Electric Charge

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Transcript Electric Charge

Electric Charge
Electric Fields
Montwood High School
Physics
R. Casao
Electric Charge
• All matter is composed of atoms which contain
negatively charged electrons, positively charged protons,
and neutrons which have no charge.
• We will concern ourselves with the relative number of
positive and negative charges because charges exert
forces of attraction or repulsion on other charges.
• The movement of charges produce electric and magnetic
fields.
• Law of electrostatics: like charges repel each other;
unlike charges attract each other
– Repel: positive & positive; negative & negative.
– Attract: positive & negative.
Attraction & Repulsion
• Attraction – rods pulled
toward each other.
• Repulsion – rods push
each other away.
Electric Charge
• The electrostatic force of attraction is the force
responsible for holding the protons and
electrons together in the atom.
• Neutral objects have an equal number of
positive and negative charge and do not exert
forces on other charges and are not effected by
other charges.
• Charge cannot be created, but can be
transferred from one object to another. Charge
is conserved.
• Everyday objects becomes charged by gaining
or losing electrons, not protons.
Electric Charge
• The protons are held tightly within the nucleus
by the strong nuclear force, even though the
electromagnetic force of repulsion would push
(repel) the protons within the nucleus.
• Other forces, like rubbing, can overcome the
electrostatic force of attraction and remove
electrons from an atom.
• Net gain in electrons results in negative charge;
net loss in electrons results in positive charge.
• When one object gains a charge, another object
loses the charge; the total electric charge on
both bodies does not change.
Electric Charge
The Transfer of Charge (Friction)
SILK
Glass Rod
Some materials attract electrons
more than others.
Electric Charge
The Transfer of Charge (Friction)
+ -
SILK
Glass Rod
As the glass rod is rubbed against silk,
electrons are pulled off the glass onto the silk.
Electric Charge
The Transfer of Charge (Friction)
+
+ -
SILK
Glass Rod
Usually matter is charge neutral, because the number of
electrons and protons are equal. But here the silk has an
excess of electrons and the rod a deficit.
Electric Charge
The Transfer of Charge (Friction)
+
+ - +
+ +
SILK
Glass Rod
•Glass and silk are insulators; charges stuck on them
stay put.
•Conductors allow charges to move through them.
Charge distributes itself over the entire surface of the
object.
Electric Charge
Electric Charge
• Gains or losses of electrons occur in whole numbers –
you cannot gain or lose half an electron.
• In chemistry, used +1, -1, +2, -2, etc. to designate the
charge on an ion. These numbers refer to the difference
in the number of electrons and protons, not the actual
charge on the ions.
• Symbol for charge is Q or q; unit is the Coulomb, C.
• Amount of charge on 1 proton: 1.602 x 10-19 C.
• Amount of charge on 1 electron: -1.602 x 10-19 C.
• The positive and negative charges will be used to
identify forces between charges as attractive or
repulsive. I don’t recommend using them in problems.
• The electroscope is the simplest device
used to determine electric charge.
– Consisting of a metal rod with a metallic bulb at
one end, the rod is attached to a solid
rectangular piece of metal that has an attached
foil “leaf” made of aluminum.
– The arrangement is insulated from its protective
glass container by a nonconducting frame.
– When charged objects are brought close to the
bulb, electrons in the bulb are either attracted to
or repelled by the charged objects.
– If a negatively charged rod is brought near the bulb, electrons in the
bulb are repelled, and the bulb is left with a positive charge.
– The electrons are conducted down to the metal rectangle and the
attached leaf, which will then swing away because they have like
charges.
Figure 15-4
The electroscope
Charging By Induction
• Bringing a positively charged rod
close to, but not touching, a neutral
conducting sphere will cause the
negative charges within the neutral
sphere to be attracted to the
positively charged rod and the
positive charges within the neutral
sphere to be repelled to the
opposite side of the neutral sphere.
• The net charge on the sphere is
still zero, but one end of the sphere
is positively charged and the other
end of the sphere is negatively
charged.
• A ground is a conducting path
between an object and the Earth to
prevent electric shock due to
excess charge.
Charging By Induction
• The Earth is a conductor and
can act as a source for extra
electrons or as a sink for
unwanted electrons.
• Electrons from the Earth will
move up the conductor and
cancel out the positive charges
on the sphere. Once the
positive charges on the sphere
have been neutralized, leaving
only the negative charges on
the sphere, the force of
repulsion prevents additional
electrons from moving onto the
sphere.
• The sphere is now negatively
charged.
Charge Neutralization
• When two charged objects are allowed to touch, charge
neutralization occurs.
• If object A has a charge of 16 C and object B has a
charge of -4 C and they are connected by a wire, charge
can flow between the objects and 4 of the positive
charges will cancel the 4 negative charges.
– A charge of 12 C will remain and due to the force of
repulsion, the positive charges will try to get as far
away from each other as possible, so 6 C will go to
object A and 6 C will go to object B.
– Remove the connecting wire and both object A and
object B have a charge of 6 C.
Coulomb’s Law
• Coulomb’s law describes the electric force between any
two charges, separated by a distance d.
F
k  Q1  Q2
d2
• k is a constant that considers the effect of the material
between the charges on the force.
• Force measured in N; for air or a vacuum, 2
Nm
9
k = 8.9875 x 10
2
C
• Many charges are expressed in micro-coulombs (C);
1 x 106 C = 1 C. Easiest solution, whenever you see
C, just add x 10-6 C to the number.
Ex. 5 C = 5 x 10-6 C; 28 C = 28 x 10-6 C.
Coulomb’s Law Example
• Two electrostatic point charges of 60 C and 50 C exert a
repulsive force on each other of 175 N. What is the distance
between the two charges?
• Q1 = 60 x 10-6 C; Q2 = 50 x 10-6 C.
F
k  Q1  Q2
d2
k  Q1  Q2
d 
F
2
F  d 2  k  Q1  Q2
d
N  m2
6
6
8.9875 x 10

