Transcript Lecture 2

Lecture 2: May 20th 2009
Physics for Scientists and Engineers II
Physics for Scientists and Engineers II , Summer Semester 2009
Electric Field due to a Continuous Charge Distribution
•
•
We can model a system of charges as being continuous (instead of discrete) if the
distance between the charges is much smaller than the distance to the point where the
electric field is calculated.
Procedure: - Divide charge distribution into small charge elements Dq.
- Add contributions to E from all charge elements.
Dq
D E  k e 2 rˆ
r
Dq
E  ke  2 rˆi
i ri
r̂
DE
r
P
Dqi
dq
ˆ
E  ke lim  2 ri  ke  2 rˆ
r
ri
D qi 0 i
Physics for Scientists and Engineers II , Summer Semester 2009
Dq
Charge Density (a useful concept when calculating E from charge distribution)
Volume charge density (if Q is uniformly distribute d throughou t a volume V) :
Surface charge density (if Q is uniformly distribute d on a surface of area A) :
Linear charge density (if Q is uniformly distribute d on a line of length l) :
Physics for Scientists and Engineers II , Summer Semester 2009
Q
V
Q

A
Q

l

 dq   dV
 dq   dA
 dq   dl
Example: Electric Field due to a Uniformly Charged Rod
dq =  dx
y
dx
x
E
x
P
a
Contributi on to E from dq :
l
dE  ke
l a
E from all dq (entire rod) :
E

a
 ke
Notice : For l  0
E
dq
 dx

k
e
x2
x2
dx
ke  2  ke 
x
l a

a
keQ
Q1
1 
 

l  a l  a  al  a) 
keQ
, the field of a point charge.
2
a
Physics for Scientists and Engineers II , Summer Semester 2009
l a
dx
 1

k

 
e
2

x
 x a
Example: Electric Field due to a Uniformly Charged Rod…..this is harder….
y
dE
a
dq =  dx
P

r
x
x
l
Contributi on to E from dq :
 dx
sin  iˆ
2
r
 dx x
 ke 2 iˆ
r r
 x dx
 ke 3 iˆ
r
 x dx ˆ
 ke
i
3
2
2 2
a  x 
d E  ke
Physics for Scientists and Engineers II , Summer Semester 2009
 dx
cos  ˆj
2
r
 dx a ˆ
 ke 2
j
r r
 a dx
 ke 3 ˆj
r
 a dx ˆ
 ke
j
3
2
2 2
a  x 
 ke
Example: Electric Field due to a Uniformly Charged Rod…..this is harder….
y
dq =  dx
P
dE
a
r

x
x
l
d E  ke
Contributi on to E from dq :
a
 x dx
2
 x2

3
2
iˆ  ke
a
 a dx
2
 x2

3
ˆj
2

 x dx ˆ
 a dx
E     ke
i

k
e
3
3

2
2 2
2
2 2
0
a x
a x
l
Total electric field at point P :

l
E x  ke  
0
a
x

l
x
2

2

3
dx
2
Physics for Scientists and Engineers II , Summer Semester 2009
E y  ke  a 
0
a

1
2
x
2

3
dx
2

ˆj 


….solving the integral for Ex
l
E x  ke  
0
a
x
2
x

2

3
dx
2
Substituti on : u  a  x
l
E x   ke  
0
2
2

1
1
x
a2  x2 a2  x2


1
1
  ke   
a2  l 2
a
 du 
2

1
2
a
x dx
2
dx   ke 
 x2
a 2 l 2

a

1
1
 1
du


k

e  
u2
 u a

Q  a 2  l 2  a 
   ke

l  a a 2  l 2 

Physics for Scientists and Engineers II , Summer Semester 2009
2
a 2 l 2
….solving the integral for Ey
l
E y  ke  
0
a
a
2
x
2

3
dx
2
Substituti on : x  a tan 
E y  ke 
 max

0
k
 e
a
 max

0
a
 dx 
a
2
 a tan 
2
2
1
 1 


2
cos



3
2

3
2
a
d
cos 2 
a
d
cos 2 
1
d
cos 2 

k
 e
a
ke 
sin 0 max  keQ sin  max
a
al
kQ
keQ
l
 e

a l l 2  a2
a l 2  a2
(I wouldn' t expect you to know that)
ke 
a
 max

0
1
1  tan 
2
3
2
1
d
cos 2 
 max
 cos  d
0

a
max
l 2  a2
l
sin  max 
Physics for Scientists and Engineers II , Summer Semester 2009
l
l 2  a2
….and the final result
Q  a 2  l 2  a 
E x   ke
l  a a 2  l 2 
In the limit
lim
l 0
lim
l 
keQ
a l 2  a2
of a very short rod :
: Ex  0
In the limit
Ey 
and E y 
keQ
a2
(field of a point charge again)
of a very long rod :
: E x  ke

