Physics 122B
Electricity and Magnetism
Lecture 2
Coulomb’s Law and the
Electric Field
March 28, 2007
Martin Savage
Recall : Coulomb’s Law
Like charges repel.
Charles Augustine
de Coulomb
(1736-1806).
Opposite charges attract.
Coulomb’s Law:
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F1 on 2  F2 on 1  K
Physics 122B - Lecture 2
q1 q2
r
2
( Magnitude of force )
2
Recall : Units of Charge
FK
q1 q2
r
2

1
q1 q2
4 0
r2
K  8.99 109 N m2 /C2  9.0 109 N m2 /C2
C  coulomb  SI unit of charge;
1.0 nC  1.0 10-9 C
(Note that in Newton’s law of gravitation, G,
which plays a role similar to K, has the value
G = 6.67  10-11 N m2/kg2.)
1
0 
 8.85 10-12 C2 /(N m 2 )  permittivity constant
4 K
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Recall : Using Coulomb’s Law
1. Coulomb’s Law applies only to point charges.
(This is particularly important because charge are
free to move around on conductors.)
1. Strictly speaking, Coulomb’s Law applies only to
electrostatics (non-moving charges).
(However, it is usually OK provided v<<c).
1. Electrostatic forces (vectors!) can be superposed.
Linear superposition !!!!
Fnet  F1 on j  F2 on j  F3 on j 
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Example: Sum of Two Forces
Two +10 nC charged particles are
2 cm apart on the x axis.
(1) What is the net force on a +1.0
nC particle midway between
them?
(2) What is the net force if the
charged particle on the right is
replaced by a -10 nC charge?
F(  ) net  F1 on 3  F2 on 3  F iˆ  F iˆ  0
FK
q1q2
r2
 (9.0  109 N m 2 /C2 )
(1.0 108 C )(1.0 108 C )
(1.0 102 m) 2
 9.0  10-4 N
F(  ) net  F1 on 3  F2 on 3  F iˆ  F iˆ
 2 F iˆ  1.8  10-3 N
8
98
-9
q1q2
9
2
2 (1.0  10 C )(1.0  10 C )
F  K 2  (9.0  10 N m /C )
r
(1.0 102 m) 2
 9.0  10-4 N
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Example: Point of Zero Force
Two positively charged
particles q1 and q2=3q1 are
placed 10. 0 cm apart. Where
(other than infinity) could
another charge q3 be placed so
as to experience no net force?
Need force vectors to be co-linear, so location must be on x axis.
Need force vectors to be in opposite directions, so location must
be between 0 and d.
Need force vectors equal in magnitude, so F=Kq1q3/x2=Kq2q3/(d-x)2.
Therefore, (d-x)2=3x2 or x=d/(1±√3) = 10 cm/(1±1.732);
x+=10.0 cm/2.732 = 3.66 cm;
x-=10.0 cm/-0.732=13.66 cm, which does not satisfy the 2nd criterion.
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Example: Three Charges (1)
Three charges with q1 = -50 nC,
q2 = +50 nC, and q3 = +30 nC, are
placed at the corners of a 10 cm x
5 cm rectangle as shown.
What is the net force on q3 due
to the other two charges?
F1 on 3  Kq1q3 / r132
 (9.0  109 N m 2 /C2 )(5.0 10-8 C)(3.0 10-8 C) /(0.10 m) 2
 1.35  10-3 N
F2 on 3  Kq2 q3 / r232
 (9.0 109 N m 2 /C 2 )(5.0 10-8 C)(3.0 10-8 C) /[(0.10 m)2 +(0.05 m) 2 ]
 1.08 10-3 N
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Example: Three Charges (2)
ˆ
ˆ
Sin  ˆ
Fnet  F2 on Cos
3 Sin i  F1 on 3 j  F2 on 3Cos j
Fx  (1.08 10-3 N)
5
 4.83 10-4 N
125
Fy  (1.35 10-3 N)  (1.08 10-3 N)
10
125
 3.84 10-4 N
Fnet  Fx 2  Fy 2  6.17 10-4 N
  ArcTan
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Fy
Fx
 38.5
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Example: Lifting a Glass Bead
A small plastic sphere is charged to
-10 nC. It is held 1.0 cm above a small
glass bead that rests on a table. The
bead has a mass of 15 mg and a charge
of +10 nC. Will the glass bead “leap up”
to the plastic sphere?
F1 on 2  Kq1q2 / r12 2
 (9.0 109 N m 2 /C 2 )(1.0 10-8 C)(1.0 10-8 C) /(0.01 m) 2
 9.0 10-3 N
w  m2 g  (1.5 10-5 kg)(9.80 N/kg)  1.5 10-4 N
Therefore, F1 on 2 exceeds w by a factor of 60.
Therefore, the sphere should indeed leap upward.
(Note that we have neglected electrical forces between
the bead and table, which could be significant.)
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Motion and Response
Question:
If charge A
moves, how long
does it take for
the force vector
on B to respond
by changing
direction?
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Isaac Newton vs.
Michael Faraday
Isaac Newton:
Forces like gravity and electricity
are examples of action at a distance.
The “influence” of A reaches across
space to affect B instantly and
without the need for contact.
Michael Faraday:
Forces like gravity and electricity
are mediated by a “field” that alters
the space around the mass or charge.
Particle A produces a field, and
particle B responds to this field by
experiencing a force. Changes in the
field require time to propagate.
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Example:
The Gravitational Field (1)
F1 on 2  F2 on 1  G
F1 on 2
m1m2
r2
 m1m2

