Download Lesson 5: The Parallel Plate System

Outcome 2 Lesson 5: The Parallel Plate System Lesson 5: The Parallel Plate System Review questions: 1. An electron with a speed of 5.0x105 m/s is decelerated through a potential difference to a speed of 1.0x105 m/s. What is the potential difference? (answer: 6.8x10-­‐1 V) 2. An alpha particle traveling 7.15x104 m/s is accelerated (speeds up) through a potential difference with a magnitude of 15.3 V. What is the final speed of the alpha particle? (answer: 8.11x104 m/s) 3. What is the speed of an electron that has 13.3 eV of kinetic energy? (answer: 2.16x106 m/s) Part 1: Introduction to the parallel plate system Activity: Field in a parallel plate capacitor 1. Go to goo.gl/h2hRtD 2. Turn off the audio intro by clicking on amber button at the bottom of the screen 3. Hover the mouse over the points shown in the diagram below and record the electric field at each location. Electric field 1. 2. 3. 4. 5. 6. 7. 8. 9. Question 1: What do you notice about the electric field at points between the plates versus points outside the plates? Outcome 2 Lesson 5: The Parallel Plate System This simulation demonstrates the experimental behavior of the parallel plate system, often referred to as a parallel plate capacitor. The parallel plate system has two flat metal conductors held parallel to each other a distance apart. One plate is given a positive charge and the other plate is given a negative charge. In Physics 30, we use a formula based on an ideal situation where the plates are very long and quite close together. In this situation, you can calculate the magnitude of the electric field between the plates using: 𝐸=
|∆!|
|∆!|
where ∆𝑉 is the electric potential difference between any two points inside the plates and ∆𝑑 is the distance between these two points. Question 2: Use the values from the simulation to calculate the expected electric field using the formula above. The plate separation distance is your ∆𝑑 and the electric potential difference between the plates is your ∆𝑉. Does this calculated value match the values you measured? Can you explain this result? Part 2: The Parallel Plate System Concepts and Formulas Example: For the parallel plate system below, the electric field between the plates is 35 N/C. a) What is the voltage (electric potential) of the battery? b) What is the electric potential difference between the positive plate and a point 3.5 cm from the positive plate? Outcome 2 Lesson 5: The Parallel Plate System Example: A Ping-­‐Pong ball of mass 2.7 x 10-­‐4 kg is hanging from a light thread 1.0 m long, between 2 parallel plates 10 cm apart (as shown to the right). When the potential difference across the plates is 720 V, the ball comes to equilibrium 1.0 cm to one side of its original position. What is the charge on the ball? Part 3: Three Common Parallel Plate Problems Activity: The Spandex Sheet Strikes Again 1. Model a parallel plate system using a spandex sheet and two meter sticks: Place each meter stick under the stretched fabric. Place one meter stick at desk height and raise the other about 50 cm above. This should create a flat incline between the two sticks representing the constant electric field. 2. The higher meter stick represents the positive plate and the lower meter stick represents the negative plate. Use chalk to draw electric field lines on the fabric. 3. Sketch the movement of the marble (positive charge) in each of the three situations: Marble starts with an initial speed against the electric field 4.
