Day 1
AP Physics - Static Electricity - phenomena due to
attractions or repulsions of electric charges but not
dependent upon their motion.
1740’s Benjamin Franklin
Neutral charge -- body had right amount of electric
fluid
Positive charge -- excess of electric fluid
Negative charge -- deficiency of electric fluid
Subatomic particles:
proton, neutron, and electron.
Fundamental law of static electricity:
Like charges repel; opposite charges attract.
Principle of Conservation of Charge - charge is not
created or destroyed, merely transferred from one system
to another.
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Negative charge (electrons) goes from one object to other
Lightning is a very large, powerful example of this.
Spark - ionized air creates path
Cool Lightning pics
4 millions joules E
Temperature -- Greater than surface of sun -- 6000 K.
100 lightning bolts strike the earth every second!
Lightning video
2
Conductors and Insulators:
Most conductors are metals (allow electrons to pass
through them) - ex.
Insulators are non metals (do not allow electrons to
pass through) – ex.
Electrolytes are solutions which have ions in them
that can carry charge.
Metals are good conductors because of the metallic bond.
"free-floating" delocalized electrons
Insulators have tightly bound electrons which are not free
to move about and transfer charge
Charging by Conduction-direct contact
Charging by Induction-without direct contact
A
B
----
+
+
+
+
+
+
A
B
--
----
+
+
+
+
+
+
A
B
--
+
+
+
+
+
+
A
B
--
3
1
2
3
4
The Electroscope: object that can detect a charge.
Foil on thread
----
+
+
+
+
+
+
-
-
---
+
+
+
+
+
+
Double pith ball
Foil leaves
Electrically polarized
The bottom of a negatively charged
cloud induces a positive charge at
the surface of the ground below.
4
why a charged rod attract bits of neutral paper.
Turn the tv set on. Then turn it off
and place a piece of paper on the
screen. Why does the paper stick to
the tv screen?
5
AP Physics - Coulomb's Law - 2
Coulomb
[C] ≡ unit for charge
symbol for charge -- q
1 C = charge of 6.25 x 10
18
electrons.
charge of one electron/proton =-/+1.60 x 10-19 C.
• How many electrons would have a charge of 211 µC ?
 6.25 x 1018 e− 
12 −
211 x 10 C 
 = 1320 x 10 e
1C


−6
=
1.32 x 1015 e−
6
Coulomb's Law:
1 q1q2
F=
4π ∈0 r 2
1
4π ∈0 ≡ Coulomb’s Constant
Use
1
4π ∈0
2
Nm
= 8.99 x 109 2
C
Coulomb force constant
1
4π ∈0
, (k) is a physical quantity
that describes how an electric field affects and is affected
by an insulating (dielectric) medium.
•Permittivity - permittivity relates to a material's ability to
transmit (or "permit") and ε0 = 8.85… × 10−12 F/m is the
vacuum permittivity.
• When using Coulomb's Law, the signs of the charges
should not be included. They are only used to determine
whether the charges are attracted or repelled.
• Two point charges are 5.0 m apart. If the charges are - 0.020 C and
0.030 C, what is the force between them and is it attractive or
repulsive?
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1 q1q2
F=
4π ∈0 r 2
Nm 2  ( 0.020 C )( 0.030 C ) 
F = 8.99 x 10


2

C 2 
( 5.0 m )

F = 0.000216 x 109 N
9
= 2.2 x 105 N
• A force of 1.6 x 10-3 N exists between 2 charges, 1.3
µC and 3.5µC. How far apart are they?
F=
1 q1q2
4π ∈0 r 2
r=
ke q1q2
F
Nm 2
8.99 x 10
1.3 x 10−6 C )( 3.5 x 10−6 6 C )
2 (
C
r=
1.6 x 10−3 N
9
r = 25.57 x 100 m 2
= 5.1 m
Electric Force and Gravity: Compare and contrast
F=
1 q1q2
4π ∈0 r 2
and
F=
G m1m2
r2
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Superposition Principle:
Resultant force on a charge is vector sum of forces
exerted on it by other charges.
What is the net force on charge q3?
q2
-
F13
F23
4.00 m
+
37.0
0
q3
3.00 m
5.00 m
+
q1
q1 = 6.00 nC
q2 = 2.00 nC
q3 = 5.00 nC
Net force on q3 is F23 + F13
Find magnitude of the 2 forces:
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F13 =
1 q1q2
4π ∈0 r 2
−9
2  5.00 x 10 −9 C
6.00
x
10
C )
(
)(
Nm
9


