Transcript Lecture 2 Presentation

Physics 1161 Lecture 2
Vectors
&
Electric Fields
Three Charges
• Calculate force on +2mC charge due to other two
charges
– Calculate force from +7mC charge
F1,3 

 3m 
6
2
  4m 
2

2
 5.04  10
3
N
6
F2 ,3 

 3m 
2
2.52  10
6
  4m 
2

2
 2.52  10
3
N
Q=+2.0mC
– Calculate force from –3.5mC charge
k   3.5  10 C   2  10 C 
3
N
– Add (VECTORS!)
Q=+7.0mC
3
N
4m
k   7  10 C   2  10 C 
6
5.04  10
6m
Q=-3.5 mC
Three Charges
• Resolve each force into x and y components
3
N  cos 53  3.03  10
3
F1,3 y  5.04  10
3
N  sin 53  4.03  10
3
F2 ,3 x  2 .5 2  1 0
3
N  co s 3 0 7  1 .5 1  1 0
F1,3 x  5.04  10
F2 ,3 y  2 .5 2  1 0
3
o
o
o
N
N
3
N
N  sin 3 0 7   1 .9 9 7  1 0
o
3
5.04  10
• Add the x-components & the y-comp.
F x  3.03  10
N  1.51  10
F y  4.03  10
3
N  1.997  10
N  4.54  10
3
3
N  2.03  10
N
3
2.52  10
3
53o
53o
N
4m
3
N
N
Q=+2.0mC
3
3
N
• Use Pyth. Theorem & Trigonometry
to express in R,θ notation
Q=+7.0mC
6m
Q=-3.5 mC
Three Charges
• Use Pyth. Theorem & Trigonometry to express
in R,θ notation
F 
2.03  10
3
N
 4.54  10
F  4.9  10
F
3
3
N

2
  2.03  10
3
N
φ
4.54  10
3
N
 Fy 
 2.03  10  3 N 
o
  arctan 

24
  arctan 

3
 4.54  10 N 
 Fx 
Since resultant is in first quadrant, θ = φ
  24
o
N

2
Electric Force on Electron by Proton
• What are the magnitude and direction of the
force on the electron by the proton?
q=1.6x10-19 C
e-
+
F 
kq1 q 2
r
F 
r = 1x10-10 m
2
 9  10
9 N m 2
 1.6  10
2
C  
10
8
F  2.30  10 N
 10
 19
m
C   1.6  10
 19
C
2
Toward the left
Comparison:
Electric Force vs. Electric Field
• Electric Force (F) - the actual force felt by a
charge at some location.
• Electric Field (E) - found for a location only –
tells what the electric force would be if a
charge were located there:
F = qE
• Both are vectors, with magnitude and
direction
Electric Field
• Charged particles create electric fields.
– Direction is the same as for the force that a + charge
would feel at that location.
E  F/q
– Magnitude given by:
• Field at A due to proton?
E 
kq
r
E 
q=1.6x10-19 C
+
2
 9  10
  1.6  10
10 m 
9 N m 2
E  1.44  10
C
2
 10
11
N
C
 19
C
r = 1x10-10 m
2
Toward the right
A
What is the direction of the electric
field at point A, if the two positive
charges have equal magnitude?
1.
2.
3.
4.
5.
Up
Down
Right
Left
Zero
A y
0%
1
+
+
B
x
0%
0%
2
3
0%
0%
4
5
What is the direction of the electric
field at point A, if the two positive
charges have equal magnitude?
1.
2.
3.
4.
5.
Up
Down
Right
Left
Zero
A y
0%
1
+
+
B
x
0%
0%
2
3
0%
0%
4
5
Two Charges
Checkpoint1
What is the direction of
the electric field at point
A?
1) Up
2) Down
A
y
3) Left
4) Right
5) Zero
+
B
x
Two Charges
Checkpoint 2
What is the direction of the electric
field at point B?
1) up
2) down
3) Left
4) Right
A
5) Zero
+
y
B
x
What is the direction of the electric
field at point C?
1. Left
2. Right
3. zero
y
+
C
-
x
0%
1
0%
2
0%
3
Electric Field Applet
• http://www.cco.caltech.edu/~phys1/java/phys
1/EField/EField.html
Checkpoint
X
A
Charge A is
Y
B
Field lines start on positive charge, end on negative.
1) positive
2) negative
3) unknown
Checkpoint
X
A
Y
B
Compare the ratio of charges QA/ QB # lines proportional to |Q|
1) QA= 0.5QB
2) QA= QB
3) QA= 2 QB
4) can’t say
Checkpoint
X
A
Y
B
The electric field is stronger when the
lines are located closer to one another.
The magnitude of the electric field at point X is greater than at point Y
1) True
2) False
Density of field lines gives E
Compare the magnitude of the
electric field at point A and B
1. EA> EB
2. EA= EB
3. EA< EB
B
A
E inside of conductor
• Conductor  electrons free to move
– Electrons feels electric force - will move until they
feel no more force (F=0)
– F=qE: if F=0 then E=0
• E=0 inside a conductor (in electrostatics)
Checkpoint
X
A
Y
B
"Charge A" is actually a small, charged metal ball (a conductor). The
magnitude of the electric field inside the ball is:
(1) Negative
(2) Zero
(3) Positive
E inside of conductor in electrostatics
• Conductor  electrons free to move
– Electrons feel electric force - will move until
they feel no more force (F=0)
– F=qE: if F=0 then E=0
• E=0 inside a conductor (in a static situation)
Recap
• E Field has magnitude and direction:
– EF/q
– Calculate just like Coulomb’s law
– Careful when adding vectors
• Electric Field Lines
– Density gives strength (# proportional to charge.)
– Arrow gives direction (Start + end on -)
• Conductors
– Electrons free to move  E=0
To Do
•
•
•
•
•
Read sections 19-6 & 19-7
Watch Prelecture 3 by 6am 1/15
Complete Checkpoint 3 by 6 am 1/15
Complete Homework 1 by 11pm 1/15
Complete Homework 2 by 11pm 1/17