60
x
10
C

50
x
10
C
2
C
175 N
9
d
k  Q1  Q2
F
d  0.3925 m
Superposition of Forces from Two Charges
Light blue charges fixed , negative, equal charge (-q)
What is force on positive charge +q at origin?
y
x
Superposition of Forces from Two Charges
Light blue charges fixed , negative, equal charge (-q)
What is force on positive charge +q at origin?
Consider effect of each charge separately:
y
x
Superposition of Forces from Two Charges
Light blue charges fixed , negative, equal charge (-q)
What is force on positive charge +q at origin?
Take each charge in turn:
y
x
Superposition of Forces from Two Charges
Light blue charges fixed , negative, equal charge (-q)
What is force on positive charge +q at origin?
Create vector sum:
y
x
Superposition of Forces from Two Charges
Light blue charges fixed , negative, equal charge (-q)
What is force on positive charge +q at origin?
Find resultant:
y
NET
FORCE
x
Superposition Principle
F31
q1
q2
F31
F
F31y
F21
F31x
F21
F21y
q3
Forces add vectorially
F21x
F = (F21x + F31x) x + (F21y + F31y) y
Everything you learned about vectors applies to the forces
between charges!
Conductors and Insulators
• Conductor
transfers charge on contact
• Insulator
does not transfer charge on
contact
• Semiconductor
might transfer charge on
contact
Charge Transfer Processes
• Conduction
• Polarization
• Induction
Metals and Conduction
•
•
Notice that metals are not only good electrical conductors, but they are also
good heat conductors, tend to be shiny (if polished), and are maleable (can
be bent or shaped).
These are all properties that come from the ability of electrons to move
easily.
This iron atom (26 protons, 26 electrons) has
two electrons in its outer shell, which can
move from one iron atom to the next in a
metal.
Path of electron
in a metal
Electric Fields
• An electric field is a region
around one charged object in
which a second charged object
will experience an electric force
of repulsion or attraction.
• If a small test charge qo is
placed near a charge Q, the
test charge qo is repelled by Q
with a force F. The electric field
strength at the location of qo is:
F
E
qo
Electric Fields
• Units for electric field: N/C.
• The direction of the electric field E depends on
the sign of the charge producing the field.
– Electric field lines point away from positive charges.
– Electric field lines point towards negative charges.
– The greater the number of electric field lines, the
stronger the electric field.
Electric Field Strength
• Electric field strength can also be determined using the
charge Q that is producing the electric field:
– The distance from the charge to the point at which we want to
determine the electric field strength is d.
E
k Q
d
2
• Electric field is a vector, so everything you learned about
vectors applies!
Electric Fields and Acceleration
• Electric fields can be used to accelerate charged
particles, such as protons and electrons.
F
E
qo
F  ma
E  qo
a
m
• Charge on proton or electron, qo = 1.602 x 10-19 C
• Mass electron = 9.11 x 10-31 kg
• Mass of proton = 1.67 x 10-27 kg
• Can use the old velocity equations to find whatever you
need to:
2
2
v f  v i  (a  t)
vf  vi  (2  a  x)
x  (vi  t)  (0.5  a  t 2 )
Electric Field Example
• A proton and an electron in a hydrogen atom are separated by about
5.3 x 10-11 m.
– What is the magnitude and direction of the electric field set up by
the proton at the position of the electron?
– Known values: k = 8.9875 x 109 N·m2/C2; Q = 1.602 x 10-19 C;
d = 5.3 x 10-11 m.
N  m2
19
8.9875 x 10