= 8.99 x 10
2
2

C 
5.00
m
(
)


F13 = 10.8 x 10 −9 N = 1.08 x 10 −8 N
F23 =
1 q1q2
4π ∈0 r 2
−9
2  2.00 x 10 −9 C
5.00
x
10
C )
(
)(
Nm
9


= 8.99 x 10
2
2

C 
( 4.00 m )


F23 = 5.62 x 10 −9 N
Add vectors in x and y
direction:
0
F13 sin 37.0
F13
37.0
F23
F13 cos 37.0
0
Fx = F13 cos 37.0o − F23
Fx = (1.08 x 10−8 N ) cos 37.0o − 5.62 x 10−9 N
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Fx = (10.8 x 10−9 N ) cos 37.0o − 5.62 x 10−9 N
Fx = 8.63 x 10−9 N − 5.62 x 10−9 N
Fx = 3.01 x 10−9 N
Fy = F13 sin 37.0o
Fy = (1.08 x 10−8 N ) sin 37.0o
Fy = 6.50 x 10−9 N
Resultant force:
2
F = Fy + Fx
F=
2
( 6.50 x 10
−9
2
N ) + ( 3.01 x 10 N )
−9
2
F = 51.31 x 10−18 N 2
F = 7.16 x 10−9 N
11
Angle of the thing:
 FY 
θ = tan  
 FX 
−1
−9

6.50
x
10
N 
−1
θ = tan 

−9
x
N
3.01
10


θ = 65.2o with the x axis
Due next class period chpt 16 pg 464
questions 1, 4,6,10,11 problems 1,4
12
Particles of charge Q1 = +68 µC, Q2 = +45 µC, and Q3 =
-80 µC are placed in a line. The center one is 0.35 m
from each of the others.
Calculate the net force on each charge due to the
other two. (State both magnitude and direction)
Three positive particles of equal charge, +12.0 µC, are
located at the corners of an equilateral triangle of
side d = 12.5 cm
Calculate the magnitude and direction of the net
force on the top of the triangle
Net force on Q1
Magnitude
144 N
Direction
90° (counterclockwise from the +x axis is positive)
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Net force on Q2
Magnitude
144 N
Direction
210° (counterclockwise from the +x axis is positive)
Net force on Q3
Magnitude
144 N
Direction
330° (counterclockwise from the +x axis is positive)
Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and
bottom left corners are +3.0 x 10-6 C. The charges in the other two corners are -3.0 x 10-6 C. What is the
net force e
n the charge in the top right corner by the other three charges?
14
To solve any problem likeIf you have the arrows giving you the
direction on your diagram, you can just drop any signs that
come out of the equation for Coulomb's law.
Consider the forces exerted on the charge in the top right by the
other three:
You have to be very careful to add these forces as vectors to
get the net force. In this problem we can take advantage of the symmetry, and combine
the forces from charges 2 and 4 into a force along the diagonal (opposite to the force from charge 3) of
magnitude 183.1 N. When this is combined with the 64.7 N force in the opposite direction, the result is a
net force of 118 N pointing along the diagonal of the square.
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The symmetry here makes things a little easier. If it wasn't so symmetric, all you'd have to do is split the
vectors up in to x and y components, add them to find the x and y components of the net force, and then
calculate the magnitude and direction of the net force from the components. Example 16-4 in the textbook
shows this process.
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