PHYSICS UNIT 7:
ELECTRICITY
ELECTRIC CHARGE


Static Electricity: electric charge at rest
due to electron transfer (usually by
friction)
+–+
 neutral: electrons equal protons (no net charge)
 positive charge: deficiency (loss) of electrons
 negative charge: excess (gain) of electrons
+ –
–
+–+
+ –
+–+
+ ––
–
ELECTRIC CHARGE

law of conservation of charge: total
charge stays constant (for every +
charge produced, there is a – charge
produced)
+
–
+
–
+
–
+
–
+
–
ELECTRIC CHARGE

law of conservation of charge: total
charge stays constant (for every +
charge produced, there is a – charge
produced)
+
–
+
–
–
+
+
–
+
–
ELECTRIC CHARGE

law of
electrostatics:
like charges
repel, unlike
charges
attract
ELECTRIC CHARGE

Charge transfer
conductor: readily transfers charge (free
electrons)
 insulator: doesn’t transfer charge
(electrons in bonds)

ELECTRIC CHARGE

Charging by
Conduction

direct
contact

same sign
permanent
charge divides
evenly
between
objects


ELECTRIC CHARGE

Charging
by
Induction

no contact

opposite
sign
temporary
unless
grounded

ELECTRIC CHARGE


Conductor that has
induced charge by
neighboring positive
wall. Free electrons
move towards the wall.
Insulator that has
induced charge by
neighboring positive
wall. Molecules are
polarized.
ELECTRIC CHARGE

Charging
by
conduction
& induction
ELECTRIC FORCE


electric force is a fundamental force of
nature: holds atoms together, holds
molecules together, causes friction &
most forces (except gravity)
Amount of charge, q or Q: measured in
coulombs, C

1.00 C = 6.25×1018 electrons

charge on one proton or electron, e = ±1.60×10–19 C
ELECTRIC FORCE

Coulomb’s Law: force between charges
depends on amounts of charge and
distance between them



inverse square law like the force of gravity
Fe = kq1q2/r2
Fe: electric force
q: charge
r: distance between charges k: 8.99×109 Nm2/C2
+Fe: repulsion, –Fe: attraction
ELECTRIC FORCE

electric fields exert force on charged
objects
electric field strength, E: force exerted on
a charge by an electric field
 E = F/q
 unit: N/C (Newtons/Coulomb), or V/m
(Volts/meter)

ELECTRIC FORCE


Electric field: region around a charge
where it exerts electric force on other
charges
field lines: show direction & amount of
force (by how close the lines are) on a
+ test charge
Electric Field Lines



E field lines are
constructed by
determining what a
positive charge
would do if placed
in the field
The denser the
lines the stronger
the field.
Lines always
emanate from
positive charge and
end at negative
charges.
Lines of Equipotential


The grey dotted
lines represent
places where the
Net E-field
magnitude is
equal.
Note how on the
parrallel plate
scenario The Efield is equal for
any point within
the plates.
ELECTRIC FORCE

constant electric fields are used to
accelerate charged particles

field is constant between parallel plates
force F = qE
 change in Kinetic Energy = Work
 Kf-K0 = Fd (Work done by the field)


d: distance traveled in electric field

K = ½mv2
Electric Potential




Imagine that a positive
charge q is released
from contact with Q and
is allowed to accelerate
to an infinite distance
away picking up KE as
it goes. The Change in
KE is the Work required
to bring the test charge
from an infinite distance
back to Q.
Electric Potential, V, is
work per unit charge
that is needed to bring q
toward Q
V = W/q
Units are
Joules/Coulomb
Q q
Examples
What is the electric potential at point A of
a 2 Coulomb charge that requires 10 J
of work to move from B to A?
A
B
V= W/q = 10/2 = 5 J/C = 5V
The Electric Potential Difference is called
VOLTAGE!
Potential Energy

Is PE increasing or decreasing as q, 5 C,
moves towards Q?
A


B
IF Vb=0 and Va=10, Then Voltage is 10 volts
at A. Potential Energy is equal to work
needed to move q to A. So…
W = qV = U = 50 Joules!
Negative Charges?

What happens if negative q, -5C, is
moved from A to B? Assume Vb=0 and
Va=10
A


B
Then as q moves to A PE decreases.
U=qVa=(-5)(10)= -50 J
Conclusion







F = qE
E = kQ/r2
W=qV
U=qV
Positive charges move naturally from high
electric potential to low electric potential
Negative charges move naturally from low
electric potential to high electric potential
All charges move from high PE to low PE
The Electron Volt




The Joule is a huge unit of energy when
dealing with electrons moving across electric
potentials.
How much energy would an electron gain if it
moved across a potential difference of 1 V?
U = qV = (1.6 X 10-19 C) (1V) = 1.6 X 10-19 J
So.. 1.6 X 10-19 J is defined as an electron
volt. This unit can be used as an energy unit
for situations dealing with small charges.
PHYSICS
UNIT 7: ELECTRICITY
ELECTRIC CIRCUITS

Basic Circuit: conductor loop for transferring
energy
load:
energy
user
(bulb,
resistor,
heater,

source:
energy
provider
(battery,
generator)
motor)
ELECTRIC CIRCUITS

Current, I: rate of flow of electric charge
unit: ampere, A
 I = Q/t
1 A = 1 C/s
 conventional current flow: positive to
negative
 (real current is electrons, flowing negative
to positive)

ELECTRIC CIRCUITS

Potential
Difference or
Voltage Drop, V:
work done per
coulomb of charge
between two
points, unit: volt, V


