Physics I
Review 3
Review Notes
Exam 3
R3-1
Newton’s Law of
Universal Gravitation

m1 m 2
F  G 2 r̂
r
The meaning of each term:

F:
G:
m1 :
m2 :
r2:
r̂ :
Gravitational force on object 1 from object 2.
–11
2
2
Universal gravitational constant = 6.673 x 10 N m /kg .
Mass of object 1.
Mass of object 2.
Center distance from object 1 to object 2, squared.
Unit vector from object 1 to object 2.
R3-2
If Gravity Varies As 1/r2,
2
Where Does g = 9.8 m/s Fit In?
Consider the force on an object near the surface of the earth.
(Assume the earth is a sphere and ignore rotation effects.)
R = radius of the earth.
M = mass of the earth.
m = mass of the object.

mM
GM

F  G 2 r̂  m 2 r̂  m g (What is the direction?)
R
R
g = 9.8 m/s2 only seems constant because we don’t go very far
from the surface of the earth.
R3-3
Class #17
Take-Away Concepts
1.
Four fundamental forces known to physics:
 Gravitational Force
 Electromagnetic Force
 Weak Nuclear Force
 Strong Nuclear Force
2.
Newton’s Law of Universal Gravitation

m1 m 2
F  G 2 r̂
r
3.
Gravitational Potential Energy (long-range form)
U g (r )  
GmM
r
R3-4
Coulomb’s Law of
Electrostatic Force

F
1 q1 q 2
( r̂ ) (Prof. B’s version.)
2
4  0 r
The meaning of each term:

F:
Electrostatic force on charge 1 from charge 2.
1
2
2
: Electrostatic force constant = 8.98755 × 10+9 N m /C
4  0
q1 : Value of charge 1, positive or negative.
q 2 : Value of charge 2, positive or negative.
r 2 : Center distance from point charge 1 to point charge 2.
r̂ :
Unit vector from charge 1 to charge 2.
R3-5
Direction of Electrostatic Force
“Opposites Attract”
R3-6
Superposition of
Electrostatic Forces
+5.0 C
+1.0 C
Y
1
X
resultant
-3.0 C
N

1 q1 q i
Fon 1  
( r̂i ) (find and add X and Y components)
2
i  2 4   0 ri
R3-7
How to Calculate a
Unit Direction Vector
From blue (x0,y0) to red (x,y):
1. Find
 the displacement in X,Y components.
d  ( x  x 0 ) î  ( y  y 0 ) ĵ  8 î  6 ĵ
2.
Find the length of this vector.
3.
Divide
 by the length to get a unit vector.

r  | d |  ( x  x 0 ) 2  ( y  y 0 ) 2  82  6 2  10


r̂  d  r  8î  6 ĵ  10  0.8î  0.6 ĵ
Y
6
0
X
0
8
R3-8
Class #18
Take-Away Concepts
1.
Charges come in positive and negative.
2.
Opposites charges attract, like charges repel.
3.
Coulomb’s Law of Electrostatic Force

F
1 q1 q 2
(r̂ )
2
4  0 r
4.
Similarities and differences with gravity.
5.
The principle of superposition.
N

1 q1 q i
Fon 1  
(r̂i )
2
i 2 4   0 ri
R3-9
The Idea of Electric Field
The electric field is a vector field that
 depends on the location of the point where we calculate it.
 depends on the unit vectors to the other charges.
 depends on the distances to the other charges.
 depends on the values of the other charges.
It does not depend on the value of any charge at that point.
In fact, it can be calculated even when there is no charge there!
N

