Transcript static elec review - Red Hook Central Schools

Electric Charge &
Electric Fields
1
Objects with excess + or - charge give
rise to electric force.
2
Acquisition of Charge
Solid objects charged by e- transfer.
+charge results from loss of e-charge fr gain of e-.
Liquids or gasses + and - ions are free
to move about.
3
Neutral objects can be polarized
Charges which are free to move are redistributed.
4
Charged objects can induce polarization.
Balloon is attracted to positive wall surface.
5
Quantity of Charge in Atoms
Outer Part
electrons e-
Q Elem
–1
Q
-1.6 x 10-19 C
Nucleus
Protons p+
neutrons no
+1
0
1.6 x 10-19 C
0
6
Conservation of Charge Q
Q can be transferred, Q cannot be
created or destroyed.
SQ in system remains constant.
7
Solids.
• Conductors – allow charges to move
around (metals).
• Insulators – hold excess charge in
place.
8
Conductor distribute Q.
9
Insulators – charges concentrated
in one spot.
10
Polarization occurs easily in
conductor.
11
Charging Objects:
• Friction – rub 2 objects. Insulators.
• Conduction (contact) – touch charged
to uncharged q shared equally.
Conductors.
• Induction – polarization & grounding
(drives off or sucks in e-). 1 conductor
1 insulator.
12
Conductors can be charged or discharged
by induction. Need to ground (Earth).
13
1. A positively charged glass wand with a charge
of +1000 is used to charge a neutral metal sphere
by induction. The resulting charge on the sphere
will be:
•
•
•
•
1) + 500
2) - 500
3) +1000
4) -1000.
14
Conservation of Charge
Charge q is quantized. There is a smallest unit.
Charge can only exist in whole number integers
of the charge on e.
15
Units of charge = coulombs (C)
Charge on e- is -1.6 x 10-19 C
Charge on p+ is +1.6 x 10-19 C
fundamental / elementary units, e:
e- has charge –1
p+ has charge +1
A. How many electrons does it take to carry a
charge of -1.0 C?
6.25 x 1018
2. Which Charge cannot exist on an object?
•
•
•
•
1) 5.6 x 10-19 C.
2) 1.15 x 1015 e.
3) 1.12 x 10-18 C.
4) 1 billion billion e.
17
Coulomb’s Law Relates Force btw.
charged objects.
Fe = kq1q2
r2
k = constant 8.99 x 109 N m2/C2.
q charge on obj in Coulombs (C)
r is dist btw centers meters.
F is force (N)
18
The constant k can be written as:
k = 1/4peo.
Where eo is the permittivity of free
space in a vacuum (air)
= 8.85 x 10-12 N m2/C2.
19
3: State Coulomb’s Law in words.
• The force between 2 point charges is
directly proportional to the amount of
charge and
• Inversely proportional to the square of
the distance between their centers.
20
Electric Force is vector quantity.
Ex 4 . Charges (3) of +1 C are located at the corners
of a 45o right triangle. What is the resultant force on
the charge located at the 90o angle.
1C
1.3 x 1010 N
45o below horz
1m
1C
1C
F2
1m
F1
Hwk Rd Kerr 6.2.1 – 6.2.4
do pg 162 #1 – 3, 7, 9,12, 14, 17 .
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Electric Field
region of space around charged
object where a charge feels an
electrostatic force.
Electric Fields-Charge alters space
around it. Charged objects feel a force.
Either repulsion or attraction.
24
Electric Field (E) defined as:
The force per unit charge at a point in
space on a small +test charge..
E = F/q.
E = Electric Field (N/C)
F is force on test charge (N).
q is amt of charge on test charge (C).
Remember gravitational field?
• Region of space where mass feels a force.
• g = force/unit mass on a small mass N/kg
or m/s2.
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5. Calculate the E field strength 0.4 m away
from a charge +20 mC.
E = F/q
Fe = kQ1q2
r2.
Sub in for F kQ1q2 where q = q2.
r2
q
E = kQ1
r2
Memorize .
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E = kQ1
r2
Fill in the numbers.
E = 9x109(20x10-6)C
(0.4)2
1.1 x 106 N/C.
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Field Lines represent electric fields.
Electric field lines show the force that a
small positive test charge feels in a field
created by a much larger charge. They
represent the strength and direction of the
field.
Sketch vectors to show force
magnitude & direction on a + test
charge at each point.
+
30
Field around positive object.
Lines start on + charge, end on –
charge.
31
The denser the field lines are, the
stronger the field.
Stronger field
near charge.
32
Field lines start on + and end on – charge.
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Field Between Parallel Plates
How would the strength of the field vary if
a charge moves from the + to the – plate?
34
Fields have strength and direction - vector.
Density of lines shows strength.
Direction shown as arrows.
Direction is determined by a + test charge.
Lines start on pos end on neg.
Electric Field lines don’t touch or cross.
35
Electric field due to more than one charge.
Field is stronger near the larger charge. Density of
lines show the increased strength.
Strength at a point is the vector sum of field strengths.