1.602
x
10
C
2
k Q
C
E 2 
2
11
r
5.3 x 10 m
9


E  5.12566 x 1011 N/C
– Direction: the proton is positive and electric field lines extend
outward away from positive charges, so the electric field
produced by the proton at the position of the electron is away
from the proton.
Electric Field Example
• If the electron could be accelerated in the electric field produced by
the proton, what is the acceleration of the electron and what is the
speed of the electron after 0.5 s, assuming the electron starts from
rest?
– Known values: mass electron = 9.11 x 10-31 kg; charge electron
= 1.602 x 10-19 C; E = 5.12566 x 1011 N/C
E  Q 5.12566 x 1011 m 1.602 x 10 19 C
a

m
9.11 x 10 31 kg
a  9.0135 x 10 22 m/s 2

vf  vi  (a  t)  0 m/s  9.0135 x 10 22 m/s 2  0.5 s
vf  4.50675 x 10 22 m/s

Forces on electron beam in a TV tube (CRT)
F = Q E and F = m g (vector equations)
Motion of a Charged Particle in a Uniform
Electric Field


F  qE



F  qE  ma

 qE
a
m
September 18, 2007

If the electric field E is uniform
(magnitude and direction), the
electric force F on the particle is
constant.

If the particle has a positive charge,
its acceleration a and electric force F
are in the direction of the electric
field E.

If the particle has a negative charge,
its acceleration a and electric force F
are in the direction opposite the
electric field E.
An Accelerated Electron
• An electron enters the region of a uniform electric
field as shown in the figure, with vi = 3 x 106 m/s
and E = 200 N/C. The horizontal length of the
plates is l = 0.1 m.
A. Find the acceleration of the electron while it is in
the electric field.
E q
a
m
a
200 N
 1.602 x 10 19 C
C
9.11 x 10 31 kg
a  3.51 x 1013 m
s2
B. If the electron enters the field at time t = 0 s, find
the time at which it leaves the field.
d l
l
vi   ; t 
t t
vi
0.1 m
t
6 m
3 x 10
 3.33 x 108 s
s
C. If the vertical position of the electron as it enters
the field is yi = 0 m, what is its vertical position
when it leaves the field?
y  vyi  t  0.5  a  t 2
vyi  0 m ; no y velocity, given velocity is horizontal
s
y  0  t  0.5  a  t 2  0.5  a  t 2
13
y  0.5  3.51 x 10
m
s
2

 3.33 x 10
8
s

2
y  0.0195 m
• Because the electron is deflected downward
between the plates by the electric field, the final y
position for the electron is -0.0195 m below its
initial y position.
TV tube with electron-deflecting charged
plates (orange)
F=QE