1 qi
(  r̂i ) Electric Field
E ( point #1)  
2
i  2 4   0 ri


Fon 1  q1 E ( point #1)
Force / Field Relationship
R3-10
The Electric Field of a Point
Charge (as a Source)
The electric field is a vector field, meaning at each point in
space the electric field has a magnitude and a direction. We
show that by drawing arrows at representative points in the
correct directions with lengths proportional to the magnitudes.
Away from positive
Toward negative
+
-
Just because we don’t draw an electric field vector at
a point doesn’t mean there is no electric field there.
R3-11
Calculating the Electric Field of a
Charge at a Point (Example)
Y
6 cm
q  1.0  109 C
0
  1 q
( r̂ )
E  
2
 4  0 r 
X
0
8 cm
First calculate the unit vector. This vector is from the point (X)
to the charge. Use the method on side R3-8:

d  (0.08)î  (0.06) ĵ
r  ( 0.08) 2  (0.06) 2  0.1
r̂  [(0.08)î  (0.06) ĵ]  0.1  (0.8)î  (0.6) ĵ
R3-12
Calculating the Electric Field of a
Charge at a Point (Example)
Y
6 cm
q  1.0  109 C
0
  1 q
 (r̂ )
E  
2
 4  0 r 
X
0
8 cm
Now calculate the number in ( ):
 1 q
9
9
2



(
9
.
0

10
)(

1
.
0

10
)

(
0
.
1
)
 900
2 
 4  0 r 
Final answer:

E  (900){[(0.8)î  (0.6) ĵ]}  (720)î  (540) ĵ N/C
R3-13
Class #19
Take-Away Concepts
1.
Electric field from point charge sources:
(Total field is the superposition of point source fields.)

1 qi
E
( r̂i )
2
4   0 ri
2.
Force on a charge in an electric field:
3.
4.
Electric field points away from + source charges.
Electric field points toward – source charges.


F  qE
R3-14
Electric Potential
Just as we noticed when we defined the electric field, we see that
there is a certain quantity that is independent of the charge
experiencing the electrostatic force. This is called the electric
potential and it is normally represented by the symbol V:
 
V ( A )    E  dx
A
0
The relationship between U and V is
U  qV
V exists everywhere that an electrostatic field exists. U exists only
where there are charges in that field.
R3-15
Calculating Electric Potential
from Point Source Charges
1 qi
1
qi
V



4   0 ri 4   0
ri
6
6



1

10

2

10
9
9 10 


1
2 

 3.7 10 3
1m
 1C
 2C
1m
R3-16
Equipotential Lines
and Field Lines
Equipotential Lines – Lines connecting points with equal electric
potential. (In 3D, these are equipotential surfaces.)
Electric Field Lines – Lines formed by taking tiny electric field
vector arrows and connecting them “tail to head.”
 Electric field lines are always perpendicular to equipotential lines.
 Electric field strength is higher where electric field lines are closer
together.
 Electric field strength is higher where equipotential lines are closer
together (assuming equal increments of potential per line).
 Electric field lines point in the direction of decreasing potential.
R3-17
Equipotential Lines
and Field Lines
+
equipotential
field line
R3-18
Calculating Electric Field
from Electric Potential
Y
X
V = 100
X=0
V = 50
X=2
V=0
X=4
dV
V
 100
Ex  


 25 V / m
dx
x
4
Ey  
dV
0
dy
R3-19
Calculating Change in K.E.
from Electric Potential
final
e
 K   U  0 or  K   U
U  q V  (e) V or  U  (e) V
 K  (e)( V)  (1.6 10 19 ) (100) 
 1.6 10 17 J
V = 100
initial
V = 50
e

