Download Power Point

Chapter 26
Electricity and Magnetism
Electric Field: Continuous Charge
Distribution
1
Find electric field at point P.
Continuous Charge Distribution
qi ˆ
Ei  ke 2 ri
ri
E   Ei
i
P








qi
dq
ˆ
ˆ
r

k

i
e 2 r
2
qi 0
ri
r
i
E  ke lim
2
Electric Field: Continuous Charge Distribution
q ˆ
E  ke 2 r
r
Electric field
E2
E1
E  E1  E2


E3
E2
E   Ei
E1
i











E2
In this situation, the system of
charges can be modeled as
continuous
The system of closely spaced charges is
equivalent to a total charge that is continuously
distributed along some line, over some surface,
or throughout some volume



E1




3
Electric Field: Continuous Charge Distribution
The total electric charge is Q. What is the electric field at point P ?
P
P
linear
volume
Q
P
Q
surface
Q
4
Continuous Charge Distribution: Charge Density
The total electric charge is Q.
Volume V
Linear, length L
Q
Amount of charge in a
small volume dl:
Q
Q


dq  dl   dl
L
L
Linear charge density
Surface,
area A
Q
Q
Amount of charge in a
small volume dV:
Q

Q
dq  dV   dV
V
V
Volume charge density
Amount of charge in a
small volume dA:
Q
dq  dA   dA
A
Q
A
Surface charge density5

Electric Field: Continuous Charge Distribution
Procedure:
– Divide the charge distribution into
small elements, each of which
contains Δq
– Calculate the electric field due to one
of these elements at point P
– Evaluate the total field by summing
the contributions of all the charge
elements
Symmetry: take advantage of any
symmetry to simplify calculations
Because the charge distribution
is continuous
qi
dq
ˆ
ˆ
r

k

i
e 2 r
2
qi 0
ri
r
i
E  ke lim
For the individual
charge elements
q
E  ke 2 rˆ
r
6
Electric Field: Symmetry
E
E
P
P


no symmetry
no symmetry
P
x
no symmetry
7
Electric Field: Symmetry
q ˆ
E  ke 2 r
r
E
q1  10 μC q2  10 μC
E1,z

E  2E1 cos φ
E1
E2
5m
E2 Electric field
E  E1  E2
E  2E1,z  2E2,z

5m


1
6m
E
E1,z

E1,z  E1 cos φ
E1
E  4E1,z  4E2 cos φ





2
The symmetry gives us the direction
of resultant electric field
8
Electric Field: Continuous Charge Distribution
What is the electric field at point P ?
P
h
90o
R

O
- linear charge density
9
What is the electric field at point P ?

- linear charge density
1. Symmetry determines the direction of the electric field.
E
P
h
90o
R
O
10
 - linear charge density
2. Divide the charge distribution into small elements, each
of which contains Δq
What is the electric field at point P ?
E

E 
all elements
z

E

all elements
E z
all elements
E
Ez  E cos   ke
E

E z
h

all elements
P

R
E
r
Ez
ke
q
cos 
2
r
q
cos 
2
r
 q   l
l
90o
O
11
What is the electric field at point P ?

- linear charge density
3. Evaluate the total field by summing the contributions of
all the charge elements Δq
E
z

ke
all elements
q
cos 
2
r
 l  R 
E
 q   l   R 
E

E z
E
P
r
h
R
90
O
ke
all elements

o

 R 
cos 
Replace Sum by Integral
l

r
2

2
E

0
d ke
R
r
2
cos 
x
12
What is the electric field at point P ?

- linear charge density
4. Evaluate the integral
2
E

d ke
0
R
r
2
cos   ke
R
r
2
z
2
cos   d  2 ke
0
cos  
E
E z
P

R
O
cos 
h
h

r
h2  R2
E  2 ke
r
h
90o
r
2
r  h2  R2
E

R
h
2
hR
R
2

3/ 2
l


x
13
What is the electric field at point P ?
P
E
E

- linear charge density
x
h
x
a
0
b
x
From symmetry
E

E
all elements
q
E  ke 2
x
 q   x
E

ke
all elements
q
x

k

x 2 all elements e x 2
Replace the Sum by Integral
b

b
dx
E  ke  dx 2   ke  2
x
x
a
a
14
What is the electric field at point P ?

- surface charge density
1. Symmetry determines the direction of the electric field.
E
P
h
90o
O
15
What is the electric field at point P ?

- surface charge density
2. Divide the charge distribution into small elements, each of which contains Δq
3. Evaluate the total field by summing the contributions of all the charge elements Δq

E
Approach A: straightforward
E z
all elements
Ez  E cos   ke
E
E z
P
at point (x,y)
r
y
x

 x, y

E
y
q
E   ke 2 cos 
r
all elements
r  h2   x2, y
q   x y
h
cos   
r
h
90o
q
cos 
2
r
h
h2   x2, y
x
O
E

all elements
ke x y

h
h2   x2, y
16

3/ 2
What is the electric field at point P ?

