Transcript File

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.
If the glass rod is suspended from a
string and a second positively charged
glass rod is brought near, a force of
electrical repulsion results. Negatively
charged objects also exert a repulsive
force on one another.
These results can be
summarized as follows:
unlike charges attract and
like charges repel.
CONSERVATION OF ELECTRIC CHARGE
In the process of rubbing two solid objects together, electrical
charges are not created. Instead, both objects contain both
positive and negative charges. During the rubbing process, the
negative charge is transferred from one object to the other
leaving one object with an excess of positive charge and the
other with an excess of negative charge. The quantity of excess
charge on each object is exactly the same.
Electrons are free to
move in metals.
Nuclei remain in place;
electrons move to bottom
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.
Charged comb attracts
neutral bits of paper.
Charged comb attracts neutral
water molecules.
The SI unit of charge is the coulomb (C).
1 C = 6.25 x 1018 electrons or protons
The charge carried by the electron is represented by the symbol
-e, and the charge carried by the proton is +e. A third particle,
which carries no electrical charge, is the neutron.
e = 1.6 x 10-19 C
melectron = 9.11 x 10-31 kg
mproton = 1.672 x 10-27 kg
mneutron = 1.675 x 10-27 kg.
Experiments performed early in this century have led to the
conclusion that protons and neutrons are confined to the
nucleus of the atom while the electrons exist outside of the
nucleus. When solids are rubbed together, it is the electrons
that are transferred from one object to the other. The positive
charges, which are located in the nucleus, do not move.
Rubber scrapes electrons
from fur atoms.
INSULATORS AND CONDUCTORS
An insulator is a material in which the electrons are tightly held
by the nucleus and are not free to move through the material.
There is no such thing as a perfect insulator, however examples
of good insulators are: glass, rubber, plastic and dry wood.
A conductor is a material through which electrons are free to
move through the material. Just as in the case of the insulators,
there is no perfect conductor. Examples of good conductors
include metals, such as silver, copper, gold and mercury.
A few materials, such as silicon, germanium and carbon, are
called semiconductors. At ordinary temperature, there are a few
free electrons and the material is a poor conductor of
electricity. As the temperature rises, electrons break free and
move through the material. As a result, the ability of a
semiconductor to conduct improves with temperature.
Charging by Contact
Some electrons leave rod
and spread over sphere.
Charging by Induction
Rod does not touch sphere. It pushes electrons out of the
back side of the sphere and down the wire to ground. The
ground wire is disconnected to prevent the return of the
electrons from ground, then the rod is removed.
INDUCED ELECTRIC CHARGE
If a negatively charged rod is brought near an uncharged
electrical conductor, the negative charges in the conductor
travel to the far end of the conductor. The positive charges
are not free to move and a charge is temporarily induced at
the two ends of the conductor. Overall, the conductor is still
electrically neutral and if the rod is removed a redistribution
of the negative charge will occur.
If the metal conductor is touched by a person’s finger or a wire
connected to ground, it is said to be grounded. The negative
charges would flow from the conductor to ground. If the
ground is removed and then the rod is removed, a permanent
positive charge would be left on the conductor. The electrons
would move until the excess positive charge was uniformly
distributed over the conductor.
CHARGE
DISTRIBUTIONS
Charge on
Metals
Metal Ball
Excess charge
on the surface
of a uniform
metal spreads
out.
Charge on Insulators Charge on Metal Points
Plastic Ball
Charge on insulating
materials doesn't
move easily.
Excess charge on a metal
accumulates at points.
Example: lightning rods.
Applications of Electrostatic
Charging
Fine mist of negatively charged
gold particles adhere to
Negatively charged paint
positively charged protein on
adheres to positively
fingerprint.
charged metal.