V = W/q
1V=1
J/C
12 V gives 12 J/C
to the electrons
ELECTRIC CIRCUITS

Sources of Potential
Difference

capacitor: stores charge
cell: stores chemicals;
reactions produce V

battery: cells
connected in series

for cells in series, battery
voltage is the sum of cell voltages

anode
cathode
ELECTRIC CIRCUITS

Resistance, R: opposition to charge flow, unit: ohm, W
 resistance limits the flow of current
 resistance turns electric energy into heat (& light)
 resistor: fixed resistance, symbol:

ELECTRIC CIRCUITS
ELECTRIC CIRCUITS

resistance of a length of wire, R = rL/A
r: resistivity (W·cm), L: length (cm), A:
cross-section (cm2)
–10
–10
 r
=1.59×10
r
=1.68×10
silver
copper
–7
 r
=3.00×10
rsilicon=0.00100
carbon
 for solids, as T increases, r increases and
vice versa

ANALYZING CIRCUITS

Ohm’s law: current is proportional to
voltage and inversely proportional to
resistance: V = IR
V: voltage, V I: current, A R:
resistance, W
 applies to circuit as a whole: V = I R
T
T T
 applies to each part of a circuit: V = I R
1
1 1
V2 = I2R2

ANALYZING CIRCUITS

Resistances in Series:
 I = I
T
1 = I2 = I3
 V = V +V +V
T
1
2
3
 R = R +R +R
T
1
2
3
R1
R2
R3
adding resistors in series increases RT,
decreases IT
 removing one resistor stops current in the
whole circuit

ANALYZING CIRCUITS

Resistances in Parallel:

IT = I1=I2+I3
R1
R2
R3
VT = V1 = V2 = V3
 1/R = 1/R +1/R +1/R
T
1
2
3
 adding resistors in parallel decreases R ,
T
increases I
 removing one resistor stops current only in
that branch

ANALYZING CIRCUITS


Kirchoff’s 1st Rule: total
current entering a
junction equals total
current leaving a
junction (conservation of
charge)
Kirchoff’s 2nd Rule: total
voltage change around
any closed loop of a
circuit is zero
(conservation of energy)
I 1 = I2 + I 3
ELECTRIC ENERGY &
POWER

Electric Power: rate of electric energy
supply or use, in Watts, W
power supplied or used, P = VI, 1 W =1
J/s
2
 power used, P = I R (appliance and light
bulb ratings)

ELECTRIC ENERGY &
POWER

Electric Energy: work done (energy
transferred) by electric current, in
Joules, J (electric companies bill for
energy, not power)
energy, E = Pt
 electric bill in kilowatt-hours, 1.00 kWh =
3.60×106 J

ANALYZING CIRCUITS
EXAMPLE CIRCUIT 1 - assume 4 V per
cell
RT=____ VT=____ IT=____ PT=____
R1= 8 W V1=____ I1=____ P1=____
R2= 8 W V2=____ I2=____ P2=____
ANALYZING CIRCUITS
EXAMPLE CIRCUIT 2 - assume 4 V per
cell
RT=____ VT=____ IT=____ PT=____
R1= 8 W V1=____ I1=____ P1=____
R2= 16 W V2=____ I2=____ P2=____
ANALYZING CIRCUITS
EXAMPLE CIRCUIT 3 - assume 4 V per
cell
RT=____ VT=____ IT=____ PT=____
R1= 8 W V1=____ I1=____ P1=____
R2= 8 W V2=____ I2=____ P2=____
ANALYZING CIRCUITS
EXAMPLE CIRCUIT 4 - assume 4 V per
cell
RT=____ VT=____ IT=____ PT=____
R1= 8 W V1=____ I1=____ P1=____
R2= 16 W V2=____ I2=____ P2=____
ANALYZING CIRCUITS
EXAMPLE CIRCUIT 5 - assume 5 V per
cell
RT=____ VT=____ IT=____ PT=____
R1= 1 W V1=____ I1=____ P1=____
R2= 6 W V2=____ I2=____ P2=____
R3= 12 W V3=____ I3=____ P3=____
CIRCUIT BOARD INTRO
CIRCUIT BOARD INTRO

Springs are
connectors for wires
and components.
Some springs are
connected to devices
on the board (like the
D-cells). If a spring is
too loose, squeeze
the coils.
CIRCUIT BOARD INTRO


When you connect a circuit to a D-cell note
the polarity (+ or –).
Only connect things long enough to make
your observations & measurements, then
disconnect one wire so the D-cells don’t run
down and resistors don’t overheat
ELECTRIC ENERGY &
POWER

Electric Hazards

effect of shock depends on location


skin: burns, muscles: spasms, nerves: pain, heart:
disruption
effect of shock depends on current




<10 mA: pain, no damage
>10 mA: severe muscle contraction, paralysis
70 mA chest: heart fibrillation
1 A chest: heart stops completely, but may restart
ELECTRIC ENERGY &
POWER

Electric Hazards


body resistance 104 to
106 W dry, 103 W wet
short circuit: low
resistance path
 low resistance = large
current  shock, fire
 fuses & circuit
breakers: disconnect
circuit above a
specific current level
UNIT 7 FORMULAS







Fe = kq1q2/r2
k = 8.99×109 Nm2/C2
e = ± 1.60×10–19 C
F = qE
K-K0 = Fd
I = Q/t
V = W/Q







R = rL/A
V = IR
P = VI = I2R
E = Pt
RT = R1+R2+R3
1/RT = 1/R1+1/R2+1/R3
1.00 kWh = 3.60×106 J