Hello everyone MrSiwWy here for today's physics scribe post, in which we actually covered quite a bit. The class began with Ms. K correcting the sheets that we had for homework over the weekend and also by reading pages 4 - 5 on the topics of Work and Kinetic Energy and Electric Potential in the handout booklet she gave to us sometime last week. This booklet's title on the cover is Grade 12 Physics: Gravitational Potential Energy.
For electric potential:
1. "Just as PE (potential energy) transforms to KE (kinetic energy) for a mass lifted against the gravitational field, the electric PE of an electric charge transforms to other forms of energy when it changes location in an electric field. When released, how does the KE acquired by each compare to the decrease in PE?"
The answer: equal (since both require work to move an object while still under the influence of the field whether it's gravitational or electric. Once work has been applied some potential energy has been transformed into kinetic energy in both cases.)
2. "Similarly, a force pushes the charge (called a test charge) closer to a charged sphere. The work done in moving the test charge is the product of the average force and the distance moved. W = FD. This work is equal to the PE of the test charge. If the test charge is released, it will be repelled and fly past the starting point. Its gain in KE at this point is equal to its decrease in PE." (note: terms in bold font are the answers since it was fill in the blanks.)
3. "complete the statements.
-Electric PE/charge has the special name Electric potential.
-Since it is measured in volts it is commonly called voltage."
4. "When a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of 2 J, its potential is = 1 V." (This is due to the fact that electric potential is the electric potential energy divided by the charge, as stated above in question 3)
5. "If a conductor connected to the terminal of a battery has a potential of 12 volts, then each coulomb of charge on the conductor has a PE of 12."
6. "If a charge of 1 C has a PE of 5000 J, its voltage is 5000 V."
7. "If a charge of 0.001 C has a PE of 5 J, its voltage is 5000 V."
8. "If a charge of 0.0001 C has a PE of 0.5 J, its voltage is 5000 V."
9. "If a rubber balloon is charged to 5000 V, and the quantity of charge on the balloon is 1 millionth coulomb (0.000001 C) then the PE of this charge is only 0.005 J.
10. "Some people get mixed up between force and pressure. Recall that pressure is force per area. Similarly, some people get mixed up between electric PE and voltage. According to this chapter, voltage is electric PE per unit charge."
Next, we moved on to a lab which consumed the majority of today's class time. The lab was essentially the assembly of a figure that embodied the main idea of increasing potential energy within an electric field (raising a charge AKA applying work to raise the charge against the electric field). The procedure and materials required for the lab are approximately similar to the following (since I don't have a green book with me right now):
Materials:
A Ruler
Clay
Steel balls (Any object can substitute for this, such as pennies in this case)
Masking tape
Procedure:
1. Insert a ruler within a molded ball of clay so that the ruler can stand upright without much independent mobility.
2. Take a 2 cm x 8 cm piece of tape and label it "3 V - 3 J/C" and repeat this process for three more equal sized pieces of tape with the labels "6 V - 6 J/C", "9 V - 9 J/C" and "12 V - 12 J/C."
3. Attach these pieces of tape to the ruler approximately at each 3 cm interval indicated on the ruler.
4. Once each piece of tape is attached to the ruler, attach the steel balls (or pennies) to each piece of tape, starting with 4 at the bottom on the 3V piece of tape, then with 3 on the 6V, and so on.
Results
Basically, we were given questions accompanying the lab as usual, and this was our overall analysis for the lab. Now essentially for the lab there was a structure which indicated that the amount of charges decreased as the voltage increased while you move farther up. If you think about the surface the structure is resting on in terms of the base of the electric field, you move the charges upwards and the voltage increases while the charges decrease. This also implicates an energy change, which can be calculated by multiplying the voltage with the amount of charges present at that level. So for this lab, the following data table was completed as the lab progressed:
For electric potential:
1. "Just as PE (potential energy) transforms to KE (kinetic energy) for a mass lifted against the gravitational field, the electric PE of an electric charge transforms to other forms of energy when it changes location in an electric field. When released, how does the KE acquired by each compare to the decrease in PE?"
The answer: equal (since both require work to move an object while still under the influence of the field whether it's gravitational or electric. Once work has been applied some potential energy has been transformed into kinetic energy in both cases.)
2. "Similarly, a force pushes the charge (called a test charge) closer to a charged sphere. The work done in moving the test charge is the product of the average force and the distance moved. W = FD. This work is equal to the PE of the test charge. If the test charge is released, it will be repelled and fly past the starting point. Its gain in KE at this point is equal to its decrease in PE." (note: terms in bold font are the answers since it was fill in the blanks.)
3. "complete the statements.
-Electric PE/charge has the special name Electric potential.
-Since it is measured in volts it is commonly called voltage."
4. "When a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of 2 J, its potential is = 1 V." (This is due to the fact that electric potential is the electric potential energy divided by the charge, as stated above in question 3)
5. "If a conductor connected to the terminal of a battery has a potential of 12 volts, then each coulomb of charge on the conductor has a PE of 12."
6. "If a charge of 1 C has a PE of 5000 J, its voltage is 5000 V."
