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Physics 1161 Lecture 2
Vectors
&
Electric Fields
Three Charges
• Calculate force on +2mC charge due to other two
charges
– Calculate force from +7mC charge
 3m   4m 
2
2
2
 5.04 103 N
2.52 103 N
– Calculate force from –3.5mC charge
F2,3 
k   3.5 106 C  2 106 C 
 3m   4m 
2
2
2
Q=+2.0mC
 2.52 103 N
– Add (VECTORS!)
Q=+7.0mC
4m
F1,3 
k   7 106 C  2 106 C 
5.04 103 N
6m
Q=-3.5 mC
Three Charges
• Resolve each force into x and y components
F1,3x  5.04 103 N  cos53o  3.03103 N
F1,3 y  5.04 103 N  sin 53o  4.03103 N
F2,3x  2.52 103 N  cos307o  1.51103 N
5.04 103 N
F2,3 y  2.52 103 N  sin 307o  1.997 103 N
Q=+2.0mC
• Add the x-components & the y-comp.
3
3
3
4m
Fx  3.03 10 N  1.5110 N  4.54 10 N
2.52 103 N
53o
53o
Fy  4.03103 N 1.997 103 N  2.0310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   4.54 10 N    2.03 10 N 
3
3
2.03 10 N
2
3
F  4.9 103 N
F
φ
4.54 103 N
 Fy
  arctan 
 Fx

 2.03 103 N 
o

24
  arctan 

3
 4.54 10 N 

Since resultant is in first quadrant, θ = φ
  24o
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
+
r = 1x10-10 m
kq1q2
F 2
r
9 N m 2
19
19
9

10

1.6

10
C

1.6

10
C



C2  
F
2
10
10 m 
F  2.30 108 N
Toward the left
e-
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?
kq
q=1.6x10-19 C
E 2
+
r
r = 1x10-10 m
9 N m 2
19
9 10
 1.6 10 C 

C2  
E
2
10
10 m 
E  1.44 1011 N
C
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