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Electrostatics
ELECTROSTATICS
the study of electric charges, forces and
fields
Static Electricity is “Stationary Electricity”
or Accumulation of charge
Fundamental Rule
Opposites attract, Likes Repel
Things don’t like having a net charge
If objects don’t like having a net charge,
then how does it happen?
When objects get Charged:
Must obey Law of Conservation of Charge
Charges may be transferred among different atoms, materials, or objects but all charge is accounted for.
NO NEW charges are created nor are any charges destroyed.
Only electrons can move
Remember:
An excess of electrons results in:
A negative charge
A shortage of electrons results in:
A positive charge
ONLY ELECTRONS MOVE
How Do Charges Behave in
Materials?
Conductor:
Allows electrons to move easily
Metals, why?
Metals lose electrons, (not held tightly)
Insulator:
Does not allow electrons to move easily
Non metals. Glass, plastic, dry wood. Why?
Electrons held tightly
Semiconductors – charges only move
freely when certain conditions are met
(i.e., heat, sufficient voltage, etc.) ex
germanium, selenium, and silicon.
Superconductors – charges move
effortlessly and cannot be stopped once
they are moving
Objects become charged by…
Friction
Induction
Conduction
Electrons are rubbed off one insulator onto another insulator
Grounding
With a credit card
9
FRICTION: e- rubbed off one
insulator to another
Objects become charged by…
Friction
Induction
Conduction Charging by CONTACT with a charged object
Grounding
Charging by Conduction
Some electrons leave rod
and spread over sphere.
Requires Contact Electrons transferred.Results in:Object with the same charge as original charged object.
Objects become charged by…
Friction
Induction
Conduction
Charging an object WITHOUT touching a charged object
Grounding
13
Induction
no contact occurs between charged
object and neutral object..
Involves temporary rearrangement of
electrons on neutral object
Neutral Object becomes “polarized”
but net charge remains the same
If neutral polarized object is
grounded, charge will become
“opposite” of the charged object and
is no longer temporary
Neutral objects can be temporarily attracted to charged objects by a process called POLARIZATON.
Charging by Induction
A negatively charged balloon is brought near a
neutral conducting sphere as shown below. As it
approaches, charge within the sphere will
distribute itself in a very specific manner. Which one of the diagrams below properly depicts the
distribution of charge in the sphere?
What is grounding?
Involves Transfer of excess electrons to
and from the ground to neutralize it
Charging by Induction AND Grounding
The rod does not touch the 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.
The charge on the object is opposite if grounded
polarization grounding permanent charge
Grounding is allowing charges to move freely along a connection between a conductor and the ground.
The Earth (the ground) is a practically infinite reservoir of electric charge.
Here a positively charge rod attracts electrons from the ground into the electroscope
Here a negatively charge rod repels electrons into the ground from the sphere
Four fundamental forces in nature
Gravity
Weak nuclear
Electromagnetic (electricity and
magnetism)
Strong nuclear
20
What exactly is CHARGE?
It is physical property of
matter.
It comes in two flavors:
“plus” and “minus.”
What is the unit for charge?
Coulombs (C)
21
Definition of Coulomb
Abbreviation: C
SI unit for charge
One coulomb is NOT equal to the charge of 1 electron!!!!
1C ~ the charge of 6.25 x 1018 electrons
It is the amount of charge to pass through a cross-section of wire in 1 second when 1 Ampere (A) of current is applied.
(We’ll cover the amp later.)
Likewise the + charge of protons is associated with 6.25 x 1018 protons
22
Elementary Particles
Particle Charge,
(Coulombs
per particle)
# of particles
in a Coulomb
electron -1.6 x 10-19 6.25 x 1018
proton +1.6 x 10-19 6.25 x 1018
23
Coulomb’s Law
Charles-Augustin de Coulomb used a torsion pendulum to establish his law.
rd
qqkF ˆ
2
21
24
Electric Force
q charge, C (coulombs)
d distance between charges, m
F electric force, N
k electrostatic constant 9.00 x 109 Nm2/C2
rd
qqkF ˆ
2
21
What happens to F as charge increases?
Increase
What happens to F as r increases?
Decreases by inverse square
Look at kc. Is this a large or small value?
large
How is q described for a proton?
positive
For an electron?
negative
26
The Product of q1and q2
If the product, q1q2 ,is
negative then the force is
attractive.