7. "If a charge of 0.001 C has a PE of 5 J, its voltage is 5000 V."
8. "If a charge of 0.0001 C has a PE of 0.5 J, its voltage is 5000 V."
9. "If a rubber balloon is charged to 5000 V, and the quantity of charge on the balloon is 1 millionth coulomb (0.000001 C) then the PE of this charge is only 0.005 J.
10. "Some people get mixed up between force and pressure. Recall that pressure is force per area. Similarly, some people get mixed up between electric PE and voltage. According to this chapter, voltage is electric PE per unit charge."
Next, we moved on to a lab which consumed the majority of today's class time. The lab was essentially the assembly of a figure that embodied the main idea of increasing potential energy within an electric field (raising a charge AKA applying work to raise the charge against the electric field). The procedure and materials required for the lab are approximately similar to the following (since I don't have a green book with me right now):
Materials:
A Ruler
Clay
Steel balls (Any object can substitute for this, such as pennies in this case)
Masking tape
Procedure:
1. Insert a ruler within a molded ball of clay so that the ruler can stand upright without much independent mobility.
2. Take a 2 cm x 8 cm piece of tape and label it "3 V - 3 J/C" and repeat this process for three more equal sized pieces of tape with the labels "6 V - 6 J/C", "9 V - 9 J/C" and "12 V - 12 J/C."
3. Attach these pieces of tape to the ruler approximately at each 3 cm interval indicated on the ruler.
4. Once each piece of tape is attached to the ruler, attach the steel balls (or pennies) to each piece of tape, starting with 4 at the bottom on the 3V piece of tape, then with 3 on the 6V, and so on.
Results
Basically, we were given questions accompanying the lab as usual, and this was our overall analysis for the lab. Now essentially for the lab there was a structure which indicated that the amount of charges decreased as the voltage increased while you move farther up. If you think about the surface the structure is resting on in terms of the base of the electric field, you move the charges upwards and the voltage increases while the charges decrease. This also implicates an energy change, which can be calculated by multiplying the voltage with the amount of charges present at that level. So for this lab, the following data table was completed as the lab progressed:
Here are the corresponding analysis questions for the lab (accompanied by my answers):"1. How much energy is required to lift each coulomb of charge from the tabletop to the 9-V level?"
The energy required to raise one coulomb of charge from the tabletop to the 9-V level is 9 J of energy. This is due to the fact that there are 2 charges at a voltage of 9 J/C accumulatively giving an energy value of 18 J, but since the question only inquires about the energy of one charge this 18 J is divided by 2 to give 9.
"2. What is the total potential energy stored in the 9-V level?"
(Note that this is basically explaining the calculations made in the table for each energy value)The total potential energy at a given location when you raise the object further up against the electric field can be found by using the electric potential of the object / charge. Since V = PE/q, and both q and V are known, solving for PE gives PE = V*q, PE = (2 C)*(9 J/C) = 18 J.
"3. The total energy of the charges in the 6-V level is not 6 J. Explain this"
The total energy of the charges in this level is not equivalent to 6 J since there are 3 charges present at a voltage of 6 J/C, rather than just one charge at a voltage of 6 J/C. Since there are 3 charges at a voltage of 6 J/C giving an overall energy for the level as 18 J.
"4. How much energy would be given off if the charges in the 9-V level fell to the 6-V level? Explain"
Now since the charges are dropping to a lower voltage at 6 J/C from 9 J/C, an energy equivalent to 3 joules would be emitted from the system since the charges are falling from 9 to 6. * not sure about this one since I can't remember what I actually put and am not in the mentality to reason out the answers momentarily.
Application question
"1. A 9-V battery is very small. A 12-V car battery is very big. Use your model to help explain why two 9-V batteries would not start your car."
Now if you think about it, the more charges that are present then the higher the energy value for that level. Now since these two cases involve voltages that are quite similar, there are obviously much more charges found in the large car battery than in the small 9-V battery, therefore allowing for quite a bit more energy available for use by applying the battery correctly.
Once all the labs were in, Ms. K went over the answers to the other sheet we were given on friday. The answers to the questions are as follows:
1. V = 31 V
2. a] PE = 3.6 x 10-14 J
2. b] V = 180 V
3. 1.9 x 10 7 m/s
4. 3.2 x 10 -9 J
5. a] w = 1.92 x 10 -18 J = 12 eV
5. b] E = 1.92 x 10 -18 J = 12 eV
5. c] V = 2.0 x 10 6 m/s
Once she finished putting that up, we were let to work on our own on either finishing up the lab or to finish the sheets she put out in the front of class. Jeez, I wish I could've put more so that anyone who doesn't quite understand this unit yet or is yearning for some help with their tribulations concerning problems in this unit. Now as for our homework, the sheets that were up front for pick up are probably going to corrected tomorrowed, and by the way, the test is on friday. Good night everyone and the next scribe for tomorrow's class will be Anthony!
1 comment:
Great post Chris. I was wondering about the results section of the lab with regards to the comment on as voltage increases, charges decrease.
I don't understand what you mean here.
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