Download Electricity and Magnetism Coulomb`s Law Electric Field

Electricity and Magnetism
Coulomb’s Law
Electric Field
Lana Sheridan
De Anza College
Sept 24, 2015
Last time
• charge
• charge interactions
• charge induction
.
Warm Up: Worksheet
.
3. Do both balloons A and B have a charge?
entry by
n entering the
rds each other
ne another
rge
opposite type
(A) yesor
r charged
(B) no, neither is charged
(C) at least 1 is charged.
ons being
.
Warm Up: Worksheet
.
3. Do both balloons A and B have a charge?
entry by
n entering the
rds each other
ne another
rge
opposite type
(A) yesor
r charged
(B) no, neither is charged
(C) at least 1 is charged.
ons being
←
charged or
Warm Up: Worksheet
5. Does this happen?
g straight
at _____.
re identical
other and both
(A) yes
(B) no
charged or
Warm Up: Worksheet
5. Does this happen?
g straight
at _____.
re identical
other and both
(A) yes
(B) no
← consider Newton’s 3rd law
Overview
• Coulomb’s Law
• The net force of several charges
• Vector review
• Charge quantization
• Charge conservation
• Current
• Forces at a fundamental level
• Electric field
• Conductors and electric fields
Electrostatic Forces
Charged objects interact via the electrostatic force.
The force that one charge exerts on another can be attractive or
repulsive, depending on the signs of the charges.
• Charges with the same electrical sign repel each other.
• Charges with opposite electrical signs attract each other.
Charge is written with the symbol q or Q.
Electrostatic Forces
For a pair of point-particles with charges q1 and q2 , the magnitude
of the force on each particle is given by Coulomb’s Law:
F1,2 =
k |q1 q2 |
r2
k is the electrostatic constant and r is the distance between the
two charged particles.
k=
1
4π0
= 8.99 × 109 N m2 /C2
Electrostatic Forces: Coulomb’s Law
F1,2 =
k |q1 q2 |
r2
Remember however, forces are vectors. The vector version of the
law is:
F1→2 =
k q1 q2
r̂1→2
r2
where F1→2 is the force that particle 1 exerts on particle 2, and
r̂1→2 is a unit vector pointing from particle 1 to particle 2.
Coulomb’s Law
Coulomb’s Law:
F1→2 =
Does this look a bit familiar?
k q1 q2
r̂1→2
r2
Coulomb’s Law
Coulomb’s Law:
F1→2 =
k q1 q2
r̂1→2
r2
Does this look a bit familiar?
Similar to this?
F1→2 = −
G m 1 m2
r̂1→2
r2
Coulomb’s Law
F1→2 =
Fields
When the charges are of the
same sign, the force is repulsive.
S
When the charges are of opposite
signs, the force is attractive.
"
q2
S
r
!
q1
k q1 q2
r̂1→2
r2
!
q2
F12
S
F12
S
rˆ12
!
q1
F21
F21
a
1
b
Figure from Serway & Jewett, Physics for Scientists and Engineers, 9th ed.
Electrostatic Constant
The electrostatic constant is: k =
1
4π0
= 8.99 × 109 N m2 C−2
0 is called the permittivity constant or the electrical
permittivity of free space.
0 = 8.85 × 10−12 C2 N−1 m−2
F12 ! " (1.15 # 10
Example
-8
N)i .
(Answer)
Sample Prob
Finding the net force due to
(a) Figure 21-8a shows two positively charged particles fixed in
place on an x axis.The charges are q1 ! 1.60 # 10 "19 C and q2 !
3.20 # 10 "19 C, and the particle separation is R ! 0.0200 m.
What are the magnitude
andisdirection
of the electrostatic force
This
the first
:
(a)
F12 on particle 1 from particle 2?
(b)
3n
has
par
T
1d
arrangement.
a
rged partiKEY IDEAS
q2
q1
q1
harges q1
x
Because
re fixed
in both particles are positively charged, particle 1 is re- Th
R magnitude given by Eq. 21-4.
pelled by particle 2, with a force
for
n an x axis.