Approach A
- surface charge density
Replace the Sum by Integral
E

ke x y
all elements
E
E z
at point
( x , y )  (  , )
P
r
y
x

 x, y

h
E  ke
O


E

dx  dy


h2   x2, y
h
h2   x2, y


3/ 2
3/ 2
( x , y )  (  , )
y
dxdy   d  d


90o


h
x

2
0
0
E  ke  d   d
h
 h2   2 
3/ 2
17
What is the electric field at point P ?

Approach A
- surface charge density
2

Evaluate the Integral
E  ke  d   d
0
0

 2 ke  d 
E
E z
at point
( x , y )  (  , )

P
r
y
x
0
 x, y

h
2

2

h
2

3/ 2
2

3/ 2




  ke 
E
0
y
h
hdt
2
 t
3/ 2
  2 ke
h
h
2
 t

1/ 2
0
 2 ke


90o
h
h
h
x
E  2 ke
O
18
What is the electric field at point P ?

Approach B
- surface charge density

E
E 
all rings
E  2 ke
E
Linear density
of the ring
E 
P
Ring of width

h

90o
O
h
2
h

2

3/ 2
Charge of
the ring

Q A  2


  
2 2
2
Length of
the ring
E

E 
all rings

2 ke
all rings

E  2 ke  d 
0
h
h
2

2

h
3/ 2
h
2
 2 
3/ 2
 2 ke
19
E
E  2 ke
P
h
2
hR
R
2
E  2 |  | ke

3/ 2
P
Q  2 R
h
E
Q0
O
R
2

3/ 2
Q  2 R
h
Q0
90o
R
h
2
hR
90o
R
O
20
E plane
σ
 2πke σ 
2ε0
E plane
σ 0
E plane
E plane
|σ |
 2πke | σ |
2ε0
Q0
Q  σS
E plane
σ 0
E plane
Q0
Q  σS
21
Electric Field: Symmetry
E
E
P
P


no symmetry
no symmetry
P
x
no symmetry
22
E
E  2 |  | ke
P
h
2
hR
R
2

3/ 2
h
90o
R
O
P
E
0
h
x
a
b
23
no symmetry, but we know E1 and E2
E
E1
E1  21ke
R2
P
h1 R1
h
2
1
90o
h2
E2  22 ke
O
E2
h

h2 R2
2
2
h1
 R1
2
 R2
2
2
R1
90o
O
E  E1  E2
1
E2
E  E12  E22
E1
E
24
3/ 2

3/ 2
E
E1
P
90o
h2
O
E2
h1
2
R1
90o
O
1
25
E
E
x
y
no symmetry

P
 We do not know the direction of
electric field at point P
r

 We need to find x and y component
of electric field
t
a
x
x
0
b
x

E
E
Then
all elements
r t x

E y
all elements
x
x

r
t 2  x2

Ex 
2
t
t
cos   
r
t 2  x2
sin  
E x
all elements
Ey 
 q   x
2

Ex 
q
sin 
r2
q
E x  ke 2 cos 
r
E x  ke
ke
q
x
sin


k
 e x2  t 2
r2
all elements
ke
q
x
cos


k
 e x2  t 2
r2
all elements
all elements
Then
Ey 

all elements
x
x2  t 2
t
x2  t 2
26
E
E
x
y
no symmetry

P
 We do not know the direction of
electric field at point P
r

 We need to find x and y component
of electric field
t
a
x
x
0
b
x
Ex 

ke
all elements
Ey 

all elements
Ey  0
E
ke
 x
x
x2  t 2
x2  t 2
 x
t
b
a
b
x2  t 2
x 2  h2
P
h
0
  ke
  ke
a
 dx
x
x2  t 2
x2  t 2
 dx
t
x2  t 2
x2  t 2
b
  ke 
a
xdx
( x 2  t 2 )3 / 2
b
  ke t 
a
dx
( x 2  t 2 )3 / 2
x
a
b
the symmetry tells us that one of the component is 0, so we do not need
27
to calculate it.
Important Example
Find electric field
    
     
    
     
E
E
σ 0
σ
E 
2ε0
σ 0
σ  0
E
σ  0
E
σ
E 
2ε0
28
Important Example
σ
E 
2ε0
σ 0
    
     
Find electric field
σ
E 
2ε0
    
     
σ  0
E  E  E 
E  E  E   0
E
E
E
E
σ
E  E  E 
ε0
σ  0
E
E
E  E  E   0
σ 0
29
Important Example
    
     
    
     
E 0
σ
E
ε0
E 0
σ 0
σ  0
σ 0
E
σ  0
30
Motion of Charged Particle
31
Motion of Charged Particle
• When a charged particle is placed in an electric field, it experiences
an electrical force
• If this is the only force on the particle, it must be the net force
• The net force will cause the particle to accelerate according to
Newton’s second law
F  qE
- Coulomb’s law
F  ma
- Newton’s second law
q
a E
m
32
Motion of Charged Particle
What is the final velocity?
F  qE
F  ma
- Coulomb’s law
- Newton’s second law
vf
|q|
ay  
E
m
Motion in x – with constant velocity
v0
Motion in x – with constant acceleration
t
l
v0
|q|
ay  
E
m
- travel time
v x  v0
After time t the velocity in y direction becomes
|q|
vy  ayt  
Et
m
then
q

v f  v02   Et 
m 
2
vy
vf
33