  G 2 , toward mass 1
r


 m

 m2  G 21 , toward mass 1
 r

 m2 g mass 1
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Example:
The Gravitational Field (2)
(1) The field strength g is
proportional to the
mass. g~m1. Larger
masses create stronger
gravitational fields.
(2) The field strength is
inversely proportional
to the square of the
distance. g~1/r2.
(3) The field produced by
mass m1 points directly
to m1.
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Fon m mg ˆj
g

  g ˆj
m
m
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Relative Force Strengths
The electric force, with a force constant of K = 8.99 x 109 N
m2/C2, is much stronger than the gravitational force, with a force
constant of G = 6.67 x 10-11 N m2/kg2. To illustrate this, let’s
consider the ratio of electric to gravitational forces between an
identical pair of protons and an identical pair of electrons.
Proton: mass = 1.67 x 10-27 kg, charge = 1.60 x 10-19 C,
Strength ratio = (K q1q2/r2)/(G m1m2/r2) = K e2/G mp2
= 1.24 x 1036
Electron: mass = 9.11 x 10-31 kg, charge = 1.60 x 10-19 C,
Strength ratio = (K q1q2/r2)/(G m1m2/r2) = K e2/G me2
= 4.16 x 1042
How is it that we notice the gravitational force at all? Because
the universe contains only positive mass, but equal amounts of
positive and negative electric charge, which tends to cancel.
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Fields
Fields:
1. The electric field permeates all of space
2. Charges interact with the electric field
inducing a non-zero value of E (subject to
boundary conditions).
3. A charge in an electric field experiences a
force F exerted by the field.
E ( x, y , z ) 
Fon q at (x,y,z)
q
3. This relation assigns a field vector to every
point in space.
4. If q is positive, the electric field vector points
in the same direction as the force vector.
5. Does E depend on q (the charge that detects
it)? No, because the force is proportional to q.
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F  qE
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Question
An electron (q = e, m = me) is placed at
the position shown by the black dot. The
force on the electron is:
(a) Zero. (b) To the right. (c) To the left.
(d) Insufficient information to tell.
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The Electric Field
of a Point Charge
Fon q'
E
 1 qq '


, away from q 
2
 4 0 r

Fon q'
q'
 1 q


, away from q 
2
 4 0 r

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Electric Field Diagrams
Points about field diagrams:
1. The diagram is only a representative sample
of the electric field. The field exists at all
points in space.
2. The arrow indicates the direction and
strength of the field at the point to which the
vector tail is attached. The length of the
vector is a relative measure of the field
strength.
3. The electric field is not a quantity that
``stretches’’ from one point to another. Each
vector represents the field at that one point
in space.
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Unit Vector Notation
Positive q
Negative q
1
q
E
rˆ
2
4 0 r
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Example:
Calculating the Electric Field
A 1.0 nC charged particle is located
at the origin. Points 1, 2, and 3 have
(x,y) coordinates in cm of (1,0), (0,1),
and (1,1), respectively.
Determine the electric field E at
each of these points.
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Example:
The Electric Field of a Proton
In a hydrogen atom, the electron orbits the
proton at a radius of 0.053 nm. Both have charges of
magnitude e = 1.60 x 10-19 C.
(a) What is the proton’s electric field strength at the
position of the electron?
(b) What is the magnitude of the electric force on the
electron?
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Question 2
Which system has the largest electric field
at the location of the numbered point?
(a)
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(b)
(c)
Physics 122B - Lecture 2
(d)
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Summary: Chapter 25 (1)
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Summary: Chapter 25 (2)
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Typical Electric Field Strengths
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End of Lecture 2
 Before the next lecture, read
Knight, Chapters 26.2 through 26.3
 This weeks homework is on Tycho,
and ready for you to think about.