:
Thus, the direction of force F12(a)
on particle 1 is away from partipar
free-body
cle 2, in the negative direction of the x axis, as indicated in the
to
free-body diagram of Fig. 21-8b.
Be
for particle
9
2
−2
−12
2
−1
−2
par
This
particle
kTwo
=elec8.99
× 10 N m
C Eq.
or21-4
0is
= the
8.85
×
10
C substituted
N m
particles:
Using
with
separation
R
ng the
rec
F12 of this force as
of interest.
forcefor
onr, we
it can write the magnitude
gra
F12 ! " (1.15 # 10
Example
-8
N)i .
(Answer)
Sample Prob
Finding the net force due to
(a) Figure 21-8a shows two positively charged particles fixed in
place on an x axis.The charges are q1 ! 1.60 # 10 "19 C and q2 !
3.20 # 10 "19 C, and the particle separation is R ! 0.0200 m.
What are the magnitude
andisdirection
of the electrostatic force
This
the first
:
(a)
F12 on particle 1 from particle 2?
(b)
3n
has
par
T
1d
arrangement.
a
rged partiKEY IDEAS
q2
q1
q1
harges q1
x
Because
re fixed
in both particles are positively charged, particle 1 is re- Th
R magnitude given by Eq. 21-4.
pelled by particle 2, with a force
for
n an x axis.
:
Thus, the direction of force F12(a)
on particle 1 is away from partipar
free-body
cle 2, in the negative direction of the x axis, as indicated in the
to
free-body diagram of Fig. 21-8b.
Be
for particle
−24
par
This
is
the
particle
Answer:
F2→1 = −1.15
10
iN
Two
particles:
Using×Eq.
21-4
with
separation
R substituted
ng the
elecrec
F12 of this force as
of interest.
forcefor
onr, we
it can write the magnitude
gra
Force from many charges
Forces from many charges add up to give a net force
This is (very grandly) called the “principle of superposition”.
The net force on particle 1 from particles 2, 3, ... n is:
Fnet,1 = F2→1 + F3→1 + ... + Fn→1
t is true
about S
the electric
forcesSon the S
objects?
S
S
5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA
Example
F BA
Consider three point charges located at the corners of a right
triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and
a = 0.100 m. Find the resultant force exerted on q3 .
y
iangle as shown in
0.100 m. Find the
q2 "
rge q 3 is near two
are exerted in difhown in the figure,
egorize this exam-
a
q1 !
Figure 23.7
S
a
S
F23
F13
!
q3
!2a
x
(Example 23.2) The
S
force
exerted
by
q 1 on q 3 is F 13. The
1
S
Figure from Serway & Jewett, Physics for Scientists
and Engineers, 9th ed.
5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA
FExample
BA
Consider three point charges located at the corners of a right
triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and
a = 0.100 m. Find the resultant force exerted on q3 .
y
iangle as shown in
0.100 m. Find the
q2 "
rge q 3 is near two
are exerted in difhown in the figure,
a
q1 !
S
a
S
F23
F13
!
q3
!2a
x
egorize this examFigure
(ExampleONLY
23.2) The
Step 1: What is the
force23.7
considering
particles 2 and 3?
S
force exerted by q 1 on q 3 is F 13. The
S
force
q 2 on for
q isScientists
F 23.
Figure
from Serway &exerted
Jewett, by
Physics
and Engineers, 9th ed.
and q on
q are
S 3
1
5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA
FExample
BA
Consider three point charges located at the corners of a right
triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and
a = 0.100 m. Find the resultant force exerted on q3 .
y
iangle as shown in
0.100 m. Find the
rge q 3 is near two
are exerted in difhown in the figure,
q2 "
a
q1 !
S
a
S
F23
F13
!
q3
!2a
x
egorize this examFigure
Step 2: What is the
force23.7
from(Example
particle 23.2)
1Son The
particle 3? It has 2
force
exerted
by
q
on
q
is
F . The
1
3
components.
S13
force exerted by q 2 on q 3 is F 23.
