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•Electrostatic Phenomena
•Coulomb’s Law: force between static charges
r
F12 =
1
q1q2
ˆ
r
12
2
4pe o r12
•Superposition
q1
F2
Ftotal
Remember: vectors
add by components!
Q
F1
Ftotal = F1 + F2 + ...
q2
Today...
•Define Electric Field in terms of force
on a test charge
•How to think about fields
•Electric Field Lines
•Example Calculation: Electric Dipoles
1
Lecture 2, Act 1
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)
Lecture 2, Act 1
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
F2
F
F = F1 + F2
Q
F1
q2
r
F=
Q  q1rˆ1 q2 rˆ2 
+ 2 

2
4pe 0  r1
r2 
test charge
- 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:
Collection of Point Charges
r
E=
1
4pe 0
+
+
+
+
-
-
-

qi
F=QE
rˆ
2 i
ri
+
-
Charge Distribution
Test
charge Q
+
r
E=
1
dq
rˆ

2
4pe 0 r
+ +++ + +
+ + +++
+
“Net” E at origin
These charges or this charge distribution are the
“source” of the electric field throughout space
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
Qq
Fx = QE x = k
2 2a 2
Qq
Fy = QE y = -k
2 2a 2
If the test charge Q is positive, the force will be
in the direction of the electric field
If the test 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 = 5 mC, a = 5 cm, and Q = 15 mC.
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
Lecture 2, Act 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
y
E
1
(b) Both charges q1 and q2 are negative
(c) The charges q1 and q2 have opposite signs
q2
x
Lecture 2, Act 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
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)
Reality of Electric Fields
•The electric field has been introduced as a mathematical
convenience, just as the gravitational field of Physics 111
•There is MUCH MORE to electric fields than this!
IMPORTANT FEATURE: E field propagates at speed of light
• NO instantaneous action at a distance (we will explain this when
we discuss electromagnetic waves)
• i.e., as charge moves, resultant E-field at time t depends upon
where charge was at time t - dt
• For now, we avoid these complications by restricting ourselves to
situations in which the source of the E-field is at rest.
(electrostatics)
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 arrows indicates the
direction of the field at each point in space
• Length of arrows is proportional to the
magnitude of the field at each point in space
Rules for Field Lines
+
graphical “trick”
for visualizing
E fields
-
• Lines leave (+) charges and return to (-) charges
• Number of lines leaving/entering charge  amount of charge
• Field lines never cross
→ Tangent of line = direction of E at each point
→ Local density of field lines ~ magnitude of E at each point
• Field at two white dots differs by a factor of 4
since r differs by a factor of 2 (Coulomb’s law, E ~ 1/ r2)
• Local density of field lines / unit area also differs by
a factor of 4 in 3D: same # lines spread over area ~ r2
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
Lecture 2, Act 3
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
Lecture 2, Act 3
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
Solution: Draw some field
lines according to our rules.
Ex
Falstad's Electrostatics Applets: 2D
Falstad's Electrostatics Applets: 3D
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
Charge distribution
within 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
Lecture 2, ACT 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
+
++
Lecture 2, ACT 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
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
(give it a try) … but we’re dealing with vectors here, not scalars!
• Note that the E 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 the force
on a “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 point on the y-axis: (0,y)
Ex = ??
Coulomb Force
Radial
Ey = ??
y
Electric Dipole
+q
a
a
-q
For points along x-axis:
For r >>a,
r
x
Case of special interest:
(antennas, molecules)
r>>a
For points along y-axis:
For r >>a