Do NOW: What do you think electrostatics means?
Electrostatics
•Electro- Electricity/charged particles
•Static-stationary/ not moving
•Electrostatics- the study of stationary charges
Three Particles
• Proton (p)
– Charge= +1.6x10-19C=+1e
BOTH
– Mass=mp=1.66x10-27kg
protons and
electrons are
• Electron (e)
elementary
– Charge= -1.6x10-19C=-1e
charges!
-31
– mass=me=9.11x10 kg
• Neutron (n)
– Charge= neutral =0e
– Mass=mn=1.67x10-27kg
The Coulomb
•
The basic unit of charge is the Coulomb
–
•
•
You MUST be in Coulombs when you plug into a
formula
1C=6.25x1018e (elementary charges)
1e=1.6x10-19C
- Elementary Charges CAN
NOT be cut in half.
- This means objects have
charges that are multiples
of 1.6x10-19C.
Practice
1) How many elementary charges are in 1.5C of
charge?  6.25 1018 e 
X

 
1C

 1.5C
 9.38  1018 e
2) How many coulombs of charge are 1.88x1019
protons?
1C
X

 



18 
19 
 6.25 10 e   1.875 10 e 
 3C
Conductors vs. Insulators
Conductors
• Allow electrons to flow
freely through the
material
• Usually materials with
loosely bound valence
electrons
– Metals
– Ionic solutions
Insulators
• Do not allow electrons
to flow freely through
the material
• Usually materials with
tightly bound valence
electrons
– Rubber
– Wood
Separation of charge
• Neutral objects have an equal number of protons and
electrons.
• If a charged object is brought near a neutral object, the
charged object can cause the charges in the neutral object to
separate.
• ONLY ELECTRONS CAN MOVE!!!!
Example: Pith Ball
-+ + +- - +
-+ + -+ -+
+
As
rod
As+the
the negative
positive rod
++
comes
comes
close, itit repels
attracts
+ + close,
- charges.
+negative
the
the
negative
- - -charges.
Because
Because opposites
opposites
- - side
attract,
attract,the
the positive
negative
side
of
of the
thepith
pith ball
ball will
will attract
attract
to
rod.
to the
thenegative
positive rod.
•NEUTRAL OBJECTS
ATTRACT BOTH
POSITIVE AND
NEGATIVE THINGS!!
Coulomb’s Law
the force between two charges
kq1q2
Fe  2
r
• Fe – Electric Force (N)
• k – Electrostatic constant (8.99x109Nm2/C2)
• q1 - the charge of the first object (C)
• q2 - the charge of the second object (C)
• r - the distance between the centers of the
two charged objects(m)
Examples
kq1q2
Fe  2
r
1. What is the electrostatic force between two protons
separated by a distance of 2m?
2. Whatkq
is the
electrostatic force between a 3C sphere and a
q
1 2separated by 50m?
F-5Csphere
e
r2
kq
1q22electrostatic
9
19
3. What(happens
to
the
between
two
Nm
Fe10
 2 2 )(1.6 10 19force
8
.
99

C
)(
1
.
6

10
C
)
if the
rC 1
Fobjects

F9
2
a. e distance is tripled?
Nm
9
 3C)(5C)
(8.99 10 2Cm)(
2
e
4 Fe

b. distance isFehalved?
2
 29
50m 
 5.8 of10one object
N is quadrupled?
c.Fe charge
7
d. charge ofFboth
objects
are

