Transcript electricitynotes revised 10

STATIC ELECTRICITY
A. Definition- Study of the forces between
charges at rest.
B. Micro structure of matter (atomic
structure)
1. Atom- 3 Parts
a. Protons-mass 1 amu, charge +1,
location nucleus
b. Neutrons-mass 1 amu, charge 0,
location nucleus
c. Electrons- very small mass, charge
-1, location outside the nucleus
C. Charged Objects
1. Ion-electrically charged particles that
result from gain or loss of electrons
a. Lose electrons- positive ions
b. Gain electrons-negative ions
•
2. Interaction of charged particles
a. Like charges are repelled by an
electrical force that acts on each particle
in opposite directions
+
+
Fe
Fe
b. Unlike charges attract each other by an
electrical force that also acts in opposite
directions.
-
Fe
+
Fe
Neutral charges are attracted to both positive and negative but are NEVER
repelled
D.Electroscope-device used to measure
charge.
KNOB
LEAVES
1. charging an electroscope by conduction
means a charged object is touched to the
top of the electroscope.
When the scope is touched by negatively
charged object the excess electrons move
from the negatively charged object
down the metal rod of the scope and to the
leaves of the eletroscope .
This causes the leaves to have like charge
(negative) and move apart (diverge).
Negatively charged rod
-
Electrons traveling
Down the rod and
Through the electroscope
Making the scope negatively charged
-
e-
ee-
b. When a positively charged rod is
touched to an electroscope the electrons
move from the scope to the rod leaving
the knob and leaves of the scope
positive and the leaves repel each other
+
Positively charged rod +
+
+
Electrons are attracted to the positively
+
charged rod and leave the scope and enter
the rod. Leaving both the knob and leaves
of the electroscope positively charged.
e-
e-
e-
2. To charge an electroscope by induction
simply bring a charged rod near the top
of the scope- do not touch it.
a. When a negatively charged object is
brought near a neutral scope the
electrons in the top are repelled and
move down into the leaves.
This results in the leaves being negatively
charged
and the top being positively charged.
Since the leaves are both negative they
repel each other and move apart
++++
e-
e- ee- e-
-
b. When a positively charged rod is brought
near a neutral electroscope electrons
from the leaves are attracted to the top,
resulting in the top having a
negative charge
and the leaves having a positive charge.
Since the leaves are both positive they
repel each other and move apart
+
+
+
+
+
-----------
e-
e-
e-
+
+
+
+
+
+
E. Grounding
1. Occurs when the earth or other large
conductor accepts or donates a large
number of electrons without significantly
affecting its own electrical state.
2. Grounding results in an object having a
net charge of zero also known as
discharging.
F. Conservation of charge
– Law In an isolated system (charge carriers
cannot enter or leave) net charge is
constant.
2. Example: 2 identical conducting spheres
A and B are separated by some distance.
Sphere A has a net charge of -16
Sphere B has a net charge of +4
total charge -12
If A and B are brought into contact electrons
will move from A to B until both have a
charge of -6,
but the net charge on the system remains at
-12
G. Measurement of Charge
1. The smallest isolated negative charge
in nature is that of an electron and for
positive charge the proton.
2. An elementary charge is symbolized by
e- (negative charge) or e+(positive
charge like a proton)
3. Unit of charge is the Coulomb
a. 1coulomb= 6.25X1018 elementary
charges
b. 1e- has a charge of -1.6X10-19C
(elementary charge)
c. 1 proton has a charge of
+1.6X10 -19C
See reference tables
d. Net charge depends on excess or
deficiency of electrons therefore a net
charge is always a whole number multiple
of the charge of one electron and can be
expressed that way.
•
2e = 2(1.6 x10-19C)= 3.2 X 10-19C
Coulomb’s Law
1. Measures the force two charged
objects exert on each other when
separated by a distance.
2.
Fe=kq1q2
r2
Where k = constant = 8.99X109 N-m2/C2
F = force in Newtons
q1 & q2= charge in coulombs
r= separation distance in meters
b., The electrostatic force of each object is directed along the line joining the
two objects and is equal in magnitude but opposite in direction for the two
objects
Examples:
1. What is the force between 2 relatively
small charged objects that carry charges
of 0.20C and -0.30C, if the distance
between them is 1m?
Fe=kq1q2 Fe= 8.99X109N-m2/C2(.2C)(-.3C)
r2
(1m)2
Fe= -539,400,000N
2. A positive charge of 6.0 X10-6 C is 0.30m
from a second positive charge of
3.0X10-6C. Calculate the force between
the charges.