If the product, q1q2 ,is
positive then the force is
repulsive.
Ex 2: Two negatively charged balloons are 0.70m apart. If the charge of each is 2.0 x 10-6C, What is the electric force between the two balloons?
q1 = q2 = 2.0 x 10-6 Cd = r = 0.70 m
F kq q
r 1 2
2
F = 9.0 x 10 9 N m2/C2 (-2.0 x 10-6 C)2
(0.70m)2
F = 0.073 NAn attracting or repelling force?
Ex.3: Two equally charged balloons repel each other with a force of 4.0 x 10-3 N. If they are 0.015 m apart, what is the charge of the each balloon?
F = 4.0 x 10-3 Nd = 0.015 m F k
q q
r 1 2
2
q2 = (4x10-3N)(0.015m)2
(9x109Nm2/C2)
q2 = Fd2
k
q1 = q2 = 1.0 x 10-8C
Ex 4:
How many Coulombs are in a µC?
1 x 10-6
Two charges are separated by 3.0 cm.
Object A has a charge of +6.0 µC.
Object B has a charge of -6.0 µC. What
is the force on Object A? Is the force
attractive or repelling?
-360N, attractive
Ex 5
Two electrons exert an electrical force of
1.0 x 10-8 N on one another.
Is this an attractive or repelling force?
Repelling
Calculate the distance between them.
Rearrange formula to solve for d
Use known charge for an electron
1.5 x 10-10 m
Two charges create a
force on one another. If
the charge of one object
is doubled, how does
the resulting force
change?
F will double
What if charge of one
object is tripled?
F will triple
rd
qqkF ˆ
2
21
Two charges create a force on one another. If the distance between the objects is increased by a factor of 2, the force changes by a factor of?
F will decrease by a factor of 4
What if distance between the objects is tripled?
F will decrease by a factor of 9
rd
qqkF ˆ
2
21
Review……….
How many electrons in one Coulomb?
6.25 x 1018 electrons
What is the charge of one electron
-1.6 x 10-19 Coulombs (C)
How many protons in one Coulomb?
6.25 x 1018 protons
What is the charge of one proton
+1.6 x 10-19 Coulombs (C).
Review……….
How many electrons in one Coulomb?
6.25 x 1018 electrons
Calculate the charge of one electron
-1.6 x 10-19 Coulombs (C)
How many protons in one Coulomb?
6.25 x 1018 protons
Calculate the charge of one proton
+1.6 x 10-19 Coulombs (C).
Force and Fields
Contact forces
What we mostly
deal with
Objects touch each
other directly
Ex. A tennis racket
hits a tennis ball
F=ma
www.CartoonStock.com.
Forces can occur without
contact!
Action at a distance
Can you think of anything that applies a
force without touching?
36
Gravity demonstrates action at a
distance
What happens if you get too far away
from the mass exerting the force?
The effects are less
37
What else applies an action at a
distance?
Magnets!
38
39
What else applies an action at a
distance?
40
Attracting and repelling forces of
charges
The space that surrounds these things is
altered
Examples:
Magnets
Sun
Planets
Electric charge
Action at a distance depends on
a field of influence An object within the field may be affected by it
Can be scalar or vector
Magnitude only
Ex. Heat
Can be vector
Magnitude and direction
Ex. Gravity (one direction only since only
attracts)
Ex. Electric (more than one direction; attracts
and repels
42
Fields are NOT Force, they exert
the force
Ex. A person pushes a box.
The person is not the force, he exerts the
force!
43
Electric field
A field that exerts force that surrounds
an electric charge or group of charges
Magnitude and direction (vector)
Electric field
How would you detect and measure an
electric field around a charge?
Place another one nearby and see what
happens!
Since all charges produce fields, come
up with a model
45
Electric field model
Source charge: charge producing the
field. Usually designated with a capital
Q
Test charge: a mathematical creation
Always positive
Symbol: q’
Doesn’t exist
Infinitely small, thus produces no field of its
own
46
What is the source charge if
The test charge q moves towards it?
Negative (attracts)
The test charge q moves away from it?
Positive (repels)
How would I draw these?
47
Where do you think the field is
strongest?
48
What if I had more than
one source charge?
What would the field lines
look like?
