Welcome to PHYS2002!
Physics I – Done! We are now all experts in mechanics.
Mechanics → Mass
M1
r
M2
Interaction:
m1m2
F = G 2 G = 6.67 ×10−11 Nm / kg 2
r
We never said what mass is, only how it behaves.
New Semester
We are concerned with something called charge.
Electricity → Charge
*And just like in mechanics, E&M will not tell us what charge
is, but only how it behaves.
Like mass, charge is an intrinsic property of matter, and it
yp
comes in 2 types:
Positive and Negative
1
Chap. 18 – Electric Forces and Electric Fields
18 1 – Electricity
18.1
El t i it
Review the standard model of the atom:
Nucleus
*The SI unit of charge is
the Coulomb [C].
e
e
Proton charge is positive: +1e = +1.602 × 10-19 C
19 C
Electron charge is negative: -1e
1e = -1.602
1 602 × 10-19
Neutrons are neutral: No charge
2
18.2 Charged Objects and the Electrical Force
*In
In nature,
nature atoms like to have equal numbers of protons and electrons
electrons,
so that their net charge is zero.
These atoms are electrically neutral.
Electrical charges can be transferred between different
objects.
For
o example:
e a p e If we
e remove
e o ea
an e
electron
ect o from
o a neutral
eut a ato
atom,, we
e
are left with a positively charged ion.
We can transfer electrons between to objects by touching them together:
Hard rubber / Animal fur → Electrons are transferred from the fur to the rod
The rod becomes negatively charged, and the fur positively charged.
glass to the silk
Glass / Silk → Electrons are transferred from the g
The silk becomes negatively charged, and the glass positively charged.
3
When objects become electrified, it results only from a transfer
of charge, not a creation of charge:
Charge can neither be created or destroyed
Law of Conservation of Electrical Charge
Or the net electrical charge of an isolated system remains constant
Or,
constant.
Electrical Force
Take any two objects: Since they have mass, they exert a
force on one another → Gravitational Force
It’s always attractive!
Objects with excess charge will also exert forces on each other:
An Electrical Force
4
But unlike the gravitational force, the electrical force can be either
attractive or repulsive.
Opposite charges attract
Like charges repel
Hanging charged pith balls:
+ +
- -
+
-
Charged objects exert forces on one another. This is a new force called
the Electrostatic Force.
Forces overcome an object’s inertia and produce accelerations.
∑ F = ma
Newton’s 2nd Law
*Electrostatic forces can accelerate charged objects
5
18.3 Conductors and Insulators
Not only can electrical charge reside on the surface of an object
object, but it
can also move through it.
Charges move through some materials more easily than others:
• Charge moves easily: Conductor
i.e. copper, silver, aluminum (metals)
• Charge can’t move: Insulator
i.e. wood, paper, rubber, plastic,
*Charge can stick on the surface of insulators, but it doesn’t really move.
Electrical Wire
Copper
Rubber
6
What determines whether a material is a good conductor or insulator?
Ultimately it’s
Ultimately,
it s the atomic structure.
structure
*The outer most electrons (valence electrons) in an atom are more weakly
bound to the nucleus. They
y can “break free” and move through
g the
material. These are called conduction electrons.
-5 C
Negatively
charged
object
Positively
charged
object
e
e
e
+13 C
Good electrical
conductor
Let’s say the object on the left starts out with a charge of -5 C, and the
object on the right starts out with +13 C.
Electrons will continue to flow until the charge on each object is….?
EQUAL!
And, each must end up with a charge of +4C, since the total (+8C) must
remain constant!
Electrons can “flow” through a good conductor.
7
*There would be no charge flow if the bridge above was an insulator.
18.4 Charging an Object
Ch
Charging
i by
b contact:
t t
Touching a metal sphere with a negatively charged rod can give
the sphere a negative charge.
------
--------
Similarly if we started with a positively charged rod:
Similarly,
++++++
+++
++++
+++
This is charging by contact.
8
Charging by Induction:
We can also charge a conductor without actually touching it
it.
Bring a negatively charged rod close to the surface of an
electrically neutral metal sphere:
+
+
+
+
------
-
-
*The free charges separate
on the sphere’s surface.
Now attach a metal wire between the sphere
p
and ground:
g
----
+
+
+
+
-
-
Grounding strap
e-
This leaves the sphere
p
with a net positive charge
+++
++++
+++
-
The electrons travel down
the strap to ground
ground.
This is charging by Induction.
