Today 3/12
-
Upload
bert-henderson -
Category
Documents
-
view
18 -
download
1
description
Transcript of Today 3/12
![Page 1: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/1.jpg)
Today 3/12
Plates if charge E-Field Potential
HW: “Plate Potential” Due Friday, 3/14
![Page 2: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/2.jpg)
(+)
How big is E?(-)
Enet = 0
02E
02
E 0
netE
Enet = 0
![Page 3: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/3.jpg)
0 = Q/A
Charged conducting plate
What’s wrong with this picture?
A = Area of one side
![Page 4: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/4.jpg)
L = 0/2
Charged conducting plate
Free charge always goes to surface of conductor.
R = 0/2
![Page 5: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/5.jpg)
L = 0/2
Charged conducting plate
R = 0/2
What is the electric field inside the conductor?
EL = L/20 = 0/40
![Page 6: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/6.jpg)
L = 0/2
Charged conducting plate
R = 0/2
What is the electric field inside the conductor?
ER = R/20 = 0/40
![Page 7: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/7.jpg)
L = 0/2
Charged conducting plate
R = 0/2
What is the electric field inside the conductor?
The electric field is zero everywhere inside the conductor.Always, any conductor, no exceptions.
![Page 8: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/8.jpg)
q=+1C
What happens if I let it go?
![Page 9: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/9.jpg)
KE = 0 KE = 100 J
A B
Assume the particle gains 100 joules of kinetic energy as it moves from A to B.
q=+1C
![Page 10: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/10.jpg)
A B
Now I stop it at B.
How much work must I do to move it back to A? +100 J
How does the potential energy change in moving from B to A? +100 J PEBA = +100 J
PEAB = -100 J
q=+1C
![Page 11: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/11.jpg)
A B
PEBA = +200 JPEAB = -200 J
What if q=+2C?
How much work must I do to move it back to A?
How does the potential energy change in moving from B to A?
+200 J
+200 J
q=+2C
![Page 12: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/12.jpg)
A B
Now we are back to our original definition.
PEBA = (+100 J/C)x(q)
VBA tells us how much PE changes when +1C is moved from B to A.
PEBA = VBA q
VBA = +100 J/C
q=+1C
![Page 13: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/13.jpg)
A B
What if q = -1C?
First I must turn my hand around.
q= -1C
![Page 14: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/14.jpg)
A B
What if q= -1C?
How much work must I do to move it back to A? -100 J
How does the potential energy change in moving from B to A? -100 J
PEBA = (+100 J/C)x(q)
PEBA = (+100 J/C)x(-1)
VBA tells us how much PE changes when +1C is moved from B to A.
PEBA = VBA q
q= -1C
![Page 15: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/15.jpg)
A B
What if q= -1C?
How much work must I do to move it back to A? -100 J
How does the potential energy change in moving from B to A? -100 J
PEBA = (+100 J/C)x(q)
PEBA = (+100 J/C)x(-1)
VBA does not depend on the sign of the point charge but PEBA does!!!!!
PEBA = VBA q
q= -1C
![Page 16: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/16.jpg)
A B
VAB = -100 J/C
VAB = -100 volts
A proton is released from rest at A. What is its speed when it reaches B?
m= 1.7 x 10-27 kg
q= 1.6 x 10-19 C
PEAB = q VAB
PEAB = q -100 J/C)
PEAB = -1.6 x 10-17 JWhat happens to the kinetic energy?
![Page 17: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/17.jpg)
A B
KEAB = +1.6 x 10-17 J
A proton is released from rest at A. What is its speed when it reaches B?
m= 1.7 x 10-27 kg
q= 1.6 x 10-19 C
mv2 = +1.6 x 10-17 J
v = 1.4 x 105 m/sWhat happens to the kinetic energy?
PEAB = -1.6 x 10-17 J
![Page 18: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/18.jpg)
A BA B
What direction is the force on an electron?
EF
![Page 19: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/19.jpg)
A B
VAB = -100 J/C
VBA = +100 J/C
An electron is released from rest at B. What is its speed when it reaches A?
m= 9.1 x 10-31 kg
q= -1.6 x 10-19 C PEBA = q VBA
PEBA = q +100 J/C)
PEBA = -1.6 x 10-17 JWhat happens to the kinetic energy?
![Page 20: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/20.jpg)
A B
KEBA = +1.6 x 10-17 J
mv2 = +1.6 x 10-17 J
v = 5.9 x 106 m/s
An electron is released from rest at B. What is its speed when it reaches A?
m= 9.1 x 10-31 kg
q= -1.6 x 10-19 C
What happens to the kinetic energy?
PEBA = -1.6 x 10-17 J
![Page 21: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/21.jpg)
B A
VAB = ?
![Page 22: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/22.jpg)
B A
How does doubling the E-field affect VAB ?
VAB = ?
VAB
doubles
![Page 23: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/23.jpg)
Bnew A
How does moving point B affect VAB ?
VAB = ?
How does VAB
depend on E and D?
D
Anything else?
VAB is halved
Bold
![Page 24: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/24.jpg)
B A
KEAB = 0
constantspeed
v0v0
KEA=1/2 mv02
KEB=1/2 mv02
How much work must I do to move the charge from A to B? D
![Page 25: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/25.jpg)
B A
How much work must I do to move the charge from A to B? FE
= qEFHq
WAB = FE x D
WAB = qED
WAB = FHq x D
![Page 26: Today 3/12](https://reader035.fdocuments.us/reader035/viewer/2022070401/56813653550346895d9dd90f/html5/thumbnails/26.jpg)
B A
What is the change in potential energy in going from A to B?
FE = qEFHq
WAB = qED
PEAB = qED
PEAB = qVAB
VAB = EDAB
WAB = qED
Only applies when the field is uniform over the distance. VAB‘s sign depends on the direction of E. In this case it’s positive.