1
q 3 are
S for Scientists and Engineers, 9th ed.
1 and q 2 on
Figure
from Serway & Jewett, Physics
Reminder about Vectors
scalar
A scalar quantity indicates an amount. It is represented by a real
number. (Assuming it is a physical quantity.)
Reminder about Vectors
scalar
A scalar quantity indicates an amount. It is represented by a real
number. (Assuming it is a physical quantity.)
vector
A vector quantity indicates both an amount and a direction. It is
represented more than one real number. (Assuming it is a physical
quantity.)
Reminder about Vectors
scalar
A scalar quantity indicates an amount. It is represented by a real
number. (Assuming it is a physical quantity.)
vector
A vector quantity indicates both an amount and a direction. It is
represented more than one real number. (Assuming it is a physical
quantity.)
There are many ways to represent a vector.
• a magnitude and (an) angle(s)
• magnitudes in several perpendicular directions
Representing Vectors: Angles
Bearing angles
Example, a plane flies at a bearing of 70◦
N
E
Reference angles x-axis, CCW
A baseball is thrown at 10 ms −1 30◦ above the horizontal.
Representing Vectors: Unit Vectors
Another useful way to represent vectors is in terms of unit vectors.
Unit vectors have a magnitude of one unit.
In this course, a unit vector r̂ is a one-unit-long vector parallel to
the vector r.
Representing Vectors: Unit Vectors
Another useful way to represent vectors is in terms of unit vectors.
Unit vectors have a magnitude of one unit.
In this course, a unit vector r̂ is a one-unit-long vector parallel to
the vector r.
In two dimensions, a pair of perpendicular unit vectors are usually
denoted i and j (or sometimes x̂, ŷ).
Representing Vectors: Unit Vectors
Another useful way to represent vectors is in terms of unit vectors.
Unit vectors have a magnitude of one unit.
In this course, a unit vector r̂ is a one-unit-long vector parallel to
the vector r.
In two dimensions, a pair of perpendicular unit vectors are usually
denoted i and j (or sometimes x̂, ŷ).
A generic 2 dimensional vector can be written as v = ai + bj,
where a and b are numbers.
um of two other component vectors A x , which is parallel to the x axis, and A y , whic
parallel
to the y axis. S
From SFigure
3.12b, we see that the three vectors form
Components
S
ight triangle
and
that
A
5
A
1
A
. We shall often refer to the “component
x
y
S
f a vector A ,” written Ax and Ay (without
the boldface notation). The compo
S A = Ax i + Ay j, where a and b
Consider the 2 dimensional vector
ent Ax represents the projection
of
A
along
the x axis, and the component A
S
are numbers.
epresents the projection of A along the y axis. These components can
be positiv
S
r negative. The component Ax is positiveSif the component vector A x points i
We then say that Ax is the i-component (or x-component) of A
he positive x direction and is negative if A x points in the negative x direction.
and Ay is the j-component (or y -component) of A.
milar statement
is made for the component Ay.
y
y
S
S
S
A
Ay
u
O
a
S
Ax
A
x
u
O
b
Notice that Ax = A cos θ and Ay = A sin θ.
S
Ax
S
Ay
x
When two vectors are added, the sum is indepe
tion. (This fact may seem trivial, but as you will s
Vectors Properties and Operations
important when vectors are multiplied. Procedures
cussed in Chapters 7 and 11.) This property, which c
construction in Figure 3.8, is known as the commut
Equality
S
S
S
Vectors
A = Blawifofand
only if the magnitudes and directions
areB the
A 1
5 B
Commutative
addition
same. (Each component is the same.)
Addition
A+B
S
S
A "
S
C
S
S
B
S
R!
S
B
A"
S
R !
S
B "
S
C "
D
S
D
S
B
S
S
A
A
S
Figure 3.6
When vector B is
S
S
added to vector A , the resultant R is
the vector that runs from the tail of
S
S
Figure 3.7 Geometric construction for summing four vectors. The
S
resultant vector R is by definition
S
1 A
mbers that
normal rules
C 1 D is the vector that completes the polygon. In other words, R is the vector
drawn from the tail of the first vector to the tip of the last vector. This technique for
adding vectors is often called the “head to tail method.”
