nˆ s E nˆ 50 kV Clean air b z r a E r b q b a E Dirty air qin 0 + - Electricity Electric Fields Electric Charge •Electric forces affect only objects with charge •Charge is measured in Coulombs (C).
Download ReportTranscript nˆ s E nˆ 50 kV Clean air b z r a E r b q b a E Dirty air qin 0 + - Electricity Electric Fields Electric Charge •Electric forces affect only objects with charge •Charge is measured in Coulombs (C).
50 kV
Dirty air Clean air
E q
in 0
E
z r a s
n
E
b
r b q b a
+ -
Electricity Electric Fields Electric Charge
•Electric forces affect only objects with charge •Charge is measured in Coulombs (C). A Coulomb is a lot of charge •Charge comes in both positive and negative amounts •Charge is conserved – it can neither be created nor destroyed •Charge is usually denoted by
q
or
Q
•There is a fundamental charge, called
e
•All elementary particles have charges that Particle Proton are simple multiples of
e
Neutron
e
1.602 10 19 C Electron Red dashed line means you should be able to use this on a test, but you needn’t memorize it -
q e
0
e
Oxygen nuc. 8
e
++ 2
e
Charge Can Be Spread Out
Charge may be at a point, on a line, on a surface, or throughout a volume •Linear charge density •Multiply by length •Surface charge density units C/m units C/m 2 •Multiply by area •Charge density units C/m 3 •Multiply by volume
Concept Question
A box of dimensions 2 cm has charge density
=
2 cm 5.0 C/cm 3 1 cm throughout and linear charge density
=
– 3.0 C/cm along one long diagonal. What is the total charge? A) 2 C B) 5 C C) 11 C D) 29 C E) None of the above
V
lwh
4 cm 3
5.0
C/cm 3
2 cm
L
l
2
w
2
h
2 2 2 2 2 2 1 cm 3 cm
q
V
L
The Nature of Matter
•Matter consists of positive and negative charges in
very
•There are nuclei with positive charges •Surrounded by a “sea” of negatively
+ +
charged electrons large quantities
+ +
•To charge an object, you can add some
+ + +
charge to the object, or remove some charge •But normally only a
very
small fraction
+ + +
•10 -12 of the total charge, or less •Electric forces are what hold things together •But complicated by quantum mechanics •Normally it is electrons that do the moving
+ + +
•Some materials let charges move long distances, others do not
+ + +
Insulators only let their charges move a very short distance Conductors allow their charges to move a very long distance
Some ways to charge objects
•By rubbing them together •Not well understood •By chemical reactions •This is how batteries work •By moving conductors in a magnetic field •Get to this in March •By connecting them to conductors that have charge already •That’s how outlets work •Charging by induction •Bring a charge
near
an extended conductor •Charges move in response •Separate the conductors •Remove the charge
+ – – – – – – – – + + + + + + + + + +
Coulomb’s Law
•Like charges repel, and unlike charges attract •The force is proportional to the charges •It depends on distance
q
1
r
q
2
F
2
k q q e
1 2
r
2
k e
2 2
Other ways of writing this formula
•The
r
-hat just tells you the direction of the force
F
2
k q q e
•When working with components, often helps to rewrite the
r
r-
1 2 2 hat •Sometimes this formula is written in terms of a quantity 0 called the
permittivity of free space
F
2
k q q e
1 2
r
3
r F
2
q q
1 2 4 0
r
2
r
ˆ 0 4 1
k e
12 2
Concept Question
What is the direction of the force on the purple charge?
A) Up B) Down C) Left D) Right E) None of the above
+2.0
C 5.0 cm
•The separation between the purple charge and each of the
L
other charges is identical •The magnitude of those forces is identical
F
k q q e
1 2
r
2 5 cm 5 cm 2 7.1 cm 9 2 2 0.071 m 2 6 2 10 C 2 •The blue charge creates a repulsive force at 45 down and left •The green charge creates an attractive force at 45 up and left •The sum of these two vectors points straight left
F
tot 7.2 N 2 10.2 N
angle
180
5.0 cm –2.0
C –2.0
C
7.2 N
Sample Problem
Three charges are distributed as shown at right. Where can we place a fourth charge of magnitude 3.0 mC such that
F
1 the total force on the 1.0 mC vanishes?
k q q e
1 2
r
2
k e
2
i
2 ˆ
i
k e
mC 2 m 2
F
2
k q q e
1 2
r
2 0 1
F
2
r
ˆ
F
3 ,
k e
2 m 2
j
F
3 2 ˆ ˆ
j
k e
ˆ
j
mC m 2 2 ,
k e
mC 2 m 2
3.0 mC ?