9
*Charging by induction doesn’t work for insulators, since the charge
can’t move through the material or down the grounding strap.
B t it does
But
d
have
h
an effect….
ff t
Bring a negatively charged rod close to the surface of an
insulating sphere:
*Even though the electrons can’t
move through
g the insulator,, the
positive and negative charge in
each atom separates slightly and
forms dipoles, since the positive
protons in the atoms are attracted
p
to the rod, and the negative
electrons are repelled.
F
This is called Polarization.
*However, due to the separation of charge,
the surface does acquire a slight positive
charge,
h
and
d is
i thus
th attracted
tt
t d to
t the
th rod.
d
This is the process that produces static cling.
D
10
Clicker Question 18-1
Object A is metallic and electrically neutral
neutral. It is charged by induction so
that it acquires a charge of -3.0 × 10-6 C. Object B is identical to object A
and is also electrically neutral. It is charged by induction so that it acquires
a charge of +3
+3.0
0 × 10-6 C.
C Which object now has a greater mass?
1.
1
2.
3.
Objectt A
Obj
Object B
They have the same mass
11
18.5 Coulomb’s Law
So charged objects exert forces on one another:
Let’s consider two positive point charges separated by some distance r:
Point charges: the charges are much smaller than the distance between them.
-F
q1
q2
+
+
+F
x
r
*Since they are like charges, they repel each other, and Newton’s 3rd Law tells
us they do this with equal and opposite force.
So what do we know about this force?
1 It grows weaker with distance
1.
distance.
2. It gets stronger for increasing charges.
3. It’s directed along the line containing the two charges.
Coulomb determined that:
q1q2
F =k 2
r
12
q1q2
F =k 2
r
This is Coulomb’s Law
k is a constant:
ε o = 8.85 ×10
-12
2
⋅
N
m
= 8.99 ×109
k=
4πε o
C2
1
C2
N ⋅ m2
This is the permitivity of free space.
Coulomb’s Law, like Newton’s Universal Law
of Gravitation, is an inverse square law.
m1m2
F =G 2
r
*Nature likes inverse square laws. They will
show up again.
13
Clicker Question 18-2
A +2.0-C point charge is located 2.0 m to the right of a -2.0-C point
charge.
h
Wh
Whatt is
i th
the magnitude
it d off th
the electrostatic
l t t ti fforce acting
ti on th
the
+2.0 C charge?
1.
1
2.
3.
4
4.
-2.0
2.0 C
-
0N
kN
2k N
16k N
2.0 m
+2.0
2.0 C
+
q1q2
(2)(2)
F =k 2 =k 2 =k
2
r
14
Clicker Question 18-3
In the previous problem, what is the direction of the force that acts on
the +2
+2.0
0 C charge?
1.
1
2.
3
3.
4.
→ (right)
← (left)
↑ (up)
( )
↓ (down)
-2.0 C
-
F +2.0 C
+
2.0 m
15
Coulomb’s Law with more than two charges:
S if we have
So
h
more than
th two
t
charges,
h
how
h
do
d we calculate
l l t the
th magnitude
it d
and direction of the force on a third (or more) charge(s)???
Let’s consider 3 positive point charges in a plane:
q1
+
q2
+
r12
r23
r13
+
q3
What is the magnitude and direction of the net force on q3?
q3 feels a force from both q1 (call it F1) and q2 (F2):
Both of these forces are repulsive since all the charges have the
same sign.
i
16
q1
+
q2
+
r12
Thus, the net force is the
VECTOR SUM of the two other
f
forces!
!
r23
r13
+
q3
F1
The magnitudes of F1 and F2 are just
calculated using Coulomb’s
Coulomb s Law:
q1q3
F1 = k 2
r13
F2
FNet
q2 q3
F2 = k 2
r23
18.6 The Electric Field
The earth exerts a force on the moon and vice
versa, even though they are 240,000 miles apart.
Likewise, two charged objects located far apart
exert forces on each other too.
How can they do this if they are not in physical
contact?
17
In the case of the earth/moon system, we say that the earth fills all space
with a gravitational field, and the moon feels the effect of this field.
Masses feel forces in gravitational fields.
Similarly, a charge creates an electric field that fills all
space Any other charge in that field will feel a force.
space.
force
Stationary charges create electric fields that fill all space.
Other
Ot
e charges
c a ges will feel
ee forces
o ces in tthese
ese e
electric
ect c fields.
e ds
Think of the electric field as a real physical entity!