When two vectors are added, the sum is independent of the order of the addition. (This fact may seem trivial, but as you will see in Chapter 11, the order is
important when vectors are multiplied. Procedures for multiplying vectors are discussed in Chapters 7 and 11.) This property, which can be seen from the geometric
construction in Figure 3.8, is known as the commutative law of addition:
Vectors Properties and Operations
f addition
S
S
S
S
(3.5)
A 1 B 5 B 1 A
A Property of Addition
S
Draw B ,
S
then add A .
S
A
A"
S
B
A + B = B + A (commutative)
S
S
S
C
A "
B
S
R !
S
S
B
S
B
S
A
S
vector B is
S
he resultant R is
from the tail of
B
S
S
C "
B "
S
S
S
S
B
R! S
B"
D
S
A!
S
D
Figure 3.7 Geometric construction for summing four vectors. The
S
resultant vector R is by definition
the one that completes the polygon.
S
A
S
Draw A ,
S
then add B .
Figure 3.8S ThisSconstruction
S
S
shows that A 1 B 5 B 1 A or, in
other words, that vector addition is
commutative.
he bookkeeping
nents
separately.
Vectors
Properties and Operations
Doing addition:
Almost always the right answer is to break each vector into
components and sum each component independently.
y
(3.14)
ant vector are
(3.15)
s, we add all the
ctor and use the
omponents with
Ry
S
By
R
Ay
S
B
S
A
x
Bx
Ax
Rx
btained from its
Figure 3.16
This geometric
S
S
S
S
S
5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA
FExample
BA
Consider three point charges located at the corners of a right
triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and
a = 0.100 m. Find the resultant force exerted on q3 .
y
iangle as shown in
0.100 m. Find the
q2 "
rge q 3 is near two
are exerted in difhown in the figure,
egorize this exam-
a
q1 !
Figure 23.7
S
a
S
F23
F13
!
q3
!2a
x
(Example 23.2) The
S
force exerted by q 1 on q 3 is F 13. The
S
1
Figure from Serway
&exerted
Jewett, by
Physics
and Engineers, 9th ed.
force
q 2 on for
q 3 isScientists
F 23.
5S2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA
FExample
BA
Consider three point charges located at the corners of a right
triangle as shown, where q1 = q3 = 5.00 µC, q2 = −2.00 µC, and
a = 0.100 m. Find the resultant force exerted on q3 .
y
iangle as shown in
0.100 m. Find the
q2 "
rge q 3 is near two
are exerted in difhown in the figure,
egorize this exam-
1
a
q1 !
S
a
S
F23
F13
!
q3
!2a
x
Figure 23.7 (Example 23.2)
The
S
Answer: Fnet,3 = force
(−1.04
i + 7.94
exerted
by q 1 j)
onN
q 3 is F 13. The
S
force exerted by q 2 on q 3 is F 23.
1
and q 2 on
q 3 are
S for Scientists and Engineers, 9th ed.
Figure
from Serway & Jewett, Physics
Charge is Quantized
quantization
A physical quantity is said to be quantized if if can only take
discrete values.
Originally, charge was thought to be a continuous fluid.
Charge is Quantized
quantization
A physical quantity is said to be quantized if if can only take
discrete values.
Originally, charge was thought to be a continuous fluid.
It is not.
Charge is Quantized
quantization
A physical quantity is said to be quantized if if can only take
discrete values.
Originally, charge was thought to be a continuous fluid.
It is not.
Just like water has a smallest unit, the H2 O molecule, charge has a
smallest unit, written e, the elementary charge.
e = 1.602 × 10−19 C
Basic Unit of Charge
The elementary charge.
e = 1.602 × 10−19 C
Any charge must be
q = ne ,
n∈Z
The charge of an electron is −e and the proton has a charge +e.
