created by the positive charge at the center.

The picture of the graph above is what I assumed the code provided by Professor mason would result in. I was incorrect though because I forgot that the number placed in the X component does not point to a location, but in fact is a direction and length. The picture should have shows all arrows pointing to the right.

The above picture shows the calculations for finding the x and y components of an electric field vector using the formula (kq/r^3)x and (kq/r^3)y. The picture of this picture shows the magnitude of the electric field vector.

Here we found the total electric field vector for the point on the y axis. Since the charges are equal and opposite they cancel out making the total electric field vector a 0 vector.

The above picture shows our excel file which has a list of radii, their corresponding energy fields with the given k and q, then the sum of those energy fields on the bottom.

This picture is similar to the one above except that it also shows the electric field in the Y and X axes, and their summations at the bottom. Since the point that we were checking was at the center of a horizontal pole below it, the electric field cancels out.

This picture shows the integral to find the electric field across a line.

3D Computer Modeling

The above picture shows how I completed the challenge from the 3D Objects Video.

Afterward, I commented out two of the arrows with #, then I shortened the length of the remaining arrow by half and flipped it so it would point in the opposite direction. As you can see the arrow is now barely visible on the orange sphere.

Note on the Syntax Errors Video: ALWAYS CAPITALIZE BOB, it may one day save your life and the lives of your loved ones.

Above is my Chipmunk themed solution for the challenge in the Variable Assignment video.

I doubled the distance that Theodore was from the y-axis and since all of the arrows were based on variables, rather than numbers, they adjusted to still be connected. The picture also shows the printed position attribute of Simon. Alvin is usually hogging the spotlight so I figured Simon and Theodore could use this win.

Electric Charges

We assumed that the balloon would stick to the glass after being rubbed on hair and then later after being rubbed with silk. Both assumptions were correct. We tried explaining charges to children but it was a bit more difficult than we expected.

The picture above is of an experiment we did in class with tape to show that there are multiple charges. When a piece of tape was stuck to the table then removed and placed near skin there was an attraction. When two pieces of tape were each stuck to the table, then another piece was stuck to the top of each. The tops would repel tops and the bottoms would repel bottoms but when top and bottom were placed near each other they would attract. This was consistent with what the hypothesis stated, that there were two kinds of attractions and more than one charge.

In the above picture, Mario shows us his expertise in free body diagrams and the equations of forces in the x and y directions. With this we were able to derive a formula for the force.

The relation between the electric force and the distance between the objects is inversely proportional.

We use Coulomb's law to find the electric force between two objects.

This is the graph from the experiment in which we mapped out a video of one metal ball on a stick pushing another ball hanging from a string through the use of charge. The graph on the right is the position graph of both balls and the graph on the left is of electrical force vs. separation distance. 

The above picture shows my free body diagram and my Newton equations and the angle and the Electric force. This number may not be accurate because I went with an assumed value at the end of the position graph rather than an actual value.

The two pictures above show the questions from the conclusion section from the logger pro experiment we conducted above. experiment.

The above picture shows and object that transfers electric energy from the base to the silver top through the use of belts that act as insulators. The paper on the silver top is being repelled as a result.

Professor Mason performed different experiments with the device, first with a pie tin which, at first, prevented the paper from lifting, right before it was ejected into the air. Then he turned on the device again with this fan on it, and the fan spun and also prevented the paper from being repelled.

Our assumptions were that the pie tin would fly off and that the propeller would spin clockwise, both of which were true..

The ratio between the force of electricity and the force of gravity.

Entropy

Formula for entropy and 2 examples of entropy. We also found out a new way to describe adiabatic processes, isentropic, since there is no change in entropy.

This is a Stirling engine. A temperature difference from the top and bottom causes it to run.

This picture shows what happens when you place a stirling motor above heated water and place an ice cube on top. The fan is spinning as a result.

The fan spins the opposite direction when hot water is placed on top and ice is used to cool the bottom.

The above picture shows the formula for efficiency of a Carnot Engine and also the efficiency of a Stirling engine. The graph is that of Temperature vs. Entropy.

The above picture shows how to solve for the Coefficient of Performance. We used that to find the heat required to operate a heat pump to warm the inside of a home.

The above picture illustrates how to find the effectiveness of heat engines. You take the actual output of the energy desired and divide it by the output of the reversible process. We took the efficiency of the theoretical actual process and divided it by the efficiency of the theoretical reversible process to find the effectiveness. 

In the picture above we were able to show the the final temperature is equal to the root of the two provided temperatures when entropy is equal to zero.

We used what we found in the previous picture to come up with this equation.

In the picture above we found the efficiency of a process then multiplied it by the max Coefficient of Performance so we could find the possible Coefficient of Performance. Then we found the Q of heat with that number since Q of heat is equal to the Work plus the Q of cold.

In the above picture we found the time it would take to freeze a beverage in a fridge that was providing a certain amount of cool energy to its interior. We did this by finding the the energy it takes to freeze the beverage and divided it by the energy output by the fridge.

In the picture above, Professor Mason showed us that bubbles would fall to the ground after being created. This was not the case with bubbles created with methane gas though. This was because bubbles he created had air on the inside while the methane gas bubbles were less dense than the air. allowing them to float.

The video above shows what would happen if methane gas bubbles were ignited. As expected the fire rose since it consisted of methane gas, which is less dense than air.

Carnot and Otto Engines

This shows a thermoelectric device which takes heat from one cup and powers the spinning disk, and the other side acts as a cold reservoir. The disc spins in whichever direction the flow goes. If the hot cup is on the left and the cold on the right the disk will go clockwise, and visa versa.

When the sensors are plugged into a power supply one side heats up and one side cools down.

We assumed that the direction of the spinning disk would change when the cups were swapped. We also assumed the heat would flow to one side of the device while cooling the other side.

We used the formulas Q=n*Cv*dT and E=3/2*n*R*T and Cv = 3/2*R to come up with the equation of 3/2*R = Cp - R, meaning that 5/2*R = Cp.

We show that n*dT = (p*dV + V*dp)/(Cp-Cv) by first finding that Cp-Cv = R and subbing that into the ideal gas formula where there is both a change in volume and pressure.

The picture above shows the simplification of the of the previous equation to show that dp/p + (dV/V)*(Cp/Cv) = 0. With that we were able to derive another equation by moving the V's to the other side, taking the integral of both sides, then take the exponential function of both sides and we then find the equation relating the pressures and the volumes.

We derive a formula for temperatures and volumes, by subbing for p values determined from the ideal gas law(nR cancel), on the left of the board and we derive a formula for work in an adiabatic process on the right.

We used the derived formula for work in an adiabatic process to find the work in this situation.

This is a table of a Carnot cycle The top and bottom are isothermal and the left and right sides are adiabatic. We found work on the adiabatic with the formula W=PiVi^5/3[(Vf^-2/3 - Vi^-2/3)/(-2/3)] and we found work in the isothermal stages with W=nRT*ln(Vf/Vi). 5/3 is gamma and -2/3 is 1 - gamma. We were able to fill out the rest of the table since change of energy is zero in isothermal processes and Q is zero in adiabatic processes.

This device is an example of how an Otto engine works. Air is brought into the cylindrical shell when the piston is drawn down. Then the piston goes up quickly to increase pressure and reduce volume to cause the temp to rise. The spark plug then ignites causing and explosion where the piston is forced down. As that happens the exhaust valve opens and releases the burnt air and fuel. The process is then repeated.

The left side of the board shows what we could do to improve the work output of an engine and the right side of the board shows the extra degrees of freedom in a diatonic gas.