5
.
39

10
N halved?
e
2
4 Fe
1
Fe
4
Electrostatic Force Graphically
kq1q2
Fe  2
r
Fe
Fe
q
r
kq1q2
Fe  2
r
Example Problems
Example Problems
1.
1.
2.
What is the magnitude of the electrostatic force
between a charge of +3.0x10-5 C and a charge of
+ 3.0x10-6 C separated by 0.3m?
Do they repel or attract?
2.
A point charge of -1.0x10-9C and a charge of +
3.0x10-9C separated by 5.0x10-2 m what is the
magnitude of the electrostatic force between
them?
Do they repel or attract?
How to charge an object?
Glass and Silk
• When a glass rod is rubbed
with silk, the silk strips
electrons from the glass.
– Silk gains electrons, so it
becomes negative
– Glass looses electrons
so it becomes positive
Rubber and Fur
• When a rubber rod is
rubbed with fur, the fur is
striped of electrons by
the rubber.
– fur looses electrons, so
it becomes positive
– rubber gains electrons
so it becomes negative
Electroscopes
(devices that show charge. They can not measure charge)
• Leaf electroscope
• Braun Electroscope
Charging by Conduction
(with a negative Rod)
Charging by conduction means there is contact
•
•
•
•
•
Charge the rod negatively by rubbing it with fur
–
What does that mean about its electrons?
Touch the electroscope with the negative rod
The excess charges from the rod will seek
equilibrium and move into the electroscope
Remove the rod
The electroscope ends up with the same
(negative) charge as the rod
–
How do I know the electroscope is charged?
–
What if I bring a negative rod back near it?
–
What if I bring a positive rod near it?
Charging by Induction
1.
2.
3.
4.
5.
6.
(with a negative rod)
No contact, the charge is induced
Charge the rod negatively
Bring the rod close to, but do not
touch, the electroscope (what do the
electrons do)
Ground the electroscope by touching
the top or the leaves (what do the
electrons do?)
The excess charges seek equilibrium
through the ground
Remove the ground first, then remove
the charged rod (why?)
The electroscope is charged opposite
the charge of the rod
+
+
++
GOAL:
How many electrons are on your balloon?
• assumptions
– Each balloon has the same charge
– The string has no mass
• Hints:
– Free body diagram of each balloon
– Use the protractor to find an angle
Calculations
• Show all work neatly on a separate sheet of
paper. Include
– Diagrams
– Formulas
– Units
– Verbal explanations
Electrostatic Force on Balloons
• Setup-uncharged
• setup- charged
Materials:
Make sure strings
are vertical
-ring stand
-2 balloons
-Tape
-String
-Protractor
-Ruler
-scale
Plan
of
Action
Goal: number of electrons (elementary charges)
- Need charge in Coulombs first (1e=1.6x10-19 C)
- How do we get the charge in Coulombs?
- Can find them using coulomb’s law
kq1q2
Fe  2
r 2
- They were assumed to be the same so… F  kq
e
2
- Need Fe and r
r
- Fe can be found using a free body diagram and the idea of equilibrium
- r can be measured
- Fe = Fx which is the horizontal component
-
We need the vertical component and the angle to find it
The angle can b
e measured
The vertical component is equal to the force of gravity
- The force of gravity can be calculated (Fg =mg)
- g=9.81m/s2
Balloon Free Body Diagram
Ft
Fe
Fg
Coulombs spheres
+5C +4C +4C +3C
Examples: What is the charge on each conducting sphere after they are brought
together and then separated. ASSUME THEY ARE EQUAL IN SIZE
1)
2)
3)
-6C
-2C
-7C
+3C
+3C
-2C
-4C
-4C
-2C
-7C
-2C
-2C
-2C
-2C
Conducting Spheres of unequal Size
-Q
-Q
• Both conducting spheres have a charge of –Q
• They are then connected with a conducting
wire. What happens?
– Charges will flow until the spheres have
• Equal electric potential
• Equal charge density
– This means electrons will flow from the smaller
sphere to the larger one.
Electric Fields
Electric Fields
(vectors)
• Use a Tiny Imaginary Proton (TIP) to see which
way it would move near the object in
question. The direction the TIP would move is
the direction of the electric field.
P+
e-
Examples-Draw the Electric Fields
1.
2.
P+
3.
P+
++++++++++
--------------
e-
4.
P+
--------------
++++++++++
Electric Field Strength
E=Fe/q
• What is the field strength if a 3.0C charge feels
a force of 12N?
• A -3.1µC charge is placed is an electric field of
2.3 N/C directed to the right.