Question
Initially, sphere A has a charge of −50e and sphere B has a charge
of 20e. The spheres are made of conducting material and are
identical in size. If the spheres then touch, what is the resulting
charge on sphere A?
(A) −50e
(B) −30e
(C) −15e
(D) 20e
Question
Initially, sphere A has a charge of −50e and sphere B has a charge
of 20e. The spheres are made of conducting material and are
identical in size. If the spheres then touch, what is the resulting
charge on sphere A?
(A) −50e
(B) −30e
(C) −15e
(D) 20e
←
Conservation of Charge
Charge can move from one body to another but the net charge of
an isolated system never changes.
This is called charge conservation.
Conservation of Charge
Charge can move from one body to another but the net charge of
an isolated system never changes.
This is called charge conservation.
What other quantities are conserved?
Conservation of Charge
One interesting phenomenon that shows the conservation of charge
is pair production.
A gamma ray (very high energy photon) converts into an electron
and a positron (anti-electron):
γ → e− + e+
New mass is created out of light, but charge is still conserved!
Current
Current is the the rate of flow of charge.
Current is written with the symbol I or i.
i=
(If you like calculus, use i =
dq
dt .)
∆q
∆t
Coulombs and Ampères
The unit for current is the Ampère, or more commonly, “Amp”.
Using the definition for current, 1 A = 1 C / 1 s.
Therefore, we can formally define the unit of charge in terms of the
unit of current:
1 C = (1 A)(1 s)
Question
574
pg 574, #4
CHAPTER 21
** View All
Solutions Her
ELECTRIC CHARGE
4 Figure 21-15 shows two charged
–3q
–q
particles on an axis. The charges are
free to move. However, a third
Fig. 21-15 Question 4.
charged particle can be placed at a
certain point such that all three particles are then in equilibrium. (a)
Is that point to the left of the first two particles, to their right, or between them? (b) Should the third particle be positively or negatively
charged? (c) Is the equilibrium stable or unstable?
5 In Fig. 21-16, a central particle of
charge !q is surrounded by two circular rings of charged particles. What
are the magnitude and direction of
the net electrostatic force on the central particle due to the other parti-
+4q
–2q
+2q
–2q
+q
–7q
r
–7q
Rank
elect
first.
9 F
charg
cles o
the m
an el
Are
Is the
great
two f
cel?
that
Question
R/2
+8Q
particles.
d
e
Question 9.
10 In Fig. 21-20, a central particle of charge !2q is surrounded by a
square array of charged particles, separated by either distance d or d/2
along the perimeter of the square. What are the magnitude and direction of the net electrostatic force on the central particle due to the
other particles? (Hint: Consideration of symmetry can greatly reduce
the amount of work required here.)
+2q
–7q
+3q
d
–3q
+4q
+4q
–3q
–5q
e
e
(4)
Fig. 21-19
pg 574, #10
(c)
p
(3)
–2q
–5q
–7q
+2q
Of the charge Q initially on a tiny sphere, a portion
cles 1 and 2 ha
L23 " L12, what
be treated as particles. For what value of q/Q will the electrostatic
••8 In Fig. 21
conducting sph
the following ch
sphere B, #6Q
Spheres A and
with a centerthat is much lar
Two experimen
touched to sphe
removed. In ex
procedure is re
(separately) to s
the electrostatic
that at the end o
•1
SSM
ILW
Question q is to be transferred to a second, nearby sphere. Both spheres can
pg 575, #2
force between the two spheres be maximized?
•2 Identical isolated conducting spheres 1 and 2 have equal
charges and are separated by a distance that is large compared with
their diameters (Fig. 21-21a). The electrostatic force acting on
:
sphere 2 due to sphere 1 is F . Suppose now that a third identical
sphere 3, having an insulating handle and initially neutral, is
touched first to sphere 1 (Fig. 21-21b), then to sphere 2 (Fig. 21-21c),
and finally removed (Fig. 21-21d). The electrostatic force that now
acts on sphere 2 has magnitude F!.What is the ratio F!/F?
