Transcript El. Fields

 E
n̂
s
E
n̂
50 kV
Clean
air
b
z
r
a
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). 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
Particle
•All elementary particles have charges that
Proton
are simple multiples of e
Neutron
Electron
e  1.602 1019 C
Oxygen nuc.
Red dashed line means you should be able to
++
use this on a test, but you needn’t memorize it
q
e
0
-e
8e
2e
Charge Can Be Spread Out
Charge may be at a point, on a line, on a surface, or
throughout a volume
•Linear charge density  units C/m
•Multiply by length
•Surface charge density units C/m2
•Multiply by area
•Charge density units C/m3
•Multiply by volume
Concept Question
V  lwh  4 cm3
2 cm
A box of dimensions 2 cm 2 cm  1 cm
has charge density  = 5.0 C/cm3
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
5.0 C/cm3
2 cm
2
2
2
L  l 2  w2  h2  2  2  1 cm  3 cm
q  V   L   5  4  3  3 C  11  C
The Nature of Matter
•Matter consists of positive and negative charges in very large quantities
•There are nuclei with positive charges
•Surrounded by a “sea” of negatively
+
+
+
+
charged electrons
+
+
+
+
•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
•Some materials let charges move long distances, others do not
•Normally it is electrons that do the moving
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
r
•It depends on distance
q1
ke q1q2
9
2
2
ˆ
k

8.988

10
N

m
/
C
F2 
r
e
2
r
ke q1q2
F2 
r2
q2
Other ways of writing this formula
•The r-hat just tells you the direction of the force
•When working with components, often helps to rewrite the r-hat
•Sometimes this formula is written in terms of a
ke q1q2
F2 
r
3
quantity0 called the permittivity of free space
r
F2 
q1q2
rˆ
2
4 0 r
0 
1
 8.854 1012 C2 /N/m 2
4 ke
Concept Question
5.0 cm
What is the direction of the force
+2.0 C 5.0 cm 5.0 cm
on the purple charge?
–2.0 C
A) Up B) Down C) Left
D) Right E) None of the above
–2.0 C
•The separation between the
purple charge and each of the L  5 cm 2  5 cm 2  7.1 cm

 

other charges is identical
•The magnitude of those forces is
2
9
2
2
6
8.988 10 N  m / C  2 10 C 
identical
ke q1q2

F

 7.2 N
2
2
r
 0.071 m 
•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
Ftot   7.2 N  2  10.2 N
angle  180
Sample Problem
Three charges are distributed as shown at
right. Where can we place a fourth
charge of magnitude 3.0 mC such that
the total force on the 1.0 mC vanishes?

1.0 mC
2.0 mC 1.0 m
2.0 m
ke q1q2
ke  2 mC 1 mC  ˆ ˆ ke mC2
F1 
rˆ 
i  2i
2
2
2
r
m
1 m 
3.0 mC
?
2
ke  4 mC 1 mC  ˆ
k
mC
-4.0 mC
ke q1q2
e
ˆ
ˆ


j
j
F2 
r
2
2
2
m
r
 2 m
2
2
k
mC
ke mC
e
ˆ
ˆ
F

2.236
F3  2i  j
,
0  F1  F2  F3 ,
3
2
2
m
m

ke q1q2 3ke mC2

F3 
2
r2
r

  tan 1  12   27
3 2.236

2
r  1.16 m
r
m2
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
r
•Generally not the way to go

dl
•The charge of the line is easy to find, Q = L
•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
ke   dl 
F  q
rˆ
•Find the separation r for each little piece
2
r
•Add them up – integrate
•For a 2D object, it becomes a double integral
ke  dA
F  q
rˆ
•For a 3D object, it becomes a triple integral
2
r
ke   dV 
F  q
rˆ
2
r
The Electric Field
•Suppose we have some distribution of charges
•We are about to put a small charge q0 at a point r
•What will be the force on the charge at r?
•Every term in the force is proportional to q0
•The answer will be proportional to q0
•Call the proportionality constant E, the electric field
F0  q0E
F0
The units for electric
E
q0
field are N/C
F  qE
q0
r
•It is assumed that the test charge q0 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 q0
present
Electric Field From a Point Charge
F0
E
q0
q
r
q0
•From a single point charge, the electric field is easy to find
•It points away from positive charges
•It points towards negative charges
+
ke qq0
F0  2 rˆ
r
ke q
E  2 rˆ
r
-
Electric Field from Two Charges
•Electric field is a vector
• We must add the vector components of
the contributions of multiple charges
ke qi
E   2 rˆi
ri
i
+
+
+
-
Electric Fields From Continuous Charges
P
r
•If you have a spread out charge, we can add up the

dl
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 =  dl for each little piece
•Find the separation r and the direction r-hat for each
ke   dl 
little piece
E
rˆ
2
r
•Add them up – integrate
•For a 2D object, it becomes a double integral
ke  dA
•For a 3D object, it becomes a triple integral
E
rˆ
2
r
ke   dV 
E
rˆ
2
r
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
ry
– It’s magnitude is r = y
– It’s direction is the minus-y direction rˆ   ˆj
•Substitute into the integral
– Limits of integral are y= a and y = 


ke dy Qy 2
dy
ˆj   k Qˆj
E 2 2

e
2
2
2

y
y

a
y

a
a
a
 
ke   dl 
E
rˆ
2
r
What is the electric
field at the origin for
a line of charge on r
the y-axis with linear
charge density
(y) = Qy2/(y2+a2)
stretching from
y = a to y = ?
y
dy
(y)
y=a
P
x
•Pull constants out of the integral
•Look up the integral

kQ
k Q   
kQ
 y
E  ˆj e tan 1    ˆj e  tan 1     tan 1 1   ˆj e     jˆ  keQ
a 2 4
a
a
aa
4a
•Substitute limits
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
r   xˆi  aˆj
P
x
r  x2  a2
•Add them up – integrate
b


c
c  xˆ
i  aˆj dx
ke  dx
r
dx
E   2 rˆ   ke  3   ke

2
2 3/2
r
r
b
b
b x  a 
 c
 •Look up
c
xdx
dx


ˆj
  ke ˆi 

a
 integrals

2
2 3/2
2
2 3/2
b x  a 


 b x  a 

c
ˆ
ˆ
 i   x a j 
 ˆi   c a  ˆj ˆi   b a  ˆj 
  ke 

   ke 

2
2 
 x2  a2 
 c2  a2
b a 

b

c
r
a

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 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
•Remaining lines
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
qE
F  qE
F  ma
a
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
m  1.676 1027 kg
•Find the acceleration
19
q


e

1.602

10
C
19
qE 1.602 10 C  100 N/C 
9
2

9.558

10
m/s
a

m
1.676 1027 kg
2
2
v

v
•Use PHY 113 formulas to get the distance f
i  2ad
2
4
2
•Solve for the distance
1.1186 10 m/s 

vf

 0.006545 m  0.65 cm
d
9
2
2a 2  9.5585 10 m/s 