Let’s say we have a positive point charge Q sitting somewhere in space:
This charge creates an electric field (E) that fills all space, so any
other charge should feel a force due to the field
field.
Bring a small positive charge qo (often called a
E
test charge), near Q:
Q +
The test charge will feel a force,
force F,
F directed
+qo
away from Q since they are like charges.
F
18
We define the electric field as:
F
E=
qo
*The electric field is a vector!
i.e. the electric field is force per unit charge:
Units?
⎡ Force ⎤ ⎡ N ⎤
⎢ Ch arg e ⎥ = ⎢ C ⎥
⎣
⎦ ⎣ ⎦
If we place a charge q in an electric field, it
will feel a force given by:
F = qE
*Note: E is created by other charges, not q.
19
Clicker Question 18-4
What is the acceleration of a proton placed in an electric
fi ld whose
field
h
magnitude
it d iis 10
10.5
5 N/C?
1. 2
1
2.5
5 m/s2
2. 5.0 × 105 m/s2
3 1.0
3.
1 0 × 109 m/s
/ 2
Thus,
a =0% 0%
mp
0%
10
9
1.
0
×
×
10
5
m
/s
2
m
/s
2
(1.602 ×10 −19 )(10.5)
9 m
a=
=
1
.
0
×
10
s2
(1.67 ×10 − 27 )
2.
5
mp = proton mass
qp = proton charge
qpE
m
/s
2
(Newton’s 2nd Law)
5.
0
F = qE = ma
20
Electric field of a point charge
Can we find an expression
p
for the electric field of a p
point charge?
g
Place a charge q at some distance r from a point charge Q:
Q
q
r
There is a force on q due to the electric field that Q creates:
F = qE
But, we can also calculate the force on q using Coulomb’s Law:
qQ
F =k 2
r
These must be equal:
qQ
qE = k 2
r
Q
E=k 2
r
This is the electric field of a point charge.
21
Clicker Question 18-5
Two identical point charges are fixed to opposite corners of a square.
What position should a third charge be placed so that it experiences a
greater force?
+
What is the force on a charge placed at the center?
0%
0%
ot
he
r
of
th
e
A
to
ne
+
e.
..
s.
..
Does the sign of the third charge matter?
ce
nt
er
of
th
e
+
he
2
2.
At the center of the
square
At one off the
th other
th
empty corners.
A
tt
1.
22
18.7 Electric Field Lines
So a charged object fills all space with an electric field. Can we visualize the field???
Yes!!! And we can do this by using a positive test charge.
+
Let’s start simple and try to map out the field lines around a positive point charge:
Move the test charge around the point charge
and note the direction of the force on it.
+
+
+
+
+
Continue this process:
*Notice that the force vector
arrows get shorter as we go
farther away from the charge
charge,
since the force gets smaller.
+
Now let’s connect the arrows!!!...
23
E
+
*The electrical field lines of a positive
point charge point radially outward
away
y from the charge
g and fill all space.
p
Spherical in 3 dimensions!!!
By a similar construction, we can show
that the field lines around a negative
point charge look like this:
i.e. the field lines around a negative
charge point radially in toward the
charge.
-
24
Electric field lines always start at and are directed away from positive
charges and always end at and are directed toward negative charges.
But what would the electric field look like in a region of
space
p
with more than one charge???
g
Consider an electric dipole:
Two equal but opposite charges
separated by some distance r.
Due to their structure, many
molecules have a dipole
moment. For example, water:
+q
r
-q
What would the
field lines look
like for the
dipole?
O-22
105o
H+
*Note: The lines start at the positive
charge and end on the negative one.
H+
25
Electric Field Lines
1. Start from positive charges and end on negative charges.
2. The electric field vector at some point in space is always tangent to the
….field
field line at that point
point.
The more field lines you have
per unit volume (density), the
stronger the field.
+q
r
-q
3. The number of field lines leaving a positive charge or going into a
….negative charge is proportional to the magnitude of the charge.
+Q
+2Q
26
Parallel Plate Capacitor
Consider a metallic plate
with a total charge
p
g +qq
distributed uniformly over its surface.
++++++
++++++
++++++
++++++
Units?
If the face of the plate has a surface area A
A, then
where σ is the surface charge density.
q
σ=
A
⎡ Charge ⎤ ⎡ C ⎤
⎢ Area ⎥ = ⎢ m 2 ⎥
⎣
⎦ ⎣ ⎦
Now let’s take two identical metal plates, one with total charge +q
and one with total charge –q.