–F
1
2
F
1
3
(a )
1
(b )
2
3
(c )
2
–F'
1
2
(d )
F'
••9 SSM WW
place, attract ea
their center-toconnected by a
the spheres rep
Of the initial c
with a positive n
the negative c
and (b) the p
other?
Forces at a Fundamental Level
Often people think about two kinds of forces: contact forces and
field forces (ie. forces that act at a distance).
In mechanics problems, all forces except gravity are from direct
contact.
Gravity is a field force.
The electric and magnetic forces are also field forces.
Forces at a Fundamental Level
Often people think about two kinds of forces: contact forces and
field forces (ie. forces that act at a distance).
In mechanics problems, all forces except gravity are from direct
contact.
Gravity is a field force.
The electric and magnetic forces are also field forces.
And actually, at a fundamental level, all forces that we know of are
field forces.
Forces at a Fundamental Level
Contact forces are a result of electrostatic repulsion at very small
scales.
Forces at a Fundamental Level
Contact forces are a result of electrostatic repulsion at very small
scales.
Fundamental forces:
Force
Gravitational
Electromagnetic
Weak Nuclear
Strong Nuclear
∼ Rel. strength
10−38
10−2
10−13
1
Range (m)
∞
∞
< 10−18
< 10−15
Attract/Repel
attractive
attr. & rep.
attr. & rep.
attr. & rep.
Carrier
graviton
photon
W +, W −, Z 0
gluons
Forces at a Fundamental Level
Contact forces are a result of electrostatic repulsion at very small
scales.
Fundamental forces:
Force
Gravitational
Electromagnetic
Weak Nuclear
Strong Nuclear
∼ Rel. strength
10−38
10−2
10−13
1
Range (m)
∞
∞
< 10−18
< 10−15
Attract/Repel
attractive
attr. & rep.
attr. & rep.
attr. & rep.
Carrier
graviton
photon
W +, W −, Z 0
gluons
Forces at a Fundamental Level
Contact forces are a result of electrostatic repulsion at very small
scales.
Fundamental forces:
Force
Gravitational
Electromagnetic
Weak Nuclear
Strong Nuclear
∼ Rel. strength
10−38
10−2
10−13
1
Range (m)
∞
∞
< 10−18
< 10−15
Attract/Repel
attractive
attr. & rep.
attr. & rep.
attr. & rep.
Carrier
graviton
photon
W +, W −, Z 0
gluons
Gravity is actually quite a weak force, but it is the only one that
(typically) matters on large scales - charges cancel out!
Fields
field
A field is any kind of physical quantity that has values specified at
every point in space and time.
Fields
In EM we have vector fields. The electrostatic force is mediated by
a vector field.
vector field
A field is any kind of physical quantity that has values specified as
vectors at every point in space and time.
Fields
Fields were first introduced as a calculation tool.
A force-field can be used to identify the force a particular particle
will feel at a certain point in space and time based on the other
objects in its environment that it will interact with.
Imagine a charge q0 . We want to know the force it would feel if we
put it at a specific location.
Fields
Fields were first introduced as a calculation tool.
A force-field can be used to identify the force a particular particle
will feel at a certain point in space and time based on the other
objects in its environment that it will interact with.
Imagine a charge q0 . We want to know the force it would feel if we
put it at a specific location.
The electric field E at that point will tell us that!
F = q0 E
Fields
The source of the field could be another charge.
We do not need a description of the sources of the field to describe
what their effect is on our particle. All we need to know if the
field!
Fields
The source of the field could be another charge.
We do not need a description of the sources of the field to describe
what their effect is on our particle. All we need to know if the
field!
This is also true for gravity. We do not need the mass of the Earth
to know something’s weight:
FG = m0 g
FE = q0 E
22
The rod sets up an
electric field, which
can create a force
on the test charge.
