Transcript No Slide Title

•Electrostatic Phenomena
•Coulomb’s Law
r
F12 =
1
q1q2
ˆ
r
12
2
4pe o r12
q1
•Superposition
F1
F
Ftotal = F1 + F2 + ...
q
F2
q2
Two balls of equal mass are suspended from the ceiling with
nonconducting wire. One ball is given a charge +3q and the
other is given a charge +q.
+3q
+q
g
Which of the following best represents the equilibrium positions?
+q
+3q
+q
(a)
+3q
+3q
(b)
+q
(c)
Which best represents the equilibrium position?
+q
+3q
+q
(a)
+q
+3q
+3q
(b)
(c)
•Remember Newton’s Third Law!
•The force on the +3q charge due to the +q charge must be equal
and opposite to the force of the +3q charge on the +q charge
•Amount of charge on each ball determines the magnitude of the
force, but each ball experiences the same magnitude of force
•Symmetry, therefore, demands (c)
P.S. Knowing the form of Coulomb’s law you can write two equations
with two unknowns (T and q )
Preflight 2:
Two charges q = + 1 μC and Q = +10 μC are placed near each
other as shown in the figure.
6) Which of the following diagrams best describes the forces
acting on the charges:
+1 μC
a)
b)
c)
+10 μC
The Electric Field
•A simple, yet profound observation
- The net Coulomb force on a given charge is always
proportional to the strength of that charge.
q1
F = F1 + F2
F1
F
q
test charge
F2
q2
r
F=
q  q1rˆ1 q2 rˆ2 
+ 2 

2
4pe 0  r1
r2 
- We can now define a quantity, the electric field, which
is independent of the test charge, q, and depends only on
position in space:
r
r
F The qi are the sources

E
of the electric field
q
The Electric Field
r
r
F
E 
q
With this concept, we can “map” the electric field
anywhere in space produced by any arbitrary:
Bunch of Charges
r
E=
+
+
r
E=
ri
+
-
 2 rˆi
4pe 0
+
-
qi
1
-
F
+
+
-
Charge Distribution
1

