Transcript Lect05

Lecture 5
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Lecture 5
• Yesterday we introduced electric field lines
• Today we will cover some extra topics on Electric
Fields before going on to Electric Flux and Gauss’
Law next week:
– Continuous Charge Distributions
– Infinite line of charge
– Motion of a charge in an electric field
Question 1
•
Consider a circular ring with total charge +Q.
The charge is spread uniformly around the
ring, as shown, so there is λ = Q/2pR charge
per unit length.
•
The electric field at the origin is
(a)
zero
(b)
2p
4p 0 R
1
(c)
y
+ +++
+
+
+
+
R
+
+
+
+
+ x
+
+
+
++ + ++
+
Question 1
77%
15%
c
b
7%
a
1. a
2. b
3. c
Question 1
•
Consider a circular ring with total charge +Q.
The charge is spread uniformly around the
ring, as shown, so there is λ = Q/2pR charge
per unit length.
•
The electric field at the origin is
(a)
zero
(b)
2p
4p 0 R
1
y
+ +++
+
+
+
+
R
+
+
+
+
+ x
+
+
+
++ + ++
+
(c)
• The key thing to remember here is that the total field at the origin is
given by the VECTOR SUM of the contributions from all bits of charge.
• If the total field were given by the ALGEBRAIC SUM, then (b) would be
correct. (exercise for the student).
• Note that the electric field at the origin produced by one bit of charge
is exactly cancelled by that produced by the bit of charge diametrically
opposite!!
• Therefore, the VECTOR SUM of all these contributions is ZERO!!
Electric Fields
from
Continuous Charge Distributions
Examples:
• line of charge
• charged plates
• electron cloud in atoms, …
r
E(r) = ?
++++++++++++++++++++++++++
• Principles (Coulomb’s Law + Law of Superposition)
remain the same.
• Only change:



Charge Densities
• How do we represent the charge “Q” on an extended object?
total charge
Q
• Line of charge:
 = charge per
unit length
• Surface of charge:
s = charge per
unit area
• Volume of Charge:
r = charge per
unit volume
small pieces
of charge
dq
dq =  dx
dq = s dA
dq = r dV
How We Calculate (Uniform) Charge
Densities:
Take total charge, divide by “size”
Examples:
10 coulombs distributed over a 2-meter rod.
10C
λ
 5 C/m
2m
14 pC (pico = 10-12) distributed over the surface of a sphere of
radius 1 μm.
12
14 10 C 14
2
σ

C/m
4π(10-6 m)2 4π
14 pC distributed over the volume of a sphere of radius 1 mm.
14 1012 C (3) 14 3
3
ρ 4


10
C/m
-3
3
π(10
m)
4π
3
Electric field from an infinite line charge
E(r) = ?
r
++++++++++++++++++++++++++
Approach:
“Add up the electric field contribution from each bit of
charge, using superposition of the results to get the final field.”
In practice:
• Use Coulomb’s Law to find the E-field per segment of charge
• Plan to integrate along the line…
– x: from  to

q: from
OR
p/2 to p/2
q
+++++++++++++++++++++++++++++
x
Any symmetries ? This may help for easy cancellations
Infinite Line of Charge
dE
Charge density = 




y
q
r
r'
++++++++++++++++ x
dx
We need to add up the E-field components
dE at the point r given by contributions from
all segments dx along the line.
q and x are not independent, choose to work
in terms of q
Write dE in terms of q, r, and 
Integrate q from -p/2 to +p/2
Infinite Line of Charge
We use Coulomb’s Law to find dE:
dE
y
q
What is dq in terms of dx?
What is r’ in terms of r ?
What is dx in terms of q ?
x  r tan q
dx  r sec 2 q dq
r
r'
++++++++++++++++ x
dx
Therefore,
Infinite Line of Charge
• Components:
Ey
dE
y
q
Ex
q
r
r'
++++++++++++++++ x
dx
• Integrate:
Infinite Line of Charge
• Now
p / 2
p

p
dE
y
q
Ex
sin q dq  0
 /2
p / 2
Ey
cos q dq  2
 /2
• The final result:
q
r
r'
++++++++++++++++ x
dx
Infinite Line of Charge
dE
y
q
r
Conclusion:
r'
++++++++++++++++ x
dx
• The Electric Field produced by an
infinite line of charge is:
- everywhere perpendicular to the line (Ex=0)
- is proportional to the charge density ()
1
- decreases as r
- Gauss’ Law makes this trivial!!
Question 2
•Examine the electric field
lines produced by the charges
in this figure.
•Which statement is true?
q1
q2
(a) q1 and q2 have the same sign
(b) q1 and q2 have the opposite signs and q1 > q2
(c) q1 and q2 have the opposite signs and q1 < q2
Question 2
1. a
2. b
3. c
86%
13%
c
b
a
1%
Question 2
•Examine the electric field
lines produced by the charges
in this figure.
•Which statement is true?
q1
q2
(a) q1 and q2 have the same sign
(b) q1 and q2 have the opposite signs and q1 > q2
(c) q1 and q2 have the opposite signs and q1 < q2
Field lines start from q2 and terminate on q1.
This means q2 is positive; q1 is negative; so, … not (a)
Now, which one is bigger?
Note that more field lines emerge from q2 than end on q1
This indicates that q2 is greater than q1
Electric Dipole: Lines of Force
Consider imaginary
spheres centered on :
a) +q
(blue)
c
b) -q
(red)
a
c) midpoint (yellow)
• All lines leave a)
• All lines enter b)
• Equal amounts of
leaving and entering
lines for c)
b
Electric Field Lines
Electric Field Patterns
Distance
dependence
Dipole
Point Charge
Infinite
Line of Charge
~ 1/R3
~ 1/R2
~ 1/R
Motion of a Charge in a Field
•An electron passes between two charged plates (cathode
ray tube in your television set)
•While the electron is between the plates, it experiences an
acceleration in the y-direction due to the electric field E
Motion of a Charge in a Field
The only force is in the y direction and the acceleration is:
F  ma
F  qE
qE ˆ
a
j
m
The initial velocity is in the x-direction, so the
velocity as a function of time (t) is:
 qE  ˆ
ˆ
ˆ
ˆ
v  v x i  v y j  v0 i  
t j
 m 
The time (T) taken by the charge to traverse the plates
is determined only by the initial velocity in the x
direction.
L1
T
v0
Motion of a Charge in a Field
•Therefore, the particle’s deflection in the y-direction is:
2
L
1
1
qE
1
y  a yT 2 
2
2 m v02
•It exits the field making an angle θ with its
original direction, where:
tan  
vy
vx


qE

 L1 
m  v0  qEL1

v0
mv02
•Exercise for the student; calculate where it hits
the screen after a distance L2
The Story Thus Far
Two types of electric charge: opposite charges attract,
like charges repel
Coulomb’s Law:
1 q1q2
F
rˆ
2
4p 0 r 
Electric Fields
•
Charges respond to electric fields:
•
Charges produce electric fields:
F  q1E
1 q2
E
rˆ
2
4p 0 r
The Story Thus Far
We want to be able to calculate the electric fields
from various charge arrangements. Two ways:
1. Brute Force: Add up / integrate contribution from
each charge.
Often this is pretty difficult.
Ex: electron cloud around nucleus
Ack!
2. Gauss’ Law: The net electric flux through any
closed surface is proportional to the charge
enclosed by that surface.
In cases of symmetry, this will be MUCH EASIER than
the brute force method.
• Finished Coulomb’s Law
• Next lecture: Electric field Flux and Gauss’
Law
• Read Chapter 23
• Try Chapter 22, Problems 28,45,70