Transcript Lectures3and4

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Answers are available immediately after the due date.
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Monday 6-8pm
Tuesday 6-8pm
Thursday 7-9pm
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Reading Quizzes
Several of you had problems remembering Newton’s second
Law, F=ma. You may wish to reread Ch. 5 on the Laws of
Motion
Keep in mind that a force is a vector, as in acceleration, so
they have magnitude and direction.
Charge Densities
Charge can be localized to discrete points (point charges), or it
may be spread out over a volume, a surface or a line
•Charge density  units C/m3
•Surface charge density  units C/m2
•Linear charge density  units C/m
A cube with side 1 cm has a charge density
of  = 1 C/m3. What is the charge of the
cube?
A) 1 C
B) 0.01 C = 10 mC
C) 10-4 C = 100 m C
1 cm
D) 10-6C = 1 mC
Coulomb’s Law
•Like charges repel, unlike charges attract
•Force is directly along a line joining the two charges
q1
ke q1q2
Fe 
rˆ
2
r
q2
r
q1q2
Fe 
rˆ
2
4 0 r
0 = 8.85410-12 C2/ (N●m2)
•Permittivity of free space
•An inverse square law, just like gravity
•Can be attractive or repulsive – unlike gravity
•Constant is enormous compared to gravity
Coulomb’s Law: Applied
A Helium nucleus (charge +2e) is separated from one of its
electrons (charge –e) by about 3.00 10-11m. What is the force the
nucleus exerts on the electron? Is it attractive or repulsive?
ke q1q2
9 Nm2/C2
k
=
8.98810
Fe 
e
2
r
-11m the force on the electron
rHow
= just
3.00
10
We
calculated
from the
q1 =of3.204
10-19C
does the acceleration
the nucleus
compare
nucleus.
does this
q2 =compare
-1.602with
10-19the
C force
to that ofHow
the electron?
theacceleration
nucleus from
the nucleus
electron?is larger
A) on
The
of the
A)
on the
nucleus
is twiceisassmaller
big
The
acceleration
of the Attractive
nucleus
FB)e=The
- force
0.513
mN
Force
B)
The
force
on
the
nucleus
is
half
as
big
C) The accelerations are equal.
C) The forces are equal in magnitude
Newton’s Laws and Kinematics
Newton’s laws and all the kinematics you learned in 113 are still
true!
Fnet  ma
F12   F21
A body in motion tends to stay in motion, therefore
changing velocity, i.e. acceleration, requires a force!
Also the same forces on different particles can lead to
different accelerations, depending on the masses.
t'
t'
dx
dv
v  ;a 
 v   adt  vt0 ; x   vdt  x0
dt
dt
t0
t0
If a does not depend on
time, then
1 2
v  at  v 0 ; x  at  v 0 t  x 0
2
Coulomb’s Law vs. Gravity
A Helium nucleus (charge +2e) is separated from one of its
electrons (charge –e) by about 3.00 10-11m, and we just
calculated the electrostatic forces involved.
ke q1q2
9 Nm2/C2
k
=
8.98810
Fe 
e
2
r
Suppose we could adjust the distance between the
nucleus (considered as a point particle) and one
electron. Can we find a point at which the
electric and gravitational forces are equal?
A) Yes, move the particles apart.
B) Yes, move the particles together.
C) No, they will never be equal.
Electric Fields
•Electric Field is the ability to extert a force at a distance on a
charge
•It is defined as force on a test charge divided by the charge
•Denoted by the letter E
•Units N/C
+ +
– ––
+ +
+
F
Small test
charge q
E  F /q
E-Field: Why?
•When we have a charge distribution, and we want to know
what effect they would have external charges, we can either
•Do many sums (or integrations) every time a charge comes
in to find the force on that charge
•Or calculate the field from the charge distribution, and
multiply the field by the external charge to obtain the force
•Simplification!
Electric Field from a Point Charge
keQq
F  2 rˆ
r
E  F /q
ke Q
E  2 rˆ
r
Point
charge
Q
Small test
charge q
Note: the field is a vector!
Quiz
Two test charges are brought separately into the vicinity of a charge +Q.
First,• test
T charge +q is brought to point A a distance r from charge +Q.
Next, the +q charge is removed and a test charge +2q is brought to point B
a distance 2r from charge +Q. Compared with the electric field of the
charge at A, the electric field of the charge at B is:
+Q
+q
+Q
A
+2q
B
A) Greater
B) Smaller
C) The same.
E-Field from two or more charges
•Each charge creates its own Electric Field
•Electric Fields must be added as a vector sum
q2 = -2 mC
10 cm
5 cm
q1 = +1 mC
ke qi
E   2 rˆi
ri
E2 = 1.797  106 N/C
Etot = 4.019106 N/C
 = 26.6o
E1 = 3.595  106 N/C
Electric Field Lines
•Graphical Illustration of Electrical Fields
•Lines start on positive charges and end on negative
•Number of lines from/to a charge is proportional to that charge
•Density of lines tells strength of field.
+
-
+
-
Electric Field from a Point Charge
ke Q
E  2 rˆ
r
+
Positive Charge
–
Negative Charge
Electric Field from two Charges
+
+
+
+
Electric Field from two Charges
+
-
+
-
Electric Field Lines
Consider the four field patterns below:
Assuming that there are no charges in the region of space depicted,
which field pattern(s) could represent electrostatic field(s)?
Electric Field Lines
Which
This
thin
of disk
the two
has surfaces
charge on(top
theand
top bottom)
and on the
hasbottom.
more charge
Whaton
type?
it?
A) Top
Positive
has more
charge on both
B) Bottom
Negativehas
charge
moreon both
C) They
Positive
arecharge
roughly
onequal
top, negative on bottom
D) Can’t
Negative
tell charge
from the
ongiven
top, positive
information
on bottom
Electric Field Lines
Which of the following is true:
A) the electric field is strongest midway between Y and Z
B) the magnitude of the electric field is the same everywhere
C) a small negatively charged body placed at X would be pushed
to the right
D) Y and Z must have the same sign
E-Field from continuous charges
ke qi
E   2 rˆi
ri
ke dq
E   2 rˆ
r
•Volume charge density : dq =  dV
•Surface charge density : dq =  dA
•Linear charge density : dq =  dl
E-Field from continuous charges
Length L
Distance R
Linear charge density 
E-field here?
R L
ke  dr
R r 2
R L
ke  L
ke  
ke 
 ke 
ˆ

r
  rˆ 

 rˆ

R

L
R
r R

 R  R  L
ke dq
ke  dr
E   2 rˆ   2 rˆ  rˆ
r
r
Electric Fields and Forces
E  F /q
F  qE
F  ma
a  qE / m
A region of space has an electric field of 104 N/C, pointing in the plus
x direction. At t = 0, an object of mass 1 g carrying a charge of
1mC is placed at rest at x = 0. Where is the object at t = 4 sec?
A) x = 0.2 m
C) x = 20 m
B) x = 0.8 m
D) x = 80 m