Force from a Field
F
+
+
+ +
Test charge q0
+ +
at point P
+ +
+ +
+ + Charged
E
22-1
P
W H AT I S
The+ physics
+ +
a +particle
1 of c
+
charge #q
+ +2. A nagging q
+
+
ence +of particle 2? Tha
+
object
2 push on particle 1 — ho
(b)
One purpose of phy
(a)
(a) A positive test charge
F = q0 EFig. 22-1
magnitude
and direction
The rod sets up an q0 placed
at point P near a charged ob:
explanation
of the
w
but also:
electrostatic
force F acts on
electric field, which ject. Andeeper
:
test charge.
(b)aThe
electric explanati
field E at
such
deeper
can create a force point P produced by the charged object
F
tance. We can answer th
E
E=
on the test charge.
q0
fieldTable
in the22-1
space surroun
space,
the particle “know
Fields
P Halliday, Resnick, Walker.Some Electric
1
Figure from
+
Electric
field on
force
at point P
Field Lines
Fields are drawn with lines showing the direction of force that a
test particle will feel at that point. The density of the lines at that
PA R T 3
point in the diagram indicates the approximate
magnitude of the
581
22-3
ELECTRIC
FIELD
LINES
force at that point.
ne the electric field of a charged
charge. The field at point P in
st charge of Fig. 22-1a was put
the presence of the test charge
arged object, and thus does not
–
–
–
–
–
–
–
F
+ Positive
test charge
e interaction between charged
ctric field produced by a given
ce that a given field exerts on a
Sections 22-4 through 22-7 for
ond task in Sections 22-8 and
oint charges in an electric field.
ic fields.
ctric fields in the 19th century,
ed with lines of force. Although
now usually called electric field
erns in electric fields.
–
(a)
–
–
–
–
–
–
–
–
E
Electric
field lines
(b)
Field Lines
The electrostatic
cally symmetric
electric
duced bysomething
a point charge
like:
field
by an
electric
dipole system looks
An caused
electric field
produced
by an
electric dipole
q
!
c
Notice that the lines point outward from a positive charge and
lue lines are intersections of these surfaces with the page) and elecinward toward
a negative
charge.
s are perpendicular
to the electric
field lines
at every point.
1
Figure from Serway & Jewett
Field Lines
C
Compare the electrostatic fields for two like charges and two
FIELDS
opposite charges:
ive
.
ng
ge.
c-
c
it
t.
tric field it represents are said to have rotational symmetry about that axis.The electric
field vector at one point is shown; note that it
is tangent to the field line through that point.
fi
n
n
positive test charge at any point near then
of Fig. 22-3a, the net electrostatic force acti
s
the test charge would be perpendicular t
s
sheet, because forces acting in all other fi
+
tions would cancel +one another as a res
the symmetry. Moreover, the net force o
E
test charge
would point away from the sh
E
2
shown. Thus, the electric
field vector at any
–
+
in the space on either side of the sheet iT
perpendicular to the sheet and directeda
from it (Figs. 22-3b and c). Because the chaF
uniformly distributed along the sheet, a
Fig.Such
22-5 an
Field
lines forfield,
a positive
point
field vectors have the same magnitude.
electric
with
the same
and a nearby
negative
point charge
nitude and direction at every point, ischarge
a uniform
electric
field.
T
that are equal in magnitude.The charges atOf course, no real nonconducting sheet (such as a flat expanse of plastic)
Gravity figure from http://www.launc.tased.edu.au ; Charge from Halliday,
Resnick, Walker
1
Page 582
qual positive
ach other.
gative
-dimenlly rotate
axis passing
of the page.
d the elecave rotahe electric
note that it
that point.
Compare the fields for gravity in an Earth-Sun system and
electrostatic repulsion of two charges:
positive test charge
of Fig. 22-3a, the net
the test charge wo
sheet, because forc
+
tions would cancel
the symmetry. Mor
E
test charge would p
shown. Thus, the ele
+
in the space on eit
perpendicular to t
from it (Figs. 22-3b
uniformly distribut
field vectors have the same magnitude. Such an elect
nitude and direction at every point, is a uniform elect
LECTRIC FIELDS
14:16
Field Lines
field lines with distance from the sphere tells us that the magnic field decreases with distance from the sphere.
of Fig. 22-2 were of uniform positive charge, the electric field
ints near the sphere would be directed radially away from
he electric field lines would also extend radially away from the
ave the following rule:
Field Lines
es extend away from positive charge (where they originate) and
harge (where they terminate).