– What is the magnitude of the force acting on the
charge?
– What is the direction of that force?
Charge on and Electric Field for a
Conducting Sphere
Electric Potential Difference
V - Electric Potential difference (Voltage): volt = J/C= eV/e
W – Electrical Energy: Joule or eV
q – charge: Coulomb or e
1. If 5 Joules of work is done on 0.5 coulombs of
charge, what is the electric potential difference?
2. A 4e charge is moved through a potential
difference of 40 volts.
a. How much energy does it gain?
b. Is that energy in eV or Joules?
Electric Potential Difference due to
a Point Charge
Ue - Electric Potential Energy (J)
k – Electrostatic Constant (9x109Nm2/C2)
q – charge: Coulomb
r - distance between the charges (m)
• How much electric potential energy is stored
between two protons that are separated by a
distance of 2.5nm?
Millikan Oil Drop Experiment
• Figured out the charge/mass ratio of an
electron. This proved that charge is quantized.
– Charge comes in multiples of 1.6x10-19 C
+++ +++ +++ +++
Fe=Fg
qE=mg
qV/d=mg
q/m=dg/V
Fe=qE
d
m
Fg=mg
----------------
V=Ed
E=V/d
Current and Resistance
An introduction to Ohm’s Law
Recall: Potential Difference
Potential Difference
• Provided by the battery
– AKA voltage measured in volts
• It is a “push” for the electrons in the wire that
causes them to flow.
• The battery DOES NOT supply the electrons,
they are already in the wire.
Current:
The rate at which electrons flow through a conductor
Conservation of charge – Junction Rule
Total current into a junction must equal
the total current out of that junction.
3 Amps
Magnitude and direction of
current flow?
6 Amps
4 Amps
In
out
6 amps
4 amps
3 amps
Total in: 9 amps
Total out: 4 amps
The unknown
current is 5 amps
out of the junction
(to the right)
Simple Circuit
• Every Circuit must have
– A source of potential difference (battery)
– A conducting material (wires)
– An electrical load (resistor/lightbulb)
Circuit Symbols
• Draw a circuit with a battery, a lamp and a switch
using the symbols above
Resistance
• A resistor is any electrical device that you plug in.
• It impedes the flow of electrons
• The resistance of an object is DIRECTLY related to
temperature.
– The hotter the wires, the more resistance they have.
• The resistance of a wire depends on
– Its length
– The material its made of (resistivity)
– The cross-sectional area of the wire
Calculating
Resistance
Resistance is measured
in Ohms (Ω)
Ohm’s Law
Slope = Resistance
V
Slope = 1/Resistance
I
I
V
R
I
Resistance and
current are
INVERSES of each
other. As one
increases the other
decreases!
Electrical Power
When resistance is constant
- “ohmic”
- “Follows ohm’s law”
- “constant temperature”
P
I
When Potential
Difference is
constant
P
I
Electrical Power
When resistance is constant
- “ohmic”
- “Follows ohm’s law”
- “constant temperature”
P
V
When current is
constant
P
V
Electrical Power
When Voltage is
constant
When current is
constant
P
P
R
R
Electrical Energy (AKA work)
Measured in Joules
Circuits Measurements
Ammeter
• Measures current (in Amps)
• Wired IN SERIES with the
resistor
• Has a very LOW resistance
Voltmeter
• Measures potential
difference (in Volts)
• Wired IN PARALLEL with the
resistor
• Has a very HIGH resistance
Types of Circuits
Series
Parallel
One complete loop
Multiple paths
Circuits
DO NOW
• With a partner, collect
– One battery
– One wire
– One lightbulb
• Working with your partner at your desk and
your knowledge of circuits, make the lightbulb
light up.
Hint #1
Hint #2
One side of the
wire in the
lightbulb touches
the metal part of
the light bulb. The
other side of the
wire is connected
to the bottom.
Solution
AP Circuit Conventions
The AP assumes a
positive current
flow. As the
electrons flow one
way, their positive
‘holes’ flow the
other way
If you focus on the electrons
(red) they look like they are
flowing counter clockwise. The
positive ‘holes’ (white) look like
they are flowing clockwise
Series Circuits
You will also have to use Ohm’s Law, but
you can NOT cross contaminate!
Totals
Only R1
Only R2
Rules
Only R3
If you add a 4th resistor, the total
resistance will INCREASE which
means the total current will
DECREASE. Total voltage is
constant
Series Circuits
Rules
1. If the current through R1 is 4 amps,