dq
4pe 0 r 2
+ +++ + +
+ + +++
+
“Net” E at origin
These charges or this charge distribution
“source” the electric field throughout space
rˆ
Example: Electric Field
What is the electric field at the origin due to this set of charges?
y
1) Notice that the fields from the top-right
and bottom left cancel at the origin?
a
+q
a
2) The electric field, then, is just the field
from the top -left charge. It points away
a
from the top-left charge as shown.
+q
3) Magnitude of E-field at the origin is:
E = kq2
2a
The x and y components of the field at (0,0) are:
kq cosq
kq sinq
=
Ex
=
E
y
2a2
2a2
kq 1
kq 1
= 2
= -2a2
2a 2
2
a
a 2
+q
a
Q
x
Example: Electric Field
Now, a charge, Q, is placed at the origin. What is the net force
y
on that charge?
a
a
+q
+q
q 1
q
1
Ex = k 2
E y = -k 2
a 2
a
a
2a 2
2a 2
x
Q
a
+q
Fx = QE x = k
Qq
2 2a 2
Fy = QE y = -k
Qq
2 2a 2
Note: If the charge Q is positive, the force will be in
the direction of the electric field
If the charge Q is negative, the force will be against
the direction of the electric field
F is
F is
Let’s Try Some Numbers...
If q = 5mC, a = 5cm, and Q = 15mC.
y
Then Ex = 6.364  106 N/C
and Ey = -6.364  106 N/C
+q
a
a
+q
a 2
a
x
Fx = QEx and Fy = QEy
So... Fx= 95.5 N and
a
a
Fy= -95.5 N
+q
We also know that the magnitude of
E = 9.00  106 N/C
We can, therefore, calculate the magnitude of F
F = |Q| E = 135 N
2
Two charges, Q1 and Q2, fixed along the x-axis as
shown produce an electric field, E, at a point
(x,y)=(0,d) which is directed along the negative
y-axis.
d
- Which of the following is true?
(a) Both charges Q1 and Q2 are positive
Q
1
(b) Both charges Q1 and Q2 are negative
(c) The charges Q1 and Q2 have opposite signs
y
E
Q2
x
Two charges, Q1 and Q2, fixed along the x-axis as
shown produce an electric field, E, at a point
(x,y)=(0,d) which is directed along the negative
y-axis.
d
- Which of the following is true?
(a) Both charges Q1 and Q2 are positive
Q
y
E
Q2
1
x
(b) Both charges Q1 and Q2 are negative
(c) The charges Q1 and Q2 have opposite signs
E
E
E
Q1
Q2
(a)
Q1
Q2
(b)
Q1
Q2
(c)
Ways to Visualize the E Field
Consider the E-field of a positive point charge at the origin
vector map
field lines
+ chg
+ chg
+
+
Rules for Vector Maps
+ chg
+
•Direction of arrow indicates direction of field
•Length of arrows  local magnitude of E
Rules for Field Lines
+
-
•Lines leave (+) charges and return to (-) charges
•Number of lines leaving/entering charge  amount
of charge
•Tangent of line = direction of E
•Local density of field lines  local magnitude of E
• Field at two white dots differs by a factor of 4 since r
differs by a factor of 2
•Local density of field lines also differs by a factor of 4
(in 3D)
3
Preflight 2:
6) A negative charge is placed in a region of electric field
as shown in the picture. Which way does it move ?
a) up
b) down
c) left
d) right
e) it doesn't move
7) Compare the field strengths at points A and B.
a) EA > EB
b) EA = EB
c) EA < EB
y
•Consider a dipole (2 separated equal and
opposite charges) with the y-axis as
shown.
-Which of the following statements
about Ex(2a,a) is true?
(a) Ex(2a,a) < 0
(b) Ex(2a,a) = 0
+Q
a
a
a
2a
x
-Q
(c) Ex(2a,a) > 0
y
•Consider a dipole (2 separated equal and
opposite charges) with the y-axis as
shown.
-Which of the following statements
about Ex(2a,a) is true?
(a) Ex(2a,a) < 0
(b) Ex(2a,a) = 0
Ex
+Q
2a
a
a
a
x
-Q
(c) Ex(2a,a) > 0
Solution: Draw
some field lines
according to
our rules.
Preflight 2:
Two equal, but opposite charges are placed on the x axis. The positive
charge is placed at x = -5 m and the negative charge is placed at x =
+5m as shown in the figure above.
3) What is the direction of the electric field at point A?
a) up
b) down
c) left
d) right
e) zero
4) What is the direction of the electric field at point B?
a) up
b) down
c) left d) right
e) zero
Field Lines From Two Opposite Charges
Dipole
Dipoles are central to our
existence!
Molecular Force Model
Basis of Attraction
Ion-dipole
Ion and polar molecule
Dipole-dipole
Partial charges of polar
molecules
Induced dipoles of
polarizable molecules
London
dispersion
y
The Electric Dipole
+Q
see the appendix for further information
a
q
a
-Q
r
E
x
E
What is the E-field generated by
this arrangement of charges?
Calculate for a point along x-axis: (x, 0)
Ex = ??
Symmetry
Ex(x,0) = 0
Ey = ??
Electric Dipole Field Lines
• Lines leave positive charge
and return to negative charge
What can we observe about E?
• Ex(x,0) = 0
• Ex(0,y) = 0
• Field largest in space between two charges
• We derived:
... for r >> a,
Field Lines From Two Like Charges
• There is a zero halfway
between the two charges
• r >> a: looks like the field
of point charge (+2q) at origin
4
•
•
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
4pe 0 R
1
(c)
y
+ +++
++
+
+
+
+
+
++
R
++
+
+
+
+ x
+
++
•
•
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
4pe 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 Field inside a Conductor
• A two electron atom, e.g., Ca
– heavy ion core
– two valence electrons
2+
• An array of these atoms
– microscopically crystalline
– ions are immobile
– electrons can move easily
• Viewed macroscopically:
– neutral
There is never a net electric field inside
a conductor – the free charges always
move to exactly cancel it out.
2+
2+ 2+
2+
2+
2+ 2+
2+
2+
2+ 2+
2+
2+
2+ 2+
2+
Summary
• Define E-Field in terms of force
on “test charge”
• How to think about fields
• Electric Field Lines
• Example Calculation: Electric Dipole
Appendix A:
Other ways to Visualize the E Field
Consider a point charge at the origin
Field Lines
+ chg
+
Graphs
Ex, Ey, Ez as a function of (x, y, z)
Er, Eq, Ef as a function of (r, q, f)
Ex(x,0,0)
x
Appendix A- “ACT”
y
Consider a point charge fixed at the origin of
a coordinate system as shown.
–Which of the following graphs best
represent the functional dependence of
the Electric Field for fixed radius r?
3A Er
r
f
x
Q
Er
Er
Fixed
r>0
0
f
2p
0
(a)
3B
f
2p
f
2p
(c)
Ex
Ex
0
0
2p
(b)
Ex
Fixed
r>0
f
0
f
2p
0
f
2p
Appendix A “ACT”
y
Consider a point charge fixed at the origin of
a coordinate system as shown.
– Which of the following graphs best
represent the functional dependence of
the Electric Field for fixed radius r?
3A Er
r
f
x
Q
Er
Er
Fixed
r>0
0
f
(a)
2p
0
f
(b)
2p
0
f
2p
(c)
• At fixed r, the radial component of the field is a constant,
independent of f!!
• For r>0, this constant is > 0. (note: the azimuthal component
Ef is, however, zero)
Appendix A “ACT”
y
Consider a point charge fixed at the origin of
a coordinate system as shown.
–Which of the following graphs best
represent the functional dependence of
the Electric Field for fixed radius r?
3B
Ex
Fixed
r>0
f
(a)
2p
f
x
Q
Ex
Ex
0
r
0
f
(b)
2p
0
f
2p
(c)
• At fixed r, the horizontal component of the field Ex is given by:
y
Appendix B: Electric Dipole
+Q
a
q
a
-Q
r
E
x
E
What is the E-field generated by
this arrangement of charges?
Calculate for a point along x-axis: (x, 0)
Ex = ??
Symmetry
Ex(x,0) = 0
Ey = ??
E
Electric Dipole
y
+Q
a
x
a
-Q
What is the Electric Field
generated by this charge
arrangement?
Now calculate for a pt along y-axis: (0,y)
Ex = ??
Coulomb Force
Radial
Ey = ??
y
Electric Dipole
+Q
a
a
-Q
For pts along x-axis:
For r >>a,
r
x
Case of special interest:
(antennas, molecules)
r>>a
For pts along y-axis:
For r >>a