Marble starts with an initial speed with the electric field Describe the behavior of the marble in each of the three situations: Marble starts with an initial speed perpendicular to the electric field Outcome 2 Lesson 5: The Parallel Plate System Problem 1: Charged particle starts out with an initial velocity parallel to the plates Example: An electron is shot through a hole in the top plate of the parallel plate system shown with an initial velocity of 5.3x106 m/s. What is the velocity of the electron when it hits the bottom plate? Step 1: Determine the potential difference the particle will be travelling through. Step 2: Determine magnitude of the change in kinetic energy for the particle Step 3: Will the particle speed up or slow down? If it speeds up ∆𝐸! is positive, and if it slows down ∆𝐸! is negative. Step 4: Calculate the final speed of the particle. Example: A proton is shot through a hole in the top plate of the parallel plate system shown with an initial velocity of 1.7x103 m/s. The electric field between the plates is 198 N/C. What is the velocity of the proton when it hits the bottom plate? Outcome 2 Lesson 5: The Parallel Plate System Problem 2: Charged particle starts out with an initial velocity perpendicular to the electric field Example: Assuming that electrons enter the following parallel plate with a horizontal velocity of 3.20 x 106 m/s, how far will the electrons travel horizontally before they crash into the positive plate? Step 1: Separate the problem into the x-­‐direction and the y-­‐direction and fill in the known and unknown variables in each direction. x-­‐direction There is no force in the x-­‐direction so the electron will travel at a constant velocity. y-­‐direction The electric force is in the y-­‐direction so there will be uniform acceleration as the electron moves towards the positive plate. Step 2: Determine how long the particle will move in the accelerated direction. Step 3: Use the time you found in step 2 to solve the problem in the constant speed direction. Outcome 2 Lesson 5: The Parallel Plate System Example: Two horizontal plates are separated by a distance of 4.00 cm. The electric potential between the plates is 120 V. A horizontal beam of electrons, with a speed of 6.50 x 106 m/s, is directed into the electric field between the plates. The electrons enter the field 1.00 cm above the negative plate. + + + + + + + + + + + + + + + + + + + + + + + + 3.00 cm Electron beam 1.00 cm – – – – – – – – – – – – – – – – – – – – – – – – – Draw the resulting path of the electron beam. Determine the horizontal distance that the electrons travel before striking the positive plate. (The diagram is not drawn to scale.) (6.94 cm) Outcome 2 Lesson 5: The Parallel Plate System Lesson 5 Summary Notes •
•
Hand in instructions: Take a picture or scan (using a scanning app) of this page when completed and submit the image file or
pdf to the dropbox folder for Outcome 2 Summary Notes on D2L.
Fill in everything you have learned about the following topics using your lesson 5 booklet
Draw and label the parallel plate system Describe the electric field between charged parallel plates and give the formula used to calculate it. Describe how the electric potential varies between the plates of a parallel plate system. How do you calculate the electric potential difference between two points between the plates? Common problem 1: A charged particle starts out with an initial velocity parallel to the electric field. Describe two situations where the particle will speed up. Explain why and give the formula(s) you would use to calculate the final velocity of the particle. Describe two situations where the particle will slow down. Explain why and give the formula(s) you would use to calculate the final velocity of the particle. Common problem 2: A charged particle starts out with an initial velocity perpendicular to the electric field. Draw a diagram of this situation Describe the steps involved to calculate how far the particle travels along the plates assuming all other relevant information is given. Outcome 2 Lesson 5: The Parallel Plate System Practice Problems 1. An electron moves between two positions with a potential difference of 4.00 x 104 V. Determine the
electric potential energy gained by the electron, in joules (J) and electron volts (eV). [6.40 x 10-15 J or
4.00 x 104 eV]
2. What maximum speed will an alpha particle reach if it moves from rest through a potential difference of
6500 V? (7.9 x 105 m/s)
3. Consider the following diagram of
equipotential lines
A.
What energy is required to move
an alpha particle from E to C?
B.
If an electron were placed at D,
what would its maximum speed
be at B?
C.
What is the energy required to
move a proton from C to A?
D.
What is the energy required to move a neutron from E to A?
2
4. The electric field strength between two parallel plates is 9.3 x 10 V/m when the plates are 7.0 cm
apart. What would the electric field strength be if the plates were 5.0 cm apart? [1.3 x 103 V/m]
5. An electron is released from rest adjacent to the negative plate in a parallel plate apparatus. A potential
difference of 500 V is maintained between the plates, and they are in a vacuum. With what speed does
the electron collide with the positive plate? (1.3 x 107 m/s)
6. An electron with a velocity of 5.0 x 106 m/s is injected into a parallel plate apparatus through a hole in
the positive plate. It moves across the vacuum between the plates, colliding with the negative plate at
1.0 x 106 m/s. What is the potential difference between the plates? (68 V)
7. Assuming that electrons enter the
following parallel plate with a
horizontal velocity of 3.20 x 106 m/ s,
how far will the electrons travel
horizontally before they crash into
the positive plate? [5.58 x 10-2 m]
8. A proton is placed in an electric field between two parallel plates. If the plates are 6.0 cm apart and have a
1
potential difference between them of 7.5 x 10 V, how much work is done against the electric field when the proton
is moved 3.0 cm parallel to the plates? [0]
9. In the previous question, how much work would be done against the electric field if the proton was moved 3.0 cm
perpendicular to the plates? [6.0 x 10-18 J]
Outcome 2 Lesson 5: The Parallel Plate System 10.