Fe=kq1q2 Fe= 8.99X109N-m2/C2(6X10 -6 C)(3X10-6C)
r2
(0.30m)2
Fe=+1.8N
Electric Fields
A. Produced by charged objects
B. Field lines are normal perpendicular to
the surface of the charge
C. Point away from positve charges
Point toward negative charges
D. Field intensity inside the charge is
zero___
+
-
E.Calculating field strength
E= Fe
q
E= electric field intensity, N/C
Fe= electrostatic force, N
q.= charge, C
Field direction is the same as the force for a
positive charge and 180 degrees opposite
for a negative charge.
Field strength indicated by closeness of
lines.
Example:
A positive test charge of 4 X 10-5 C is placed
in an electric field . The force on it is
0.60N acting at 10 degrees. What is the
magnitude and direction of the electric
field?
• E=Fe
E= 0.6N
E=15,000N/C
q
4x10-5C
G. Parallel Plates
1. The field that exists between two
oppositely charged parallel plates is
uniform as long as their separation
distance is small
H. Electric Potential
1. The work needed to bring a unit
positive charge toward a positive object.
+
WORK
Test charge
++++
+
++++
Positively charged object
2. Conversely the work done to pull to
oppositely charged objects apart
WORK
--------
+++
+++
+++
WORK
Potential Difference
A. The change in energy of a positive test
charge as it is moved closer to a positively
charged object
B. The volt
1. Unit of potential difference
2. 1volt=1joule/1coulomb
3. Definition: In an electric field when 1
Joule of work is required to move a one
coulomb of charge between two points
•
4. V= W
q
V=potential difference, volts, V
W= work, joules, J
.q =charge, Coulombs, C
Example: What is the potential difference
between two points in an electric field if
10J of work is required to move a 2C
charge between them?
V= W
q
V= 10J
2C
V=5J/C = 5volts
What is the work required to move 5
coulombs of charge through a potential
difference of 10V?
V= W
q
10V= W
5C
W=50J
6. Electronvolt (eV)
a. Definition-Amount of energy
required to move one electron through
a potential difference of 1volt. It is a
unit of work/energy just like the joule.
1eV = 1.60 X10-19 joule
III. Current electricity
A. Definition- Flow of electrons from
high potential (V) to low potential (V)
B. Required conditions
a. A potential difference supplied by a
battery or generator.
b. A complete or closed conducting
path connecting regions of high
potential to low potential (like a wire)
2. Unita. 1 amp of current is measured when
one coulomb of charge flows past a point
each second
3. Calculated
I = ∆q
t
I = current in ampere,A
q= charge in coulombs
t= time in seconds
Example:
What is the current if 15C of charge pass a
point in 3 seconds?
I=q
t
I=15C/3s
I= 5C/s or 5A
What is the amount of charge passing
through a lamp in 10 seconds if it draws
0.5Amps of current?
I=q
t
0.50A= q
10s
q=5C
4. An Ammeter is used to measure current
D. Resistance, R
1. Determines the amount of current that
can be produced by a conductor.
2. Causes there to be a potential
difference between the ends of a
conductor when a current is passed
through it
Resistance depends on the length, crosssectional area, and resistivity of the
conductor and temperature
•
4. Resistivity- Physical property of
material to resist or oppose the
movement of charge through a material
5. Calculating Resistance
R=ρL
A
R = resistance, in ohms (Ω)=V/A
ρ = (greek letter rho) resistivity, ohm-meter
found on ref. tables page4
L= length, meters
A= cross-sectional area, m2
Example: Calculate the resistance at 20oC
of an aluminum wire that is 0.200m long
and has a cross-sectional area of
1X10-3 square meter.
R=ρL
A
R= 2.82x10-8 Ω-m(0.200m)
1x10-3m2
R= 5.64X10-6 Ω
6. Temperature
a. Resistance is directly related to
temperature
b. Low temp, low resistance and vice
versa
E. Ohm’s Law.
R= V
I
R= resistance, ohms
V= potential difference, volts
I=Current, amps
Graph
PD
OI
T F
EF
NE
T R
I E
A N
L C
E
V
• Symbols
CURRENT, I (AMPS)
AMMETER(CURRENT)
VOLTMETER(POTENTIAL DIFF)
A
V
RESISTANCE, Ω
Example:
If the current across a wire is 2A and the
potential difference across the wire is 10V,
R= V
I
R=10V
2A
R=5 Ω
A potential difference of 12V is applied
across a circuit which has a 4 ohm
resistance. What is the magnitude of the
current in the circuit?