what is the current through
a. R2?
b. R3?
c. The total current coming out of
the battery?
2. If R1 is 3Ω, R2 is 7Ω, and R3 is 4 Ω,
what is the equivalent resistance of
the circuit?
3. What is the potential drop across
a. R1
b. R2
c. R3
4. What is the total potential difference
supplied by the battery?
Parallel Circuits
You will also have to use Ohm’s
Law, but you can NOT cross
contaminate!
Totals
Rules
Only R1
Only R2
Only R3
Req will ALLWAYS be less than
the smallest resistor in the
circuit!
If you add a 4th resistor, the total
resistance will DECREASE
which means the total current
will INCREASE. Total voltage is
constant
Parallel Circuits
Rules
1. If the potential difference over R1 is
12 volts, what is the potential
difference
a. R2?
b. R3?
c. The total potential difference
supplied by the battery?
2. If R1 is 3Ω, R2 is 7Ω, and R3 is 4 Ω,
what is the equivalent resistance of
the circuit?
3. What is the current through
a. R1
b. R2
c. R3
4. What is the total current coming
from the battery?
2. What is the equivalent resistance for the circuit?
3. What is the total current coming out of the battery?
Equivalent Resistance of a Combined
Circuit
These two resisters
are in series, so
their equivalent
resistance is the
sum of them.
10Ω
Finding the Equivalent Resistance
of a Combined Circuit
These two resisters
are in parallel, so
their equivalent
resistance is the
sum of them.
5Ω
Finding the Equivalent Resistance
of a Combined Circuit
These two resisters
are in series, so
their equivalent
resistance is the
sum of them.
10Ω
Finding the Equivalent Resistance
of a Combined Circuit
This is now a simple
circuit that has a
current of 4amps
10Ω
Finding the Equivalent Resistance
of a Combined Circuit
These two resisters are in
series, so they must have
the same current of
4amps, so the voltage
drop across each one is
20V
Finding the Equivalent Resistance
of a Combined Circuit
These two resisters
are in parallel, so
they must have a
voltage drop of 20V
which means they
both have a current
of 2amps
Finding the Equivalent Resistance
of a Combined Circuit
These two resisters
are in series, and
they each have a
current of 2 amps
which means the 8Ω
has a voltage drop
of 16V and the 2Ω
has a voltage drop
of 4V
Find the Equivalent Resistance
Find the Equivalent Resistance
Find the Equivalent Resistance
Find the current across and the
potential drop across each resistor.
Find the current across and the
potential drop across each resistor.
Kirchhoff’s Circuit Rules
Junction Rule
• The total current flowing into a junction must be equal to
the total current
Loop Rule
• The total potential gained or lost in a circuit loop must be
equal to zero.
– Batteries:
• Across to the positive side is a potential gain
• Across to the negative side is a potential loss
– Resistors
• Crossing the resistor in the direction of the current flow is a potential loss
• Crossing the resistor against the current flow is a potential gain.
Simple Loop Rule Example
1. Label all batteries: V1, V2…
2. Label all Resistors: R1, R2, R3…
3. Label all the currents: I1
R1
I1
R2
V1
V2
R3
R1
I1
R2
V1
V2
R3
• Pick a starting point.
• Choose a Loop direction
• CREATE a formula for the potential gains and losses
R1=15Ω
I1=?
V1=20V
R2=20 Ω
V2=35V
R3=10 Ω
• Plug in the values given and solve for the unknown.
• If the current comes out positive, you chose the
correct direction. If it comes out negative, if flows
in the opposite direction chosen.
R1=15Ω
I1=?
V1=20V
R2=20 Ω
V2=35V
R3=10 Ω
• Plug in the values given and solve for the unknown.
The negative sign
means the current
flows in the other
direction
Create two Loop Rule Equations
Find the magnitude and direction of
the current coming from each battery.
If
V1
V2
R1
R2
R3
= 50V
= 75V
= 45Ω
= 30Ω
= 20Ω
Find the magnitude and direction of
the current coming from each battery.
I1
I2
I3
V1
V2
R1
R2
R3
= 50V
= 75V
= 45Ω
= 30Ω
= 20Ω
Find the magnitude and direction of
the current coming from each battery.
V1
V2
R1
R2
R3
= 50V
= 75V
= 45Ω
= 30Ω
= 20Ω
distribute
Combine like
terms
Mult by 2
Mult by 5
Add the two expressions
Sub back in
All of the currents came out
positive so we picked the correct
directions!!
Kirchhoff's Loop Rule
Find the current flowing through
each resistor