[1.5x10-16J]
11. An electron is accelerated through a potential difference of 500 V. With what speed does the electron collide with
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the positive plate? [1.3 x 10 m/s]
12. What potential difference would accelerate an alpha particle from rest to a kinetic energy of 1.9 x10
3
[5.9 x 10 V]
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J?
13. With what velocity would an alpha particle reach the negative plate of a parallel plate apparatus with a potential
difference of 2.0 x 103 V, if it started at rest,
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a) next to the positive plate [4.4 x 10 m/s]
5
b) at the midpoint between the plates [3.1 x 10 m/s]
14. The electric field strength between two parallel plates is 930 V/m when the plates are 7.0 cm apart What is the
electric field strength when the plates are moved to a point where they are 5.0 cm apart? [1302 V/m]
15. An alpha particle with an initial speed of 7.15 x 104 m/s enters through a hole in the positive plate between two
plates that are 0.090 m apart as. If the electric field between the two plates is 170 N/C, what is the speed of the
4
alpha particle when it reaches the negative plate? [8.11 x 10 m/s]
16. An alpha particle is placed between two parallel plates set 4.0 cm apart with a potential difference of 7 500 V
5
across them. What is the maximum speed that the alpha particle could achieve in this field? [8.5 x 10 m/s]
17. A proton is placed in an electric field between two parallel plates set 6.0 cm apart with a potential difference of 75
V across the plates.
A. How much work is done if the proton is moved 3.0 cm parallel to the plates? (0)
B. How much work is done against the electric field if the proton is moved 3.0 cm perpendicular to the
-18
plates? (6.0 x 10 J)
C. If the proton is released at the positive plate, what will its speed be when it impacts on the negative
5
plate? (1.2 x 10 m/s)
Outcome 2 Lesson 5: The Parallel Plate System 18. An alpha particle is placed at the positive side of two parallel plates. The electric field in between the plates is 1.50
x 103 N/C. If it takes the alpha particle 1.2 x 10-6 s for the alpha particle to reach the negative side, what is the
separation of the two plates? [0.052 m] What is the potential difference between the plates? [78 V]
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19. A metallic Ping-Pong ball, of mass 0.10 g, has a charge of 5.0 x 10 C. What
potential difference, across a large parallel plate apparatus of separation 25 cm, would
be required to keep the ball stationary? (49 V)
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20. A Ping-Pong ball of mass 3.0 x 10 kg is hanging from a light thread 1.0 m long,
between 2 parallel plates 10 cm apart (as shown to the right). When the potential
difference across the plates is 420 V, the ball comes to equilibrium 1.0 cm to one side
of its original position.
a)
b)
c)
d)
1.0 m 3
What is the electric field intensity between the plates? (4.2 x 10 N/C)
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What is the tension in the thread? (2.9 x 10 N)
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What is the magnitude of the electric force deflecting the ball? (2.9 x 10 N)
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What is the charge on the ball? (7.0 x 10 C)
1.0 cm 10 cm 3
21. An electron, travelling at 2.3 x 10 m/s, enters perpendicular to the electric field
2
between two horizontal charged parallel plates. If the electric field strength is 1.5 x 10 V/m, calculate the time
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taken for the electron to deflect a distance of 1.0 x 10 m towards the positive plate, ignore any gravitational
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effects. [2.7 x 10 s]
22. An electron is fired horizontally at a velocity of
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1.5 x 10 m/s in between two oppositely
charged plates as shown below. The plates
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have a separation of 1.0 x 10 m and a
2
potential difference of 3.0 x 10 V across them.
How far along the positive plate does the
electron collide? (0.0029 m)
23. Two horizontal plates are separated by a distance of 4.00 cm. The electric potential between the plates is 120 V.
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A horizontal beam of electrons, with a speed of 6.50 x 10 m/s, is directed into the electric field between the
plates. The electrons enter the field 1.00 cm above the negative plate. Draw the resulting path of the electron
beam. Determine the horizontal distance that the electrons travel before striking the positive plate. (The diagram
is not drawn to scale.) (6.94 cm)