Imagine an infinite sheet of charge. The lines point outward from
the positively charged sheet.
shows part of an infinitely large, nonconducting sheet (or plane)
stribution of positive charge on one side. If we were to place a
electrostatic force
harge near a very
g sheet with uniositive charge on
:
ctric field vector E
test charge, and
s in the space
eld lines extend
vely charged
of (b).
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
F
+
+
+
+
+
(a)
1
Positive test
charge
+
+
+
+
+
+
+
+
+
+
+
+
(b)
Figure from Halliday, Resnick, Walker.
E
+
+
+
+
+
+
+
+
+
(c)
Field from a Point Charge
We want an expression for the electric field from a point charge, q.
Using Coulomb’s Law the force on the test particle is
0
F→0 = k rqq
2 r̂.
F
1 k q
q
0
E=
=
r̂
2
q0
q
r
0
The field at a displacement r from a charge q is:
E=
kq
r̂
r2
charge and a nearby negative point charge
that are equal in magnitude.The charges attract each other.The pattern of field lines and
the electric field it represents have rotational
symmetry about
axis passing
through
both
displacement
r an
from
a charge
q is:
charges in the plane of the page.The electric
field vector at one pointkisqshown; the vector
is tangent to the E
field
the point.
r̂
=line through
Field from a Point Charge
The field at a
r2
This is a vector field:
+
The electric field vectors at
various points around a positive point
charge.
Fig. 22-6
:
The direction of F is d
toward the point charg
:
The direction of E is
directly away from th
Because there is
gives the field at eve
point charge is shown
We can quickly fin
charge. If we place a p
from Eq. 21-7, the net f
Therefore, from Eq. 2
:
Here Ei is the electri
Equation 22-4 shows
fields as well as to ele
Charges and Conductors
Excess charge sits on the outside surface of a conductor.
The electric field lines are perpendicular to the surface.
1
Figure from OpenStax College Physics.
ormly
Conductors and Electric fields
charge
reachConsider a neutral conductor placed in an electric field:
d, proushes,
arges,
ch cirwhere
+ + +++
–– – –
––
++
–
made
––
+
20, all
er the
–
Fig. 24-20
–
–
E=0
–
–
–
+
+
+
+
+
++
An uncharged conduc-
Conductors and Electric fields
Electric fields exert forces on free charges in conductors.
Each charge keeps moving until:
1
the charges reaches the edge of the conductor and can move
no further OR
2
the field is cancelled out!
Inside a conducting object, the electric field is zero!
12
E (kV/m)
om the curve of Fig. 24-18a by differentiating with respect to r,
Faraday
recall
that theCages
derivative of any constant is zero). The curve of
A
conducting
can of
shield
interior
from even very
be derived
from theshell
curves
Fig. the
24-18b
by integrating
withstrong
electric fields.
g Eq. 24-19.
8
4
0
Fig. 24-1
rge spark
dy and then
ross the
t tire (note
ving the pered. (Courtesy
ric
1
Photo from Halliday, Resnick, Walker
inside and
cal shell o
E(r) for t
Faraday Cages
1
Photo found on TheDailySheeple, credits unknown.
)
Charges Inside Conductors: The Faraday Ice Pail
Gaussian
surface
_
_ +
+
_
+
+
_+
+
_ +
+
_
+
_
+
_
_
+
_+
+_
+
_
(b )
_
_
ch
ou
ge
un
sk
no
ut
tiv
ap
th
Summary
• Coulomb’s law
• Quantization of charge
• Charge conservation
• Current
• electric field
• field of a point charge
Homework
worksheets:
physicsclassroom.com/getattachment/curriculum/estatics/...
• ...static5.pdf
• ...static7.pdf
Halliday, Resnick, Walker:
• Ch 21, onward from page 573. Questions: 1; Sec Qs: 3, 9, 27,
42 & 43