ELECTROSTATICS:

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ELECTROSTATICS The study of the behavior of
stationary charges ELECTRIC CHARGE There are
two types of electric charge, arbitrarily called
positive and negative. Rubbing certain
electrically neutral objects together (e.g., a
glass rod and a silk cloth) tends to cause the
electric charges to separate. In the case of the
glass and silk, the glass rod loses negative
charge and becomes positively charged while the
silk cloth gains negative charge and therefore
becomes negatively charged. After separation, the
negative charges and positive charges are found
to attract one another.
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When a rubber rod is rubbed against fur,
electrons are removed from the fur and deposited
on the rod.
Electrons move from fur to the rubber rod.
The rod is said to be negatively charged because
of an excess of electrons. The fur is said to be
positively charged because of a deficiency of
electrons.
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When a glass rod is rubbed against silk,
electrons are removed from the glass and
deposited on the silk.
Electrons move from glass to the silk cloth.
The glass is said to be positively charged
because of a deficiency of electrons. The silk
is said to be negatively charged because of an
excess of electrons.
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Laboratory devices used to study the existence of
two kinds of electric charge.
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1. Charge the rubber rod by rubbing against fur.
2. Transfer electrons from rod to each pith ball.
The two negative charges repel each other.
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1. Charge the glass rod by rubbing against silk.
2. Touch balls with rod. Free electrons on the
balls move to fill vacancies on the cloth,
leaving each of the balls with a deficiency.
(Positively charged.)
The two positive charges repel each other.
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Note that the negatively charged (green) ball is
attracted to the positively charged (red) ball.
Opposite Charges Attract!
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Like charges repel unlike charges attract.
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Charging by Contact  
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Charging by Induction    
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The law of conservation of electric charge "The
net amount of electric charge produced in any
process is zero." Another way of saying this is
that in any process electric charge cannot be
created or destroyed, however, it can be
transferred from one object to another.
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Applications of Electrostatic Charging           
                            
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The quantity of charge (q) can be defined in
terms of the number of electrons, but the Coulomb
(C) is a better unit for later work.
The Coulomb 1 C 6.25 x 1018 electrons
The charge on a single electron is
1 electron e- -1.6 x 10-19 C
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The coulomb (selected for use with electric
currents) is actually a very large unit for
static electricity. It is common to use the
metric prefixes.
1 mC 1 x 10-6 C
1 nC 1 x 10-9 C
1 pC 1 x 10-12 C
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COULOMBS LAW   Coulombs Law states that two
point charges exert a force (F) on one another
that is directly proportional to the product of
the magnitudes of the charges (q) and inversely
proportional to the square of the distance (r)
between their centers. The equation is
F electrostatic force (N) q charge (C) k
9x109 Nm2/C2 r separation between charges (m)
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The value of k can also be expressed in terms of
the permittivity of free space (eo)
9x109 N. m2/C2
The proportionality constant (k) can only be used
if the medium that separates the charges is a
vacuum. If the region between the point charges
is not a vacuum then the value of the
proportionality constant to be used is determined
by dividing k by the dielectric constant (K).
For a vacuum K 1, for distilled water K 80,
and for wax paper K 2.25
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Problem-Solving Strategies
1. Draw and label a figure indicating positive
and negative charges along with the given
distances. 2. Draw the force of attraction or
repulsion on the given charge on a neat, labeled
FBD. 3. Find the resultant force. Important Do
not use the signs of the charges when applying
Coulomb's law!
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11.1 Two charges q1 - 8 µC and q2 12 µC are
placed 120 mm apart in the air. What is the
resultant force on a third charge q3 - 4 µC
placed midway between the other charges?
FR
F2
q1 - 8x10-6 C q2 12x10-6 C q3 - 4x10-6 C r
0. 120 m
F1
- q1
q2
- q3
0.06 m
0.06 m
80 N
120 N
FR 80 120 200 N, to the right
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11.2 Three charges q1 4 nC, q2 -6 nC and q3
-8 nC are arranged as shown. Find the
resultant force on q3 due to the other two
charges.
F1
FR
q1 4x10-9 C q2 -6x10-9 C q3 -8x10-9 C
F2
37
?
 
2.88x10-5 N
6.75x10-5 N
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FR
F1
From the FBD
F2
37
?
S Fy F1 sin 37 (2.88x10-5)(sin 37)
1.73x10-5 N
S Fx F2 - F1 cos 37 (6.75x10-5) -
(2.88x10-5)(cos 37) 4.45x10-5 N
4.8x10-5 N
FR (4.8x10-5 N, 21)
? 21
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ELECTRIC FIELD An electric field is said to exit
in a region of space in which an electric charge
will experience an electric force. The magnitude
of the electric field intensity is given by
Units N/C
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The direction of the electric field intensity at
a point in space is the same as the direction in
which a positive charge would move if it were
placed at that point. The electric field lines or
lines of force indicate the direction. The
electric field is strongest in regions where the
lines are close together and weak when the lines
are further apart.
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11.3 The electric field intensity between two
plates is constant and directed downward. The
magnitude of the electric field intensity is
6x104 N/C. What are the magnitude and direction
of the electric force exerted on an electron
projected horizontally between the two plates?
E 6x104 N/C qe 1.6x10-19 C
F qE 1.6x10-19 (6x104) 9.6x10-15 N,
upward
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11.4 Show that the gravitational force on the
electron of example 16-3 may be neglected.
me 9.11x10-31 kg
FG mg 9.11x10-31 (9.8) 8.92x10-30
N
The electric force is larger than the
gravitational force by a factor of 1.08x1015!
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The electric field intensity E at a distance r
from a single charge q can be found as follows
Units N/C
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11.5 What is the electric field intensity at a
distance of 2 m from a charge of -12 µC?
r 2 m q -12 µC
27x103 N/C, towards q
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When more than one charge contributes to the
field, the resultant field is the vector sum of
the contributions from each charge.
Units N/C
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11.6 Two point charges q1 -6 nC and q2 6 nC,
are 12 cm apart, as shown in the figure.
Determine the electric field a. At point A
q1 -6 x10-9 C q2 6 x10-9 C
E1
ER
E2
3.38x104 N/C, left
8.44x103 N/C, left
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E1
ER
E2
ER E1 E2 3.38x104 8.44x103
4.22x104 N/C, left
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11.6 Two point charges q1 -6 nC and q2 6 nC,
are 12 cm apart, as shown in the figure.
Determine the electric field b. At point B
E2
q1 -6x10-9 C q2 6x10-9 C
37º
?
ER
E1
6.67x103 N/C
2.4x103 N/C
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E2
From vector diagram
37º
S Ex - E2cos 37 - (2.4x103)(cos 37)
-1916.7 N/C
?
ER
E1
S Ey E2 sin 37- E1 (2.4x103)(sin 37) -
(6.67x103) - 5225.6 N/C
5566 N/C
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70
E2
37º
?
180 70 250 ER (5566 N/C, 250)
ER
E1