Transcript sound - Dr. Robert MacKay

Electrostatics
+
+
-
-
-
+
Conservation of Charge
Charge can neither be created nor destroyed
Positive ions ---- fewer electrons than protons
Negative ions ---- fewer protons than electrons
Electric Charge is measured in Coulombs
6.3x1018 electrons make -1.0 C of charge
6.3x1018 protons make +1.0 C of charge
Conservation of Charge
Charge can neither be created nor destroyed
rub electrons from a bar with fur
bar becomes positively charge by
the exact amount that fur becomes
negatively charged.
bar becomes a tiny bit less massive
Coulomb’s Law
q1q 2
Fk 2
d
The interaction force between two charges is:
directly proportional to the size of each charge (q1 and q2)
and
inversely proportional to the square of their separation
distance (d)
k= 9.0 x 109 N/m2/C2
Fe>>>Fg
Coulomb’s Law
q1q 2
Fk 2
d
The interaction force between two charges is:
directly proportional to the size of each charge (q1 and q2)
Double either q1 or q2 then F doubles.
Double both then F quadruples
Coulomb’s Law
q1q 2
Fk 2
d
The interaction force between two charges is:
inversely proportional to the square of their separation
distance (d)
Double the separation distance then F is reduced to (1/4)
Halve the separation distance then F is quadrupled (4x)
Coulomb’s Law
q1q 2
Fk 2
d
The interaction force between two charges is:
inversely proportional to the square of their separation
distance (d)
triple the separation distance then F is reduced to (1/9)
(1/3) the speration distance then F is increased 9 fold (9x)
Coulomb’s Law
q1q 2
Fk 2
d
The interaction force between two charges is:
inversely proportional to the square of their separation
distance (d)
If separation distance is increased by 10 then F
(Reduces/increases) by _________________
Coulomb’s Law
q1q 2
Fk 2
d
+
+
As the charges above are released the force on each
(increases or decreases)
Coulomb’s Law
q1q 2
Fk 2
d
+
+
As the charges above are released the speed of each
(increases or decreases)
Coulomb’s Law
q1q 2
Fk 2
d
+
+
As the charges above are released the acceleration
of each
(increases or decreases)
Coulomb’s Law
q1q 2
Fk 2
d
?
+
As the charges above are released the speed of each
increases. Thus the green object has a __________
charge. Positive, negative, can’t tell
Coulomb’s Law
q1q 2
Fk 2
d
?
+
As the charges above are released the force on each
increases. Thus the green object has a __________
charge. Positive, negative, can’t tell
Conductors
Conductors have very loosely bound electrons.
That is electrons that are not really attached to one
particular nucleus. These electrons are sometimes
called free electrons because they move freely when
exposed to an electric field
Gold
Copper
Silver
Ionic solutions (salt water)
Insulators
Insulators have very tightly bound electrons. That
is electrons that are firmly attached to one
particular nucleus. These electrons are very hard
to set in motion throughout the material
Glass
Dry wood
Plastic
Semiconductors
Semiconductors have moderately bound electrons.
These electrons can be set into motion throughout
the material when a moderately strong electric
field is established in the material.
Carbon
Silicon
Superconductors
Superconductors have no electrical resistance to
charge flow (infinite electrical conductivity)
Very cold silver (-269 °C)
Charging
Friction
Contact
Induction
Charge Polarization
+- - - - --- -- -- -- -- - - -- -
F
+++-
Electric Field
E=F/q
+
+
q
Electric Field
E=F/q
+
q
Electric Field
E=F/q
or
F=q E
F
Uniform Electric
Field between
two charged plates
-
+
-
+
-
+
q
+
-
+
-
+
-
+
Electric Shielding
E=0 inside metals
Electric Shielding
E=0 inside metals
-
+
-
+
-
+
-
+
-
+
-
+
Electric Potential (Volts)
Electric Potential energy (J)
Charge (C)
EP = EPE / q
Volt=Joule/Coulomb
F
-
+
-
+
-
+
q
+
-
+
-
+
Electric Potential Energy = Work Electric Potential Energy = Charge x Volts
1 Joule= Coulomb x Volt
+
Physical science 101 stop here
Electric Potential (Volts)
Electric Potential energy (J)
Charge (C)
F
What is the electric
potential between two
plates when it takes 2.0 J
of work to move a 0.001 C
charge from - to + plate?
-
+
-
+
-
+
q
+
-
+
-
+
-
+
What is the electric potential between
two plates when it takes 2.0 J of work
to move a 0.001 Ccharge from - to +
plate?
Given: W=EPE=2.0 J Charge=0.001 C
Want: EP
Solution: Electric Potential=
Electric Potential energy (J)
Charge (C)
Given: W=EPE=2.0 J Charge=0.001 C
Want: EP
Solution: Electric Potential=
Electric Potential energy (J)
Charge (C)
=2.0 J/0.001 C = 2000 Volts
1 Volt=1J/C
Capacitors & Energy
Storage
-
F
-
+
q
+
+
+
-
+
-
+
-
+