a
and E y  ke
Physics for Scientists and Engineers II , Summer Semester 2009

a
Visualizing Electric Fields with Electric Field Lines
•
•
•
The electric field vector is always tangent to the electric field line.
The electric field line has a direction (indicated by an arrow). The direction is
the same as that of the electric field (same direction as force on a positive test
charge).
The number of lines per unit area through a normal plane (perpendicular to
field lines) is proportional to the magnitude of the electric field in that region.
Example: Electric field lines of a point charge
N field lines
+
Surface density of field lines at an
N
imagined sphere of radius r is
4 r 2
Electric field strength is proportional to
Physics for Scientists and Engineers II , Summer Semester 2009
1
r2
Visualizing Electric Fields with Electric Field Lines
•
•
•
•
•
For a single positive point charge: Electric field lines go from the positive charge to
infinity.
For a single negative point charge: Electric field lines go come from infinity and end at the
negative point charge.
For multiple point charges: Lines can start at the positive charges and end at the negative
charges.
Electric field lines can never cross (think about why that is so).
For two unequal point charges of opposite sign with charges Q1 and Q2 , the number N1 of
field lines terminating at Q1 and the number N2 of field lines terminating at Q2 are related
by the equation
N 2 Q2

N1 Q1
Physics for Scientists and Engineers II , Summer Semester 2009
Motion of a Charged Particle in a Uniform Electric Field
•
•
Assume particle has charge q, mass m.
Particle experiences a force
•
The force results in an acceleration (according to Newton’s second law):
Fe q E
a 
Fe
qE

m
m
•
•
•
For positive charges: Acceleration is in the same direction as electric field.
For negative charges: Acceleration is in a direction opposite to the electric field.
A uniform electric field will cause a constant acceleration of the particle.
 You can use equations of motion for constant acceleration.
•
Work is done on the particle by the electric force as the particle moves.
W  Fe D x
Physics for Scientists and Engineers II , Summer Semester 2009
Example (similar to Ex. 23.10 in book)
+ + +
-
vi  ?
+ + +
+ + + +

E
- - - - - - - - - L = 0.100 m
Electron: m = 9.11x10-31 kg ; q = 1.60x10-19 C
Electric Field: E = 800 N/C
The electron leaves the electric field at an angle of  = 65 degrees.
Q1: What was the initial velocity of the electron?
Q2: What is the final velocity of the electron (magnitude)?
Q3: How low would the electric field have to be so that the net force on the electron is zero?
Q4: Were we justified in neglecting the gravitational force in Q1 and Q2?
Physics for Scientists and Engineers II , Summer Semester 2009
Question 1 :
vy f


F
qE
 ay t  e t 
t
m
m
vy f 
qE L
m vi

L
t
vi
vi tan  
;
vy f
vi
 tan 
qE L
m vi
N

 1.60 10 19 C    800   0.100m
qE L
C

6 m
vi 


2
.
5

10
m tan 
9.10 10 31 kg  tan 65 deg 
s
Question 2 :
2.5 106 m
vi
s  6.1106 m
vf 

cos 
cos65
s
Physics for Scientists and Engineers II , Summer Semester 2009
Question 3 :
qE
m
 g

E 
gm

q
9.8 m
2
 9.10 10 31 kg
s
1.6 10 19 C
 5.6 10 11
N
C
Question 4 :
Yes, except for extremely small electric fields, the electric force on an electron
is much larger tha n the gravitatio nal force.
Physics for Scientists and Engineers II , Summer Semester 2009
Gauss’s Law – An alternative procedure to calculate electric fields of highly
symmetric charge distributions
The concept of “Electric Flux”:
Area = A
E
Electric flux :  E  A E
for constant E and E being perpendicu lar to area.
Physics for Scientists and Engineers II , Summer Semester 2009
Area perpendicu lar to E is A

A  A cos 
E
Area not perpendicu lar to E is A



The electric flux through the two surfaces is the same
 E  E A  E A cos 
Physics for Scientists and Engineers II , Summer Semester 2009
Normal to green surface



The electric flux through the two surfaces is the same
Normal to green surface
 E  E A  E A cos 
 To calculate the flux through a randomly oriented area you need to know the angle
between the electric field and the normal to the area.
Physics for Scientists and Engineers II , Summer Semester 2009
How to treat situations where the electric field is not constant over the area?
• Divide area into small areas over which E is constant.
• Calculate flux for each small area.
• Add fluxes up.
Ei
i
Ai
Area vector:
magnitude = area
direction = perpendicular to area
Electric flux throu gh surface element : D E  Ei DAi cos i  E i  D Ai
Electric flux throu gh entire surface :
 E   E i  D Ai
i
....and in the limit of infinitesi mally small surface segments :  E 
Ed A
surface
Physics for Scientists and Engineers II , Summer Semester 2009
“surface integral”
Flux through a closed surface:
•Convention: Area vectors always point outwards.
Field lines that cross from the inside to the outside of the surface :   90
(positive flux because cos  is positive)
Field lines that cross from the outside to the inside of the surface: 90    180
(negative flux because cos  is negative)
Flux throu gh closed surface :  E   E  d A   En dA
Physics for Scientists and Engineers II , Summer Semester 2009
Example: Cube in a uniform field
dA3
E
dA1
dA6
dA2
dA5
dA4
E   E  d A   E  d A   E  d A   E  d A   E  d A   E  d A   E  d A
1
2
3
  E d A  E d A 
1
4
0
5

0
6

0

0
2
  E cos180dA   E cos 0dA   E  dA  E  dA   EL2  EL2  0
1
2
Physics for Scientists and Engineers II , Summer Semester 2009
1
2