1.0 mC 2.0 mC 1.0 m -4.0 mC
F
3 2.236
k e
mC 2 m 2 tan 1 2
F
3
k q q e
1 2
r
2
k e
2
r
2 3
r
2 2.236
m 2
r
1.16 m Angle 360 333
Forces From Continuous Charges
•If you have a spread out charge, it is tempting to
q
start by calculating the total charge •Generally not the way to go •The charge of the line is easy to find
, Q
=
L
r
dl
•But the distance and direction is hard to find •To deal with this problem, you have to divide it up into little segments of length
dl
•Then calculate the charge
dQ
=
dl
for each little piece •Find the separation
r
for each little piece •Add them up – integrate •For a 2D object, it becomes a double integral •For a 3D object, it becomes a triple integral
F
q
F
k e r
2
q
k e
dA
r
ˆ
r
2
F
q
k e
dV
r
ˆ
r
2
The Electric Field
F
F
•Suppose we have some distribution of charges •We are about to put a small charge
q
0 at a point
r
•What will be the force on the charge at
r
?
•Every term in the force is proportional to
q
0 •The answer will be proportional to
q
0 •Call the proportionality constant
E
, the electric field 0
q
0
q
E
E
E
F
0
q
0 The units for electric field are N/C •It is assumed that the test charge
q
0 is small enough that the other charges don’t move in response •The electric field
E
is a function of
r
, the position •It is a
vector field
, it has a direction in space everywhere •The electric field is assumed to exist even if there is no test charge
q
0 present
r
q
0
Electric Field From a Point Charge
E
F
0
q
0
q
r
q
0 •From a single point charge, the electric field is easy to find •It points away from positive charges •It points towards negative charges
F
0
k qq e
0
r
2
r
ˆ
E
k q e r
2
r
ˆ
+
Electric Field from Two Charges
•Electric field is a vector • We must add the vector components of the contributions of multiple charges
E
i k q e i r i
2
r
ˆ
i
+ + +
Electric Fields From Continuous Charges
P
•If you have a spread out charge, we can add up the contribution to the electric field from each part •To deal with this problem, you have to divide it up into little segments of length
dl
•Then calculate the charge
dQ
•Find the separation
r
=
dl
for each little piece and the direction r-hat for each little piece •Add them up – integrate •For a 2D object, it becomes a double integral •For a 3D object, it becomes a triple integral
r E
k e r
2
E
dl
r
ˆ
k e
dA
r
ˆ
r
2
E
k e
dV
r
ˆ
r
2
Sample Problem
•Divide the charge into little segments
dl
– Because it is on the
y
-axis,
dl
=
dy
•The vector
r
points from the source of the electric field to the point of measurement – It’s magnitude is
r
= •Substitute into the integral – Limits of integral are
y
– It’s direction is the minus-y direction
y= a
and
y
= 2
E
a
k dy e y
2
y
2
Qy
a
2
j
k Q e
ˆ
j
a y
2
dy
a
2
r
y
r
ˆ ˆ
j E
k e r
2
r
ˆ What is the electric field at the origin for a line of charge on the
y
-axis with linear charge density (
y
) =
Qy
2 /(
y
2
+a
2 ) stretching from
y
=
a
to
y
= ?
•Pull constants out of the integral •Look up the integral
E
ˆ
j
k Q e
tan
a
•Substitute limits 1
y
a
ˆ
j
k Q e a
tan 1 tan 1 ˆ
j
k Q e a
2
r
y dy
(
y
)
y=a P
4 ˆ
j
k Q e
4
a x
Sample Problem
•Divide the line charge into little segments •Find the charge
dQ
=
dx
for each piece •Find the separation
r
for each little piece
P a
r
x
ˆ
a
ˆ
j
r
x
2
a
2 •Add them up – integrate
E
b c
k e
dx r
2
k e
k e
k e c b
r
dx r
3
k e c b
x
2
x
ˆ
a
2
j
3/ 2 ˆ
i
c b
ˆ
i
x
2
x
2
xdx
a
2
a
2 ˆ
j
b c
3/2
k e a
ˆ
j
c b
x
2
dx a
2 3/2 ˆ
i
c
2
a
2
j
i
b
2 •Look up integrals
a
2 ˆ
j
b x
r
dx c
What is the electric field at the point
P
for a line with constant linear charge density and the geometry sketched above?
Electric Field Lines
•Electric field lines are a good way to visualize how electric fields work •They are continuous oriented lines showing the direction of the electric field •They start on positive charges and end on negative charges (or infinity)
+
•They never cross •Where they are close together, the field is strong •The bigger the charge, the more field lines come out
Sample Problem
Sketch the field lines coming from the charges below, if
q
is positive •Let’s have four lines for each unit of •Remaining lines
q
•Eight lines coming from red, eight going into green, four coming from blue •Most of the “source” lines from red and blue will “sink” into green must go to infinity
+2q -2q
+q
Acceleration in a Constant Electric Field
•If a charged particle is in a constant electric field, it is easy to figure out what happens
F
q
E
F
m
a
a
q
E
m
•We can then use all standard formulas for constant acceleration A proton accelerates from rest in a constant electric field of 100 N/C. How far must it accelerate to reach escape velocity from the Earth (11.186 km/s)?
•Look up the mass and charge of a proton •Find the acceleration
a
qE m
19 27 kg 9
m
q
2 27 kg 19 C •Use PHY 113 formulas to get the distance •Solve for the distance
d
v
2
f
2
a
v f
2 4
v i
2 9 2
ad
2 2 0.006545 m 0.65 cm