Day 9 - Electric Flux and Dipole moments

We simulated our own electric field with a square of nails, and the amount of nails we had to encompass within the wire determined our angle theta. We had angles ranging from 0 to 90 degrees, and the amount of nails came in multiples of 7.

Graph of # of Nails vs Angles between normal and electric field vectors; a manual fit of a sine/cosine curve works best-

Section 11.7 Activphysics - Electric Flux

Question 1: Flux Into or Out of an Oval
Design your own experiments to see how electric charge affects the flux into or out of the oval. In your experiments, you can:

• change the shape of the oval;

• move the center of the oval so that it surrounds the charge or does not surround the charge; and

• change the magnitude and sign of the electric charge.

When finished, develop in words a qualitative rule to determine the electric flux flowing into or out of the oval. Give examples to support your statements. When finished, compare your thinking with that of the Advisor.

There was a directly proportional relationship, since the flux followed the charge value of the inside of the ring consistently, whether it was zero, positive, or negative.

Question 2: Electric Flux with Two Charges
In the simulation, click the "two charges" configuration. You can now adjust the sign, magnitude and separation of two electric charges. Repeat the experiments such as done in Question 1 to see if your rule applies for this two charge system. You can:

• change the shape of the ring;

• change the position of the center of the ring (move it all over); and

• change the magnitudes and signs of the electric charges.

When finished, compare your thinking with that of the Advisor.

The dual charge examples seem very similar to the single charge examples. The flux follows the ring's inside charge whether it is positive or negative. If the charge was zero there was no flux.

Question 3: First way to determine electric flux
Change the simulation back to "one charge." The meter indicates the electric flux Φ into or out of the oval and the net electric charge Q inside the oval.

• What happens to the flux if you double or triple the positive electric charge inside the oval?

• What happens to the flux if you double or triple the negative electric charge inside the oval?

• Find an equation with a proportionality constant that relates the electric flux into or out of the oval and the electric charge inside the oval.

After answering the questions, compare your thinking with that of the Advisor.

When the charge is doubled or tripled the flux also increases accordingly. This is true for the positive and negative aspects. Thus we know that flux has a direct propotional relationship to the net charge in the ring.

Question 4: Second way to determine electric flux
The green electric field lines represent the electric field surrounding the source charges. Develop a rule for the electric flux passing out of or into the oval by counting the electric field lines passing out of or into the oval. After answering the questions, compare your thinking with that of the Advisor.

-The amount of field lines traveling out or inside of the oval/ring is proportional to the flux.

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Day 8 - Electric Fields, activphysics, and online game

ActivPhysics 11.4:Electric Field -- Point Charge

Open the electric force simulation.

Electric field and electric force:
Consider charge Q₁ = 10.0 x 10^-8 C in the simulation as the source of an electric field, just as the earth's mass is the source of a gravitational field. This electric field exerts an electric force on any charge, such as Q₂, that is placed in the field (just as the earth's gravitational field exerts a force on the mass of your body when placed in that field).

We can probe the field produced by source charge Q₁ by using a positive test charge Q₂. The field that the source charge Q₁ produces at some position points in the direction of the force

that Q₁ exerts on Q₂, divided by Q₂:

 .

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Question 1: Explore the electric field

Suppose that Ball 1 at the bottom of the simulation screen is the source of the electric field. Increase the charge of Q₂ to +4.0 x 10^-8 C. Use it to probe the electric field produced by Q₁. To see this field, grab Q₂ and move it around. Notice that if Q₂ gets closer to the location of Q₁, the magnitude of the electric force exerted on Q₂ increases--the electric field is greater. If Q₂ is moved farther away from Q₁, the magnitude of the electric force exerted on Q₂decreases--the electric field is less. After systemmatically moving Q₂ around the region surrounding Q₁, summarize in words your observations about the direction of the electric field produced by Q₁ at different points in space and how the field's magnitude varies. When finished, compare your observations to those of the Advisor.

Answer: As seen from the above picture the electric field of the two charges point away from each other. This makes sense because the two positive charges are repelling each other. As they were moved closer together the electric field increased in strength and decreased when spread further apart.

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Question 2: Field due to single positive charge
Set Q₁ = +10.0 x 10^-8 C. Use Q₂ = +4.0 x 10^-8 C as the test charge to measure the electric field due to source charge Q₁. Move Q₂ to a point 1.0 m to the right side of the source charge Q₁ (when r₁₂ = 100 cm--2.5 divisions on the screen). Observe the direction of the electric force exerted by Q₁ on Q₂. Use the force shown in the simulation

to calculate the electric field caused by Q₁ at that point. Then, compare your field calculation with that of the Advisor.

Answer: Using the given force(F12) and charge(q2) we were able to determine the magnitude of the electric field to be 900 N/C.

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Open the electric field representation simulation.

Question 3: Representing an electric field
The traditional method for representing an electric field is shown for a single point charge. You can press the pointer down on the screen to see the magnitude of the electric field  at a point. Vary the value and the sign of the charge that is the source of the field (use only one point charge for this activity) and develop some rules for the way that the electric field is represented. Consider in particular:

• the direction of the lines;

• where the lines start or end relative to the sign of the charge causing the field;

• the separation of the lines in a particular region relative to the magnitude of the field in that region;

• the magnitude of the field on a line and in the dark region next to the line, and

• the number of field lines that eminate from or terminate on a point charge relative to the magnitude and sign of that charge.

When finished, compare your observations to those of the Advisor.

 (Positive Charge)

(Negative Charge)

Answer: Based on the two above pictures we can see that when the charge is positive the electric field points away from it and when the charge is negative the electric field points towards it. Another conclusion we can see from these pictures is that as the electric charge increases (positive or negative) the density of the electric field also increases. 

