Transcript PPT

Physics 102: Lecture 02
Coulomb’s Law
and Electric Fields
Today we will …
• get some practice using Coulomb’s Law
• learn the concept of an Electric Field
Physics 102: Lecture 2, Slide 1
Recall Coulomb’s Law
• Magnitude of the force between charges q1 and q2
separated a distance r:
F = k q1q2/r2
k = 9x109 Nm2/C2
• Force is
– attractive if q1 and q2 have opposite sign
– repulsive if q1 and q2 have same sign
• Units:
– q’s have units of Coulombs (C)
• charge on proton is 1.6 x 10-19 C
– r has units of m
– F has units of N
Physics 102: Lecture 2, Slide 2
Coulomb Law practice:
Three Charges
• Calculate force on +2mC charge due to other two charges
Draw forces
Calculate force from +7mC charge
Calculate force from –7mC charge
Add (VECTORS!)
F+7
Q=+2.0mC
4m
–
–
–
–
Q=+7.0mC
Physics 102: Lecture 2, Slide 3
6m
F-7
Q=-7.0 mC
Three Charges – Calculate forces
• Calculate force on +2mC charge due to other two charges
–
–
–
–
Draw forces
Calculate force from +7mC charge
Calculate force from –7mC charge
Add (VECTORS!)
F+7
Q=+2.0mC
• Calculate magnitudes
𝐹+7
4m
𝑞1 𝑞2
𝐹=𝑘 2
𝑟
−6
−6
2
×
10
|7
×
10
|
9
= 9 × 10
52
𝐹−7 = 9 × 10
9
2 × 10
Physics 102: Lecture 2, Slide 4
−6
−6
| − 7 × 10 |
52
Q=+7.0mC
6m
F-7
Q=-7.0 mC
Three charges – Adding Vectors F+7+F-7
• Calculate components of vectors F+7 and F-7:
𝐹+7,𝑦
𝐹−7,𝑥
3
= 𝐹−7 cos 𝜃 = 5 × 10 𝑁
5
𝐹−7,𝑦
4
= −𝐹−7 sin 𝜃 = −5 × 10 𝑁
5
F+7
Q=+2.0mC
4m
𝐹+7,𝑥
3
= 𝐹+7 cos 𝜃 = 5 × 10 𝑁
5
4
−3
= 𝐹+7 sin 𝜃 = 5 × 10 𝑁
5
−3
−3
Physics 102: Lecture 2, Slide 5
F-7
−3
Q=+7.0mC
6m
Q=-7.0 mC
Three charges – Adding Vectors F+7+F-7
• Add like components of vectors F+7 and F-7:
𝐹𝑥 = 𝐹+7,𝑥 + 𝐹−7,𝑥 = 6 × 10−3 𝑁
F+7
𝐹𝑦 = 𝐹+7,𝑦 + 𝐹−7,𝑦 = 0
Q=+2.0mC
𝐹=
2
2
4m
• Final vector F has magnitude
and direction
−3
𝐹𝑥 + 𝐹𝑦 = 6 × 10 𝑁
𝐹
−1 𝑦
𝜑 = tan
=0
𝐹𝑥
Q=+7.0mC
• Double-check with drawing
Physics 102: Lecture 2, Slide 6
6m
F
F-7
Q=-7.0 mC
Electric Field
• Charged particles create electric fields.
– Direction is the same as for the force that a + charge
would feel at that location.
– Magnitude given by:
E  F/q = kq/r2
Qp=1.6x10-19 C
+
r = 1x10-10 m
E
E = (9109)(1.610-19)/(10-10)2 N = 1.41011 N/C (to the right)
Physics 102: Lecture 2, Slide 7
Preflight 2.3
What is the direction of the electric field at point B?
71% 1) Left
“it is closer to the negative charge, and the field lines point toward
negative charges .”
“B only has the charge from the negative which is pushing away from
15% 2) Right itself .”
13% 3) Zero
“electric fields of equal magnitudes but opposite directions are
present due to the positive and negative charges .”
Since charges have equal magnitude, and point B is closer
to the negative charge net electric field is to the left
A
y
B
x
Physics 102: Lecture 2, Slide 8
Preflight 2.2
What is the direction of the electric field at point A?
7%
1) Up
7%
2) Down
2%
3) Left
53% 4) Right
32% 5) Zero
Physics 102: Lecture 2, Slide 9
A
y
B
x
ACT: E Field
What is the direction of the electric field at point C?
A. Left
B. Right
C. Zero
Away from positive charge (right)
Towards negative charge (right)
Net E field is to right.
A
y
C
B
x
Physics 102: Lecture 2, Slide 10
E Field from 2 Charges
• Calculate electric field at point A due to two unequal charges
–
–
–
–
Draw electric fields
Calculate E from +7mC charge
Calculate E from –3.5mC charge
Add (VECTORS!)
A
4m
Note: this is similar to (but a bit
harder than) my earlier example.
You try this at home!
Ex = 2.2510+3 N/C
Ey =
1.010+3
Physics 102: Lecture 2, Slide 11
N/C
Q = +7.0mC
6m
Q = –3.5 mC
E Field from 2 Charges
• Calculate electric field at point A due to charges
– Calculate E from +7mC charge
– Calculate E from –3.5mC charge
– Add*
E7
• E = k q/r2
Physics 102: Lecture 2, Slide 12
E3
4m
(9 109 )(7 10 6 )
E7 
N/C
25
E7  2.5 103 N/C
(9 109 )(3.5 10 6 )
E3 
N/C
25
E3  1.25 103 N/C
A
Q=+7.0mC
6m
Q=-3.5 mC
32
Adding Vectors E7+E3
• Decompose into x and y components.
E7
E7y=E7 (4/5)
A