R= V
I
4Ω =12V
I
I= 3 amps, 3A, 3C/s
F. Power
P=VI= I2R = V2/R
P =power, watts (J/s)
V = potential difference, volts
I = current, amp
R= resistance, Ω
Example: A lamp operates at 10volts and
draws a current of 0.5amps for 60
seconds. What power is developed in the
lamp?
P=VI= I2R = V2/R
P=VI
P=10V(0.5A)
P= 5watts
While operating at 120V, an electric toaster
has a resistance of 15 ohms. What is the
power used by the toaster?
P=VI= I2R = V2/R
P= V2/R
P= (120V)2/15 Ω
P= 960 watts
G. Work-Electrical Energy
W= Pt=VIt= I2Rt = V2t/R
P =power, watts
V = potential difference, volts
I = current, amps
R= resistance, t = time, s
W = work , J
Examples:
• A toaster dissipates 1500 watts in 90
seconds. What is the amount of electrical
energy used by the toaster?
W= Pt=VIt= I2Rt = V2t/R
W=Pt
W=1500watts(90s)
W=135,000J
An iron has a current of 10Amps when 120V
of potential difference is applied for 60s.
What is the total energy dissipated during
the 60seconds?
W= Pt=VIt= I2Rt = V2t/R
W=VIt
W=120V(10A)(60s)
W=72,000J
H. Conservation of charge and energy
1. Conservation of charge
Kirchhoff’s first rule: For any point in
a circuit, the total current arriving at a
point must equal the total current
leaving the point.
2. Conservation of energy
Kirchhoff’s second rule: The algebraic
sum of all the voltage drops and
applied voltage (battery) around a
circuit is zero
Example:
2A
3A
?
6A
Current into the juncture: 3A+6A=9A
Current leaving the juncture: 2A+?
Since current in must equal current out ?=7A
Series Circuit
1. Only one path for the current to follow
Sketch
2. Current
a. Same through each resistor
b. b. IT = I1= I2 = I3 …..
c. Read by an ammeter
d. Connected in series
SymbolsVOLTAGE (BATTERY)
Resistance
a. The total resistance is equal to the
sum of the components (resistors)
b. RT = R1 + R2 + R3 ….
– Symbol See previous notes
Voltage-(potential drops)VT= V1 + V2+V3 +…
• REMEMBER: R=V
I
A 10 ohm and a 20 ohm resister are
connected in series with a 60V battery.
a. Make a sketch of the circuit
10Ω
60V
20Ω
b. What is the total resistance of the circuit?
RT = R1 + R2 + R3 ….
RT = 10Ω +20Ω
RT= 30Ω
c. What is the total current in the circuit?
R=V/I
30Ω = 60V
I
I= 2 amps
Since it is a series circuit the current is the
same everywhere
d,. What is the voltage drop across the 10
ohm resistor?
R=V/I
10Ω = V
2A
V= 20V
e. What is the voltage drop across the 20
ohm resistor?
R=V/I
20Ω = V
2A
f. What is the current through the 10 ohm
resistor?
2A-CURRENT IS THE SAME
EVERYWHERE IN A SERIES
g. What is the current through the 20 ohm
resistor?
2A-CURRENT IS THE SAME
EVERYWHERE IN A SERIES
Parallel Circuit
1. Circuit in which the current can flow
through various paths
Current
a. The current is divided among the
different branches.
b. The total current is equal to the source
(battery)
c. IT = I1+ I2 + I3 …..
Potential Drops (voltage)
a. The potential difference across each
branch (resistor) is equal to the total
(battery voltage)
b. VT = V1= V2 = V3
4. Resistance
1 = 1 + 1 + 1
RT R 1 R 2 R 3
• REMEMBER: R=V/I APPLIES TO
BOTH SERIES AND PARALLEL
CIRCUITS.
• Using the circuit diagram below, answer
the following questions:GO TO
OVERHEAD
• What is the total resistance of the circuit?
• What is the total current in the circuit?
• What is the reading on ammeter A1?
• What is the reading on ammeter A2?
• What is the reading on ammeter A3?
• What is the reading on voltmeter V1?
• What is the reading on voltmeter V2?What
is the total resistance of the circuit?