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Open the electric field representation simulation.

Force on a charge in the electric field

The force  that an electric field  exerts on a charge q in that field is:

 .

In this simulation, the source charges that produce the field are positive charges distributed uniformly on one metal plate and an equal number of negative charges on the other plate. The electric field produced by these charges is represented by the green lines between the plates. (Be sure that the "Equipotential lines" button is off.) The pointer can grab the charge q and move it to different places between the plates. Before moving the charge, answer these questions.

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Question 4: Uniform Field
Explain why the word "uniform" was used in describing the electric field in the middle region between the plates. When finished, compare your thinking to that of the Advisor.

Answer: The word uniform implies that the electric field is the same everywhere between the two plates.

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Question 5: Force on a charge in a uniform field
Adjust the plate charge per unit area to 0.6 x 10^-8 C/m² and the value of the charge q between the plates to +0.5 x 10^-8 C. Suppose you place +q in the middle near the top plate . Note the arrow showing the force on q and the magnitude of that force. Is the force greater in magnitude, the same, or less if qis moved directly below its present position so that it is near the bottom plate? Justify your choice. After your prediction, move q down and compare the force on it when in this new position. When finished, compare your thinking to that of the Advisor.

Answer: The force at the top of the plate and the bottom of the plate turned out to be the same. This fits with what we expected to happen because the electric field between the two plates is uniformly distributed and spaced. This means that the electric field does not depend on its location. 

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Question 6: Force on a negative charge
If you leave q in the middle half way between the plates and change the sign of q from +0.5 x 10^-8 C to -0.5 x 10^-8 C, what happens to the force exerted by the field on q. After your prediction, adjust the charge of q to check your thinking. Try different magnitudes of q and different signs. Move q around to different places between the plates and observe the direction and magnitude of the force. Then, move the charge to the sides of the plates where the electric field is not uniform. Are your observations consistent with the equation that relates the field and the force:

 

Answer: The force stays the same whether the charge is positive or negative.

- We found the electric field from the two point charges at variable distances. This table shows us the various values for the electric fields.

In class calculations for the two point charges-

Excel data for the electric field for each element -

Our group was able to determine the value of the total electric field for the parallel axis, which was 5.965*10^5. The Y component was 1.273*10^6. The Y component is much larger because of the summation of forces in its direction.

Online Electric Hockey Game-

We were assigned to play a hockey game which would make us score a goal by orienting positive and negative charges. I was never able to play the game because my computer could not support the game and had technical issues, so here is a diagram of what one would do regarding placement of the charges if one were to play the game.

Day 7 - Electrostatic Forces

Video experiment analysis

For the video presented in class, our task was to analyze the video such that we could determine the charges of each ball in the video. We found the charge for each ball by means of Coulumb's law using the values for the distance and masses that we knew.

Using Loggerpro we were able to manually plot the movement of the ball for each specified frame. The initial starting point of the second ball was used as our frame of reference for the origin. After playing the video, we were able to approximate the values for x_1 and x_2, the distance between the ball on the rod and the origin, and the traveled distance of the hanging ball from the origin, respectively.

Above is our free body diagram done in class which we used to determine the forces affecting the hanging sphere, and thus the electric force. The resulting electric force was a nostalgic F_e=mgtan(theta).

We rewrote this equation as: F_e=mg((x_2)/(L^2-x_2^2))

We then calculated this formula in logger pro below:

Electric Force vs Separation Graph:

- Based on the graph we found the relationship to be inversely proportional.



Conclusion Questions:

1. It was evident that there is an inversely proportional relationship between the electric force and the squared distance of the two charges. The graph shows that the force of the charge decreases as distance increases, which confirms what we would intuitively think.

2. A) The percent difference of the experimental and theoretical exponent was 13.2 percent.

% difference = 1- Experimental/Theoretical = 1-1.763/2.0 = 13.2%

B) When we assumed that the charges were the same we were able to calculate the charge by manipulating Coulumb's Law, which resulted in a charge of 3.91*10^-6.
F= kq^2/r^2
q=((Fr^2)/k)^0.5 = .00000391

C) When we assumed that the charge of the hanging sphere was half that of the other sphere, we multiplied q_2 by 0.5 since the first was equal to half of the second, and then substituted q_1's new form into the equation. This resulted in values of q_1 and q_2 being 2.76*10^-8 and 5.53*10^-8, respectively.
q_1=0.5q_2
F=K(0.5q_2)^2/r^2
q_2=(2Fr^2/k) = .000000053
q_1= .0000000276

3. The charges can not be determined, since if they were both negative or both positive they would still both repel each other, and since coulumb's law has the two charges multiplied, the end result will always be positive.

4. The largest source of error/uncertainty in this lab was the fact that we had to manually trace the distances shown in the video in logger pro. Other groups likely had slightly different values compared to ours. Deviating measurement values would ultimately give different graph values for each group. A comparison of all other groups would give insight to what the proper data set should look like.

Day 6 - Diesel Cycle

Diesel Cycle Work-

We had a quiz on the diesel cycle during class. Most of us did not get it entirely correct because we did not know the correct formulas to apply. We found the efficiency to be approximately 62%.

Extra Credit - Chinese Institue of Engineers Presentation

Our class was asked to attend the program for CIE in order to get extra credit on our first exam. I was fortunate enough to see presentations from fellow Mt SAC students regarding pancreatic cancer cells, a quadcopter drone, and a 3d printer. I found the pancreatic cancer presentation to be the most interesting and technical, as the student had to reproduce the cells in an incubator.