V=0
R3-20
Class #20
Take-Away Concepts
Physicist General’s Health Warning:
Confusing electric potential (V) with electric potential energy
(U) will be hazardous to your score on the next exam.
1.
Electric potential (V) and electric potential energy (U):
2.
Electric potential from point source charges:
U  qV
V
3.
4.
5.
1 qi
4   0 ri
Relationship of electric field and electric potential.
Electric field lines and equipotential lines.
Calculating change in kinetic energy using electric potential.
R3-21
How to Solve Problems Asking
“Where is Net Electric Field = 0?”
+X
Region I
Region II
+ charge
Region III
- charge
Many things to consider:
 The number of regions (N+1) is one more than number of charges (N).
 The total electric field is the sum of the individual fields from each charge.
(Principle of Superposition)
 In each region, the direction of electric field from each charge is constant in that region.
(Away from +, toward –.)
 The magnitude of electric field depends on the value of charge and inverse distance squared
to the charge.
 In any region, if you get close enough to one of the charges at the end points, the inverse
distance squared dependence will make the field from that charge much larger than the
fields from all of the other charges.
 If you get far enough away from all the charges in Region I or Region N+1, the inverse
distance term is about the same for all points, so the relative sizes of the electric fields from
each charge will be determined by the sizes of the charges.
 In a region where the total electric field is zero at some location, it will point in one direction
at one end of the region and the other direction somewhere else in the region – the middle or
the opposite end.
R3-22
How to Solve Problems Asking
“Where is Net Electric Field = 0?”
+X
Region I
Region II
Region III
+7 C
-4 C
Region I:
E field from +7C direction is to the left.
E field from –4C direction is to the right.
Every point is closer to +7C than to –4C, so net field is always to the left.
No points where E = 0.
Region II:
E field from +7C direction is to the right.
E field from –4C direction is to the right.
Net field is always to the right.
No points where E = 0.
Region III:
E field from +7C direction is to the right.
E field from –4C direction is to the left.
Close to –4C, net E field is to the left.
Far from –4C, net E field is to the right because 7 > 4 and far enough away the inverse
distance squared will be about the same for both.
There is one point where E = 0.
R3-23
How to Solve Problems Asking
“Where is Net Electric Field = 0?”
+X
Region I
Region II
+7 C
d
Region III
-4 C
x-d
x
1 7
1
4

0
4   0 x 2 4   0 ( x  d) 2
7
4

x 2 ( x  d) 2
7
2

x x d
7 ( x  d)  2 x
x
7
d  4.097 d
72
R3-24
How to Solve Problems Asking
“Where is Electric Potential = 0?”
+X
Region I
Region II
+ charge
Region III
- charge
This is similar to finding zero electric field, but V is a scalar, not vector!
 The total electric potential is the sum of the individual potentials from each charge.
(Principle of Superposition)
 The sign of an individual contribution is always the same as the sign of charge.
 The magnitude of potential depends on the value of charge and inverse distance to the charge –
not squared in this case.
 If you get close enough to one of the charges, the inverse distance dependence will make the
potential from that charge much larger (in magnitude) than the potentials from all of the other
charges.
 If you get far enough away from all the charges, the inverse distance term is about the same for
all, so the relative sizes of the electric potentials from each charge will be determined by the
relative sizes of the charges.
 On any line or curve segment, not passing through a charge, where the electric potential is
negative at one end and positive at the other, there will be at least one point on that segment
where the electric potential is zero.
 The electric potential can be zero at a point not at infinity only if there are at least two charges in
the problem, one positive and one negative. In that case, there will be infinitely many points not
at infinity where the electric potential is zero.
R3-25
How to Solve Problems Asking
“Where is Electric Potential = 0?”
+X
Region I
Region II
+7 C
Region III
-4 C
Region I:
Every point is closer to +7C than to –4C, so net V is
always positive.
No points where V = 0.
Region II:
On the line from +7C to –4C, V switches sign.
There is a point where V = 0.
Region III:
Close to –4C, net V is negative.
Far from –4C, net V is positive.
There is a point where V = 0.
R3-26
How to Solve Problems Asking
“Where is Electric Potential = 0?”
+X
Region I
Region II
+7 C
Region III
-4 C
Region II
1 7
1
4