COULOMB’S LAW
Coulomb’s 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:
q1q2
Fk 2
r
F = electrostatic force (N)
q = charge (C)
k = 9x109 N. m2/C2
r = separation between charges (m)
The value of k can also be expressed in terms of the permittivity
of free space (εo):
k
1
4 o
 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
16.1 Two charges q1 = -8 μC and q2= +12 μC are placed 120mm 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 = -8 μC
F1
q2= +12 μC
q1q2
q3 = -4 μC
+
Fk 2
q
q
q
r = 0. 120m
1
6
6
(
8
x
10
)(
4
x
10
)
9
F1  9 x10
(0.06) 2
6
6
(
12
x
10
)(
4
x
10
)
9
F2  9 x10
(0.06) 2
3
2
= 80 N
= 120 N
FR = 80 + 120
= 200 N, to the right
r
16.2 Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are arranged as
shown. Find the resultant force on q3due to the other two charges.
q1 = +4 nC
q2= -6 nC
q3 = -8 nC
F1
FR
37˚
9
9
(
4
x
10
)(
8
x
10
)
9
-5 N
=
2.88x10
F1  9 x10
(01
. )2
9
9
(
6
x
10
)(
8
x
10
)
9
-5 N
=
6.75x10
F2  9 x10
(8x102 ) 2
θ
F2
q1q2
Fk 2
r
FR
F1
37˚
θ
F2
From the FBD:
Σ Fy = F1 sin 37˚
= (2.88x10-5)(sin 37˚)
= 1.73x10-5 N
Σ Fx = F2 - F1 cos 37˚
= (6.75x10-5) - (2.88x10-5)(cos 37˚)
= 4.45x10-5 N
F  (4.45x105 ) 2  (173
. x105 ) 2
5
173
.
x
10
  tan 1
4.45x105
θ = 21˚
= 4.8x10-5 N
FR (4.8x10-5 N, 21˚)
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:
F
E
q
Units: N/C
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.
16.3 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
16.4 Show that the gravitational force on the electron of example 16-3 may
be neglected.
me = 9.1x10-31 kg
Fg = mg
= 9.1x10-31 (9.8)
= 8.92x10-30 N
The electric force is larger than the gravitational force by a
factor of 1.08x1015!
The electric field intensity E at a distance r from a single
charge q can be found as follows:
kq
E 2
r
Units: N/C
16.5 What is the electric field intensity at a distance of 2 m from a charge
of -12 μC?
r=2m
q = -12 μC
kq 9 x109 (12 x106 )
E 2 
2
2
r
= 27x103 N/C, towards q
When more than one charge contributes to the field, the
resultant field is the vector sum of the contributions from each
charge.
kq
E 2
r
Units: N/C
16.6 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 nC
q2 = +6 nC
E1
E2
kq
E 2
r
9 x109 (6x109 )
4 N/C, left
E1 
=
3.38x10
2 2
(4 x10 )
9 x109 (6 x109 )
3 N/C, left
E2 
=
8.44x10
(8 x102 ) 2
ER
E1
ER
E2
ER = E 1 + E2
= 3.38x104 + 8.44x103
= 4.22x104 N/C, left
16.6 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
q1 = -6 nC
q2 = +6 nC
E2
37º
θ
ER
E1
9 x109 (6x109 )
E1 
(9 x102 ) 2
= 6.67x103 N/C, down
9 x109 (6 x109 )
3 N/C at 37˚
E2 
=
2.4x10
2 2
(15x10 )
16.6 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
From FBD
E2
37º
θ
Σ Ex = - E2cos 37˚
= - (2.4x103)(cos 37˚)
= -1916.7 N/C
ER
E1
Σ Fy = E2 sin 37˚- E1
= (2.4x103)(sin 37˚) - (6.67x103)
= - 5225.6 N/C
E R  (1916.7) 2  (5225.6) 2
= 5566 N/C
16.6 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
37º
θ
ER
5225.6
  tan
= 70˚
1916.7
1
180˚ + 70˚ = 250˚
ER (5566 N/C, 250˚)
E1