Electrostatic Equilibrium
Fields produced by more that a single charge will have
spots where the forces on a charge in the field will be
balanced.
F net = 0.
E field inside a conductor is zero!
• Why?
• If an E field existed inside a conductor, and q are
free to move, the E field would exert Fnet on all q
present. Fnet would accelerate the q until
equilibrium were reached and E goes to zero.
Superposition
To find the force or field on a charge q, due to
the presence of more than one other charge,
you must use vector addition.
39
6: Two 10 mC charges separated by 30 cm.
What is the field strength 10 cm to the right
of A?
10 cm
10 mC
20 cm
10 mC
• B
• 6.75 x 10 6 N/C
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1909 Millikan measured charge on e-
Drops suspended when Fg = Fe.
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Film Mech Universe E fields
Hwk Kerr pg 162 #4 – 6, 8 10, 11,
15 16, 18 -21, 23
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Work & Energy
Charges in an E field can have PEelc.
It takes work to bring charges near a
repelling charge causing charge to gain
PE.
Charges that are attracted by opposite
charges are said to “fall” toward them
losing PE. Work is done by the field.
45
The convention:
• If the charge gains PE then work is positive.
• If it loses PE work is negative.
46
Where will a charge feel no
force from electric field?
+
Infinity!
47
Electric Potential
• Wk per Coulomb to bring a charged
particle to point in field from infinity.
48
Electric Potential
It requires energy to bring +q fr. infinity to point
P in E field. They repel. Each point in a field has
electric potential, like a height.
+q
P
Must put a force
on +q & push it
charge thru a
distance.
49
The amt of work done on every coulomb of
charge moving it is called electric potential, V.
V = W/q.
W work in J
q is charge in C.
V is Volts = J/C.
V defines the potential at P also called voltage.
P is like a particular height in a gravity field.
Note: Wk also = DE.
50
Note: the work done by an electric field is
conservative!
It is independent of the path taken and
depends only on the endpoints.
51
Since work is done to move charge,
the PEelc of q must change:
V = W/q
but
W = DPE.
W = DPE =qV.
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Potential Difference
Imagine moving a 2 C charge from A to B in field
where A and B are at dif potentials (heights). What is
the dif in potential?
pd = 28V – 15V = 13 V.
B = 28 V
A = 15 V
In Uniform Field
Constants between plates:
F on q.
field intensity E= F/q,
work done (Fd) to move a charge
potential difference, V.
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Prove that for uniform field:
E= V
d
V=W
q
V = Fd
q
F =E
q
V = Ed
E = V/d
55
Since: E = V
d
for a constant Electric field
Units of E can be V/m
56
Energy of Moving Charges in Fields.
• As a charge moves thru a field, its total E is constant.
By consv of E:
• If a charges “falls” toward the oppositely charged plate
its PEelc decreases, but it will accelerate, its KE
increases.
• Work done by field will accelerate charge
W = DKE = qV.
• So:
•
ET
before
=
after
ET .
57
There is a potential difference – voltage
between the two plates based on their
charge & distance between them.
A proton near the positive plate is at a
high potential (energy).
What is the potential of a proton stuck to
the negative plate?
(0)
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The electron-volt: a unit of
work & energy.
For very small changes in PEelc (on the order of
10-19J) eV is used.
The electron-volt, eV, is the work & E required to
push 1 e- (or p+) through a voltage of 1V.
W = qV = (1.6 x 10-19 C)(1V) = 1.6 x 10-19 J = eV.
59
If 1 e- is pushed across 1V then 1 eV of work is
done.
If a charge of 2e- is pushed across a 1V pd then
work = 2 eV.
If 2e- pushed across 6V then work is 12 eV.
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What if 3e- move across 12 V?
36 eV
To find eV (# elm charges) (voltage)
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7. How many joules of energy are
represented by 6.9 x 1029 eV.
6.9 x 1029 eV x 1. 6 x 10-19 J. = 1.1 x 1011 J
eV
Review: Voltage or Electric Potential
Wk per Coulomb to bring a charged particle to
point in field from infinity.
Potential / Voltage difference
Wk per Coulomb to move charge between two
points at different potentials.
Charges in field have PEelc.
High PE charge near point with same charge.
Low PE charge near point with opposite charge.
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Charges set loose in E
fields will accelerate!
By conservation E
PE
lost
= KE
gained.
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Some typical voltages
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Plates with battery
d = 1 cm
-
+
AC Delco
12 volts
DVAB  Ed
E  DVAB / d
A
B
E  12 / 0.01
E  12000 N/C
d
Batteries are meant to maintain the constant
potential difference & electric field.
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Mech Uni Potential Dif
67
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If a p+ is released from rest near
the positive plate in a uniform field,
it will accelerate toward the
negative plate changing PE to KE
so:
PE elc = KE
qV = ½ mv2.
A proton is released from rest in a pd =
1200 V. What will be its maximum speed?
PE elc = KE
qV = ½ mv2.
(1.6 x 10-19 C)(1200 J/C) = ½ (1.67x10-27kg)(v)2.
v = 4.8 x 10 5 m/s