0
4   0 x 4   0 (d  x )
Region III
1 7
1
4

0
4   0 x 4   0 ( x  d)
7
4

x (d  x )
7 (d  x )  4 x
7
x
d  0.6364 d
74
7
4

x ( x  d)
7( x  d )  4 x
7
x
d  2.333 d
74
R3-27
Class #21
Take-Away Concepts
To find points where E = 0, consider the following:
1. Consider N+1 regions for N charges on a line.
2. Superposition Principle: Add E from each charge.
3. E points away from + charge, toward – charge.
4. E depends on charge value and inverse distance squared.
5. Very close to a charge, E from that charge dominates.
6. Far away right or left, all distances are about the same.
7. Look for reversal of total E direction in a region.
Finding where V = 0 is similar but easier, since V is a scalar.
Look for regions where V changes sign.
R3-28
Magnetic Field Lines
From N to S.
In direction of
compass
needle.
Try this web site:
http://home.a-city.de/walter.fendt/phe/mfbar.htm
R3-29
Class #22
Take-Away Concepts
1.
The magnetic force was known in ancient times.
2.
Magnets have two poles, N and S.
3.
Opposite poles attract, like poles repel.
4.
N and S poles are always in pairs, never alone.
5.
Electric currents also cause magnetic fields.
6.
Magnetic field lines start at N and go to S,
following the direction of a compass needle.
7.
Units of tesla (T) and gauss. 1 T = 10,000 gauss.
R3-30
Magnetic Force on a
Moving Charge

 
F  q vB
charge of the particle (C; + or –)
v : velocity of the particle (m/s)
B : magnetic field (T)
q:

 Force is at a right angle to velocity.
 Force is at a right angle to magnetic field.
Important: If q is negative, that reverses the direction of force.
R3-31
Analysis of the Magnetic Force
Y
X

 
F  q vB
Z
F
We will evaluate this expression before
the electron starts turning.
 
First, evaluate v  B . In this case,
 they are 90° apart, so all we

need is the direction. v is +X, B is –Z, so v  B is +Y.
Next, we need to account for q. This is an electron, so q is
negative. Therefore, the magnitude of the force is (e v B) and the
direction is –Y.
R3-32
The Radius of the Circle
F
r
v
Although the directions of the vectors are
changing, the magnitudes stay the same.
v2
F  ma  m
r
F  qvB
v2
qvB  m
r
v2
mv
rm

qvB qB
R3-33
Class #23
Take-Away Concepts
1.
The Lorentz
(magnetic)
force on a moving, charged particle:



F  q vB
2. The Lorentz force cannot change a particle’s speed, only the
direction of its velocity.
3. Radius and angular frequency of a charged particle in uniform
circular motion in a magnetic field:
mv
r
qB
qB

m
R3-34
Apparatus for Measuring e/m
We will set
 Potential in tube (V).
 Current in coils (I).
We will observe
 Radius of circular
electron path (r).
We will calculate
 e/m.
R3-35
Analysis of Electron Acceleration
cathode
anode
Electron Stream
pot = -V
pot = 0
 K   U
 U  (e) V
 K  (e)[ V ] 
e [0  ( V )]  e V
Electrons have very low kinetic
energy when they leave the
cathode – essentially zero.
1
2
m v2  e V
or
2eV
v
m
R3-36
Derivation of e/m Formula
r
mv
eB
e
v

m rB
2eV
m
2eV
e
2V
e
e 2V
m
m



m
rB
rB
m rB
v
2V
e

m
rB
2V
e

or
m r 2 B2
We set or observe all of the variables
on the right side of the equation.
R3-37
Mass Spectrometer Diagram
Adjusting the accelerating
potential (V) and/or the
magnetic field (B) “tunes”
the spectrometer to detect
only ions with a specific
ratio of charge to mass.
R3-38
Class #24
Take-Away Concepts
1.
2.
The ratio of e/m is determined by accelerating a stream
of electrons through an electric potential difference and
observing their path in a magnetic field.
The formula for e/m (know how to derive it):
2V
e
 2 2
m r B
3.
Mass spectrometers work by the same principle.
R3-39
Relationship of Energy and
Electric Potential Changes
Memorize these tables or know how to derive them!
K = –U
U=qV
or U = q V
Potential Energy Change
+ charge
– charge
V > 0 (+)
+ (increase)
– (decrease)
V < 0 (–)
– (decrease)
+ (increase)
Kinetic Energy Change
+ charge
– charge
V > 0 (+)
– (decrease)
+ (increase)
V < 0 (–)
+ (increase)
– (decrease)
R3-40