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PY212
Electricity and Magnetism
I. Electrostatics
07. 07. 2003
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I-1 Electric Charge
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Main Topics
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Why Electrostatics?
Demonstration of Electrostatic Effects.
Electric Charge and its Properties.
Coulomb’s Law.
Some Applications of the C. L.
Electric Field and Electric Intensity
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Why Electrostatics?
• Many important properties of the Nature
exist due to electric interactions of charged
particles.
• We shall first deal with fields and charges
which are static = do not move.
• It is for simplicity but such fields really
exist, if some equilibrium can be reached!
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Demonstration of Electrostatic
Effects
• A comb after it has been run through hair attracts
little pieces of paper. The force is a long-distance
one. It can be attractive or repulsive.
• We attribute these forces to the existence of a
property we call the electric charge.
• Bodies can be charged by conduction via contact
with other bodies but even remotely by induction.
• Using some materials we can easily discharge
charged bodies, these are conductors. By others it
is slow or even impossible, they are insulators
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Main Properties of Charges
• Since both the attractive and the repulsive forces
exist, charges must be of two kinds, positive and
negative. Unlike charges attract and like charges
repel themselves.
• Charges are quantized. They can only be isolated
in integer multiples of the elementary charge
e = 1.602 10-19 C
• In all known processes charges appear or
disappear only in pairs (+q and -q), so the total
charge is conserved
• Charge is invariant to the Lorentz transformation
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Main Properties of Electrostatic
Interactions
• Charged particles act by a force on each
other. Forces :
• are Long-distance – mediated by electric field
• obey the principle of superposition
• Mutual interaction of two static point
charges is described by Coulomb’s Law
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Coulomb’s Law I
• Let us have two point charges Q1 and Q2 at the
distance r apart. Then the magnitude of the
interaction force is:
F = k Q1 Q2 / r2
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The SI unit of charge is 1 Coulomb [1 C]
k = 9 109 Nm2/C2
k = 1/40
0= 8.85 10-12 C2/ Nm2 is the permitivity of vacuum
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Coulom’s Law II
• Since forces are involved the directional
information is as important as the magnitude.
• To get the full information let’s place Q1 into the
origin and let describe the position of Q2 by the

radius vector r . Then the force acting on Q2 is :

 
kQ1Q2 r kQ1Q2  0
F21 (r ) 

r
2
2
r
r
r
• Forces act in the straight line joining the charges .
• Positive force is repulsive .
• Forces acting on Q1 and Q2 are action and reaction of
each other .
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Coulomb’s Law III
• The most general formula we get if we define the
position of each
 charge Qi (i=1, 2) by its own
radius vector ri . Then the force acting on Q2 is :
 
 
kQ1Q2 (r2  r1 )
F21 (r2 ) 
  3
| r2  r1 |
• Since the force depends only on the difference
between the radius vectors, the position of the
origin is arbitrary .
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Comparison with the Force of
Gravity
• Formally, Coulomb’s Law is Analogous to
Newton’s Gravitational Law
• But electrostatic force is ~ 1042 (!) times
stronger
• Such a weak force still dominates the universe
because matter is usually neutral
• Charging something means to break very
slightly the great equilibrium
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The Concept of the Field
• If a charge is located in some point in the
space it sends around an information about
its position, polarity and magnitude. The
information spreads with the speed of light.
It can be “received” by another charge . The
interaction of a charge with field produces a
force.
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Electric Intensity I
• Electrostatic field could be described by taking
some test
Q and recording the vector of the
 charge

force F (r ) acting on it in every point of interest.
• This description would, however, depend on the
magnitude and polarity of the test charge and these
properties would have to be provided as the
additional information to make the description
unique.
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Electric Intensity II
• By dividing of the force by the magnitude of the
test charge the electric intensity
is defined :
 
 
F (r )
E (r ) 
Q
• It is a unique property of the described field, now.
• Numerically it is equal to the force which would
act in the particular point on a unit positive charge,
but be careful with dimensions.
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Electric Intensity III
• It is important to realize that by dividing by the
magnitude of the charge, the information, how the
charge ‘feels’ the field, becomes an objective
property of the field.
• The same field acts on various charges by various
forces which can be even opposite due to the
existence of two polarities of charges.
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Electric Field Lines I
• Electric field is a three dimensional vector
field which is in general case difficult to
visualize.
• In cases of simple symmetry, it is possible
to use electric field lines which are lines
tangent to vectors of the electric intensity in
every point. The magnitude of the field can
be illustrated by their length or density.
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Electric Field Lines II
• A positive charge of a very little mass
would move along a certain field line in the
direction of the electric field while negative
charge would move in the opposite
direction.
• Field lines can’t cross!
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Homework 1
• The homework is selected from “problem”
sections that are in the end of each chapter.
• Due Wednesday !
• 21- 2, 7, 9, 14, 15
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Things to Read :
• This lecture covers:
Giancoli: Chapter 21, Sections 1-8 (ex. 7)
• Advance reading :
Giancoli:
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Chapter 22
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Two electrons 1 m apart
They are repelled by electric force but
attracted by the force of gravitation. Which
force will prevail?
Fe  9 10 (1.6 10
9
Fg  6.67 10
11
19 2
)  2.31 10
(9.1 10
Fe
31 2
28
N
)  5.54 10
 71
N
 4.2 10 !!!
42
Fg
^
One electron and one proton 0.53 10-10 m
apart
This corresponds to their distance in hydrogen
atom.
Fe  9 10 (1.6 10
9
 8.2 10
8
19
/ 0.53 10
10 2
)
N
Force of this magnitude can be in principle
measured macroscopically! This is the
secret why matter holds together.
^
Let us separate protons and electrons
from one gram of H and put each group
on each pole of the Earth
1 g is 1 gram-molecule of H, so we have
NA=6.02 1023 of both types of particles.
Fe  9 10 (1.6 10
9
19
6.02 10 / 12.7 10 )
23
6 2
 5.2 10 N (!)
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^
Two 1g iron spheres, 1 m apart are attracted
by the force of 10 N. What is their excess
charge compared to the total charge?
The excess charge:
5
q  10 / 9 10  3.3 10 C  2 10 e
9
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The total charge:
6.02 10  26
23
qt 
 2.8 10 e
55.8
9
q / qt  10 (!)
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^