+
+ + +
So the magnitude of σ is q/A for each plate.
+
+
E
+q
- - -
*In between the plates there exists a uniform electric field that
points from the positive plate to the negative one.
This is called a parallel plate capacitor.
-q
q
The electric field between the plates is given by:
q
σ
E=
=
εo A εo
Notice: The field does not depend on the distance from the charged plates! It’s uniform!
27
Clicker Question 18-6
An electron is released from rest at the negative plate of a parallel plate
capacitor What happens?
capacitor.
1.
2.
Nothing
The electron moves
right.
The electron moves
up.
The electron moves
down.
3.
4.
-
+
e-
1.5 × 10-2 m
28
Clicker Question 18-7
If the charge per unit area on each plate is σ = 1.8 × 10-7 C/m2,
and
d the
th plates
l t are separated
t db
by a di
distance
t
off 1
1.5
5 × 10-22 m, how
h
fast is the electron traveling when it strikes the positive plate?
1.
2.
3
3.
4.
1.0 m/s
1.0 × 103 m/s
1 0 × 105 m/s
1.0
1.0 × 107 m/s
-
+
e-
1.5 × 10-2 m
29
Clicker Question 18-7
(continued)
0
From kinematics we know that:
To solve for vf we need a:
v = v + 2ax = 2ax
2
f
2
o
qE
F = qE = ma ⇒ a =
m
E of a capacitor.
2qσx
2qEx
7
=
⇒ v f = 1.0 ×10 m / s
⇒v =
ε om
m
2
f
30
One more note on parallel plate capacitors: The field is only uniform near the middle of the plates.
+
E
*This is a fringing
effect. Also called
the fringe field.
-
18.8 Electric Field Inside a Conductor
Remember that charges can easily move through the interior of a conductor
conductor.
Let’s place a bunch of electrons at the center of a solid conducting sphere:
Repulsion
epu s o sp
spreads
eads the
t e electrons
e ect o s out!
out
- -- -- -
They move as far away from each other as
they can. Thus, they move to the surface.
*At equilibrium under electrostatic
conditions, any excess charge resides on
the surface of a conductor.
31
But what about the interior of the conductor?
There are still electrons there, but they are compensated exactly by the
positive charges in the interior. Thus, the interior is electrically neutral.
Since the charges are static, i.e. not moving, the force on them is zero, thus
E must be zero inside the conductor!
E = 0 iinside
id a conductor.
d t
Let’s place a conducting sphere in a uniform electric field:
+
+
+
+
+
E
-
- +
E=0
- +
+
+
+
-
No field lines penetrate the sphere, since E=0
inside a conductor.
-
The field lines end or begin
at the surface charges!
g
Cut out a cavity in the sphere interior:
E=0
E
0 iinside!
id ! Whatever
Wh t
we putt inside
i id the
th
conductor will be shielded from electric fields.
32
This is a static condition, so once the surface charges are induced, they don’t move.
g special
p
about the electric field at the
This tells us something
surface of a conductor: It must be perpendicular to the surface!
E
If the electric field was not perpendicular to the surface, then
there would be a component
p
of the field along
g the surface,, and
thus a force on the charges along the surface.
Perpendicular
- - - This is not correct!
E┴
E
E║
-
E must be perpendicular to the conductor’s surface!
-
-
-
-
33
Let’s look at a solid conductor with a hallowed out cavity inside:
Now let’s p
place a p
positive p
point charge
g inside the cavity:
y
+
+
This induces a net negative surface charge of
–Q on the inside surface of the cavity.
+
+
+
-
+
+
- - +Q
- - - -
-
+
+
+
+
+
*Thus, a net positive surface charge is induced
on the outer surface of the conductor.
The electric field lines would
then look like the following:
The electric field lines look
j t like
just
lik the
th field
fi ld lines
li
off a
positive point charge if the
sphere wasn’t there!
18.9 Electric Flux
Flux is a measure of how much field passes
perpendicularly through a surface.
34
Given a surface of area A, the electric flux
passing through the surface is given by:
Φ E = E⊥ A = ( E cos φ ) A
where
h
ΦE is
i the
th
electric flux.
Units?
⎡ N ⋅ m2 ⎤
[(Electric Field)(Area)] = ⎢
⎥
C
⎣
⎦
**Note: φ is the angle between E and the
normal to the surface!
Thus, Ecosφ gives us the
perpendicular component of E
passing through the surface.
Normal
E
Ecosφ
φ
A
35
,