49
Think: Where is the electrical potential energy of a
positive test charge (q+) higher, at the point A or B?
Why?
Point A. Because of it’s
location, it is not where
it “wants” to be. It took
work to get it there!
The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.
Threads floating on oil bath become polarizedand align themselves with the electric field.
These fields can be detected in lab…
54
55
56
How do I measure the strength of the electrical field
around a source charge (Q)?
What factors do you think the
electrical field strength is dependent
on?1. Force (Push or Pull) of Source Charge
on Test Charge
2. Distance Between Source Charge
and Test Charge
First let’s consider effect of force
58
Electric Field Intensity (Strength)
E - Electric Field Strength or Intensity (N/C)
F - Force experienced by a test charge at that
location (N)
q’ - magnitude of the test charge placed at that
location (C).
'q
FE
A test charge has a magnitude of 1 x 10-10C. It experiences a force of 2N in an electrical field. What is the Intensity of the field?
E = F/q’Variables:F = 2Nq’ = 1 x 10-10C
E = 2N/ 1 x 10-10C
E = 2 x 1010 N/C
Ex 6
A test charge of 6 x 10-26 uC is placed 200 mm from a proton (this is the source charge). What is the electrical force between them? What is the Field strength at this point? What is the direction of the field?
Variables:q’ = 6 x 10-26µCp+= 1.6 x 10-19Cd = 200 mm
Variables:q’ = 6 x 10-32Cp+= 1.6 x 10-19Cd = 0.2m
Now we can solve. This is a 2 step problem.
Step 1: Solve for force using Coulombs Law
Step 2: Use the calculated force and solve for Field Intensity
Ex 7
A test charge of 6 x 10-26 µC is placed 200 mm from a proton (this is the source charge). What is the electrical force between them? What is the Field strength at this point? What is the direction of the field?
Variables:q’ = 6 x 10-32Cp+= 1.6 x 10-19Cd = 0.2m
F = (9 x 109C)(6 x 10-32C)(1.6 x 10-19C)0.2m2
F = 2.16 x 10-39N
E = F/q’
F = kqq
d2
E = 2.16 x 10-39N / 6 x 10-32C
E = 3.6 x 10-8 N/C
Ex 7 cont
Remember: The direction of the electric field 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.
Electric field line flow Out of positive charges and into Negative charges.
+ -Q
The electric field intensity E at a distance d from a source charge Q can be found without knowing the test charge!:
Units: N/C
EX 8: What is the electric field intensity at a distance of 2 m from a source charge of -12 μC? Include direction.
d = 2 mq = -12 μC
9 10 12 10
2
9 6
2
x x( ) = 2.7x104 N/C, towards q
or to the left
q = -12μC
To determine the direction of the field, ask
If the source charge is negative do the
field lines go out or in?
2d
kQE
How do I determine the field strength if
there are multiple charges?
66
When more than one charge contributes to the field, the
resultant field is the vector sum of the contributions from
each charge.
Where k : 9x109Nm2/C2
Units: N/C
Note we will look at direction of the field to know whether fields add or subtract at a point.
2d
kQE
Remember this?
Electric field line flow Out of positive charges and into Negative charges.
+ -Q
Ex 9: Two charges q1=-8 μC and q2=+12 μC are placed 120 mm apart in the air. What is the electric field at the midpoint between them?
q1 = -8 μCq2 = +12 μCr = 0. 120m -
q1
+q2
E1 E2
ET
X
= kq1 + kq2
r2 r2
= (9 x 109)(8 x 10-6) + (9 x 109)(12 x 10-6)(0.06)2 (0.06)2
E= 2.0 x 107 + 3 x 107 = 5.0 x 107 N/Cto the left
2d
kQE
Ex.10: Two charges q1=+8 μC and q2=+12 μC are placed 120 mmapart in the air. What is the electric field at the midpoint between them?
q1 = + 8 μCq2 = +12 μCr = 0. 120m +
q1
+q2
E1E2
ET
X
Ekq
r
2
= kq1 - kq2
r2 r2
= (9 x 109)(8 x 10-6) - (9 x 109)(12 x 10-6)(0.06)2 (0.06)2
The fields are in opposite directions so they subtract
E= 2.0 x 107 - 3 x 107 = -1.0 x 107 N/C
E = 1.0 x 107 N/C to the left