E7x=E7 (3/5)
4m
3

E7 x  E7 cos( )  E7   1.5 103 N/C
 5
4

E7 y  E7 sin( )  E7    2 103 N/C
 5

Q=+7.0mC
Physics 102: Lecture 2, Slide 13
6m
Q=-3.5 mC
34
Adding Vectors E7+E3
• Decompose into x and y components.
• Add components.
3
E7
3
E7 x  1.5 10 N/C E7 y  2 10 N/C
3
Etotal
A
3
4m
E3 x  0.75 10 N/C E3 y  110 N/C
E3
Ex = 2.2510+3 N/C
Ey = 1.010+3 N/C
6m
Q=+7.0mC
Q=-3.5 mC
3

2
.
5

10
N/C
E  E E
2
x
Physics 102: Lecture 2, Slide 14
2
y
35
Comparison:
Electric Force vs. Electric Field
• Electric Force (F) – the actual force felt by a real
charge at some location
• Electric Field (E) – found for a location only (any
location) – tells what the electric force would be if
a charge were located there:
F = Eq
• Both are vectors, with magnitude and direction.
Ok, what is E actually good for?
Physics 102: Lecture 2, Slide 15
Electric Field Map
• Electric field defined at any location (we did
three: A, B, C)
A
y
C
B
x
Physics 102: Lecture 2, Slide 16
Electric Field Lines
•
•
•
Closeness of lines shows field strength (lines never cross)
Number of lines at surface  Q
Arrow gives direction of E (Start on +, end on –)
6
5
4
4
3
2
2
1
0
0
-1
-2
-2
-3
-4
-4
-6
-6
-4
-2
Physics 102: Lecture 2, Slide 17
0
2
4
6
-5
-5
-4
-3
-2
-1
This is becoming a mess!!!
0
1
2
3
4
5
Preflight 2.5
X
A
Charge A is
Y
B
Field lines start on positive charge, end on negative.
1) positive
93%
Physics 102: Lecture 2, Slide 18
2) negative
4%
3) unknown
3%
Preflight 2.6
X
X
A
A
Y
Y
B
B
Compare the ratio of charges QA/ QB # lines proportional to Q
1) QA= 0.5QB
17%
Physics 102: Lecture 2, Slide 19
2) QA= QB
9%
3) QA= 2 QB
61%
Preflight 2.8
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
14%
Physics 102: Lecture 2, Slide 20
2) False
86%
Density of field lines gives E
ACT: E Field Lines
B
A
Compare the magnitude of the electric field at
point A and B
1) EA>EB
Physics 102: Lecture 2, Slide 21
2) EA=EB
3) EA<EB
E inside of conductor
• Conductor  electrons free to move
– Electrons feels electric force - will move until
they feel no more force (F=0)
– F=Eq: if F=0 then E=0
• E=0 inside a conductor (Always!)
Physics 102: Lecture 2, Slide 22
Preflight 2.10
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
9%
Physics 102: Lecture 2, Slide 23
(2) Zero
68%
(3) Positive
23%
Demo: E-field from dipole
y
A
C
B
x
Physics 102: Lecture 2, Slide 24
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
Physics 102: Lecture 2, Slide 25
To Do
• Do your preflight by 6:00 AM Wednesday.
• Homework 1 due Tuesday, Jan 25 8AM!
Physics 102: Lecture 2, Slide 26