Hi everyone, here's what happened in class on Wednesday,
But before I start, Professor Mason advised me to use humor in the blog.
This is why I am going to do the exact opposite!
Just kidding...
Anyway, here's the first thing we did: the mighty quiz:
It was a question similar to the problem of the Geiger counter that we did on mastering physics.
a) The first part was to find the electric field due to a wire inside of a cylinder if I remember correctly of potential difference V = 50000 Volts:
V = 2 k λ ln (rb/ ra)
Then we used E = (2 k λ) / r
2 k λ = E . r ( use this equation to substitute in the potential equation)
V = E . r ln (rb/ ra)
E = V / (r ln (rb/ ra)) = 50000/ (0.07 * ln (0.14/ 88*10-6) = 96891 N/C.
b) the Second part was to evaluate the charge of a dust particle if there was an electric force 10 times as the weight of that particle (I don’t really remember if it was a dust particle, but the mass was 30.5 * 10-9 Kg and that’s what matters in this problem)
W = mg = 30.5 * 10-9 * 10 = 3.05 * 10-7 N (Use g = 10 m/s2)
Fe = 10 * W = 10 * 3.05 * 10-7 = 3.05 * 10-6 N = q . E
q = F / E = 3.05 * 10-6 / 96891 = 3.15 * 10-11 C
c) The last part of the question was asking for its direction:
The answer was going towards the wall. (The answer was given inside the description of the problem)
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Now the part of the quiz was over and of course as usual, most of the students asked for a make-up quiz.
Request was denied!
Moving on to Unit 6 in our lab notebook, which is chapter 25 in the book, Professor Mason started out by explaining what an electric current is flow is.
According to Professor Mason's definition, "The rate of flow of electric charge is more commonly called electric current. If Charge is flowing through a conductor, then the official mathematical definition of the average current is giving by:
Iavg = ∆Q / ∆t
Instantaneous current is defined by I = dQ/ dt.
The Unit of current is called the ampere (A). One ampere represents the flow of one coulomb of charge through a conductor in a time interval of one second. "
According to the book's definition, "A net charge dQ flows through an area in a time dt, the current I through the area is I = dQ/dt. (Current is not a vector)"
Here's a video done by UCI addressing this issue.
http://www.physics.uci.edu/outreach/demos/electricity/foamp.php
I thought so too.. !!
So back to our labs, we had to do our first experiment: lighting a bulb.
Activity 6.3.1
a) Given a light bulb, a wire, and a battery, we had to think of 2 different ways to light a bulb. Sketch it.
b) We also had to think of two others in which the bulb doesn't light. Sketch it.
I have to say, there were marvelous drawings from most physics students.
That was a good time for some of us to think of switching majors to art.
I thought that was funny, anyway, the final conclusion was:
c) - The circuit has to be closed for the bulb to light.
- The wire has to be touching the bottom of the light bulb where electricity can travel through.
- The wire has to be touching the opposite ends of the battery simultaneously.
Activity 6.3.2
a) List some materials that allow the bulb to light:
- Keys
- Human beings
- Coins
b) List some materials that prevent the bulb from lighting:
- Rubber
- Paper
c) Categories of materials
- Metals are conductors.
- Non-metals are non-conductors.
- Semi-metals are semi-conductors.
We moved on to Activity 6.4.1 (Action of the switch)
a) The bulb socket is made of a plastic base and a piece of metal attached to the top of it in a certain shape where you can screw a bulb. It provides a conducting path for electricity to run through the metal part. The bulb must be screwed properly for it to light. If you unscrew the bulb while it is still connected to the metal piece, the circuit is open and it won't light.
b)The filament of the bulb is made of tungsten as its coefficient of linear expansion is relatively high. It is shaped in a "zigzag" way so that the two rods that are holding the filament won't break if the metal expands.
c)The light bulb will light with the switch closed (contact) as it would be a complete closed circuit, where electricity flows through all conducting materials.
d) Observations:
- The light bulb lights with the switch closed.
- The light bulb doesn't light with the switch open.
e) As current flows through the bulb, the bulb gets warmer. Some electric energy is being lost or transferred into heat energy.
f) The current path must:
- be a complete closed path.
- be made of all electric conducting material.
- produce enough energy for the bulb to glow.
I think it was time for the best part in class: THE SHORTEST BREAK FOR THE LONGEST CLASS!
Activity 6.5.1
a) We had to connect three light bulbs, a switch, a battery and a lot of wires. We had to do a circuit drawing on paper also describing this circuit. Each bulb is connected separately so that if one of the light bulbs is disconnected, the rest will still light.
b) We had to use a tunnel switch and draw it on paper also. Whichever side of the switch it is closed on, the bulb will light and the other one will not and vice versa.
c) CROSS IT OUT ! I am so happy when we get a chance to do that.
After our great drawings once again, Professor Mason explained that it was impossible for scientists to communicate in that way. That's why we had to learn the circuit diagrams, and use the electric circuit symbols indicated in the lab notebook.
Activity 6.6.1
a) The long line represents the positive terminal while the short line represents the negative terminal.
b) We had to redraw the circuits designed in Activity 6.5.1.
Activity 6.7.1
The first answer for most groups in class was C: there will be less current in the return wire.
However, the right answer was discovered later on in another activity that it was D.
The current will be the same anywhere along the wire indicated in Model D.
The current is a scalar quantity and it's never lost anywhere in the wire. It's different than the concept of some of the electrical energy being transferred to heat energy.
Activity 6.8.1
a) If I recorded this correctly, the value was + 150 mA.
b) If the leads going into and out of the ammeter are reversed, the indicator needle would go in the opposite direction trying to measure a negative value below 0 mA.
Activity 6.9.1
a) Different circuits were achieved by displacing the ammeter and drawings of the circuits were included on paper in the lab notebook. Our group only had one ammeter where we could displace it.
b) Both models seem to work as the current is not different at any location in the circuit by displacing the ammeter. We reached this conclusion by recording the values in both cases and they were consistent. This will verify our conclusion for choosing Model D in activity 6.7.1.
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Lastly, Professor Mason went over his notes to explain the current using different quantities
I = ρn q vd A
Professor Mason went over this equation explaining each quantity and especially the drift velocity by poking some students with a long meter stick.
He also explained the idea that if you have a bunch of electrons stuck next to each other in a wire and if somehow you push one of the electrons in one end, an electron will get loose from the other end and it's not the same electron! The rate at which the electrons travel is called the drift velocity.
Everything is nicely explained in his notes posted. We didn't spend too much time at that last topic as the class was almost over and we didn't go over Resistance and Ohms Law.
Quick note: Chapter 22 and 24 are skipped for now and we'll come back to them later before the test!
The class stopped at that point and we finished 10 minutes early.
Another great moment in our Physics 4B experience this semester.
I hope this post was helpful. If there's anything that I missed or gave wrongful information Please let me know and I'll fix it. Especially in Activity 6.7.1. & 6.9.1
See you Monday!
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