Transcript 23_Electricity

Introduction to Electricity
Electric charges come in two varieties.
We have named these positive and negative.
To be mathematically consistent all of electricity is defined
from the perspective of positive charges.
When there are two or more charges in a region, they will
experience a force.
The electric force is a vector.
To determine the force on one charge in the vicinity of
one or more other charges, you must first determine
the force on that one charge from each of the other
charges in the region and add them as individual
vectors.
The force can also be determined if the charge is in a
known electric field. The force will be the product of
the charge and the strength of the field.
Charges repel if they have the same sign ( + + or - - ), and
they attract if the signs are opposite ( + - ).
Coulomb’s Law is the equation that is used to determine the
force between a pair of charges:
F =
k q1 q2
r2
k = Coulomb’s Constant
k = 8.987 x 109 Nm2/C2 (on your equation sheet)
q1 and q2 are the two charges (in units of Coulombs, C)
r is the distance between the charges
What is the force between a proton and an electron if they are
separated by 1.50 x 10-10 m?
k q 1 q2
F =
r2
(8.987 x 109 Nm2/C2) (1.602 x 10-19 C)(1.602 x 10-19 C)
F =
(1.50 x 10-10 m)2
F = 1.025 x 10-8 N = 1.03 x 10-8 N
Michael Faraday developed the idea that there is a field that
surrounds each charge and that this field is what causes the
force on other charges in the region.
The electric field (E) is the identified as the force per unit
charge from one or more charges in a region.
E = F/q
The electric field can be written in the same form as Coulomb’s
Law:
kq
This is not a UNIFORM field.
E =
It varies with distance
r2
The electric field is also a vector. The direction of the electric
field is the direction the force would be if a POSITIVE charge
were the field.
The electric field is always away from positive charges and
toward negative charges.
What is the electric field 0.250 m from a 15.0 mC (15.0 x 10-6 C)
charge?
kq
E =
r2
E =
(8.987 x 109 Nm2/C2) (15.0 x 10-6 C)
(0.250 m)2
E = 2.157 x 106 N/C = 2.16 x 106 N/C
What is the force on an electron at that location?
E = F/q
F = q E = 1.602 x 10-19 C (2.157 x 106 N/C)
F = 3.45 x 10-13 N
There is energy stored in electric fields much like
gravitational potential energy in gravitational fields.
This electric potential energy (like all energy) is NOT a vector.
Electric Fields around Simple Arrangements
Point Charges
+
-
A Dipole Field
+
-
Like Charges
+
+
Two negative charges will have the same field pattern
except the arrows would be on the other end (toward the
charges).
When the position of a mass determines the force on the mass,
there is a potential energy involved ( W = DE = F d ). Moving in a
field of force means there is a change in potential energy.
The electric potential energy ( U ) is determined with an equation
similar to Coulomb’s Law and will have units of Joules, but it is
NOT A VECTOR.
In the force equation ( Coulomb’s Law ), we ignored the signs of
the charges to find the magnitude and used the signs to
determine direction. In the energy equation, SIGNS MUST BE
INCLUDED!
A negative potential energy means that it takes energy to escape
the effects of the forces (zero energy is at a distance of
).
The equation for electric potential energy is:
U =
k q1 q2
r
The amount of potential energy per unit charge is called ELECTRIC
POTENTIAL ( V ) which has units of VOLTS.
1 Volt = 1 Joule per Coulomb
The equation for electric potential from one charge is:
V = U/q
OR
V =
kq
r
The equation for electric potential from multiple charges is:
V =
S
kq
r
Two charges are placed 0.400 m apart. The one on the left has a
charge of + 245 mC, and the one on the right has a charge of
– 125 mC. What is the electric field midway between the
charges?
E1 =
E2 =
k q1
r12
=
8.987 x 109 Nm2/C2 ( 245 x 10-6 C )
( 0.200 m )2
8.987 x 109 Nm2/C2 ( 125 x 10-6 C )
= 5.50454 x 107 N/C
= 2.8084x 107 N/C
( 0.200 m )2
The direction of the field from both charges will be to the right, so they
add together.
E = 8.31 x 107 N/C
Now determine the electric potential of a point midway between
the two charges.
V1 =
k q1
r1
V2 =
=
8.987 x 109 Nm2/C2 ( + 245 x 10-6 C )
= 1.1009 x 107 V
( 0.200 m )
8.987 x 109 Nm2/C2 ( - 125 x 10-6 C )
( 0.200 m )
V = 1.1009 x 107 V + ( - 5.6169 x 106 V )
= - 5.6169 x 106 V
=
5.39 x 106 V
Two charges of the same magnitude and opposite signs are
placed near each other. The one on the left has a positive
charge, and the one on the right has a negative charge. What
is the electric field midway between the charges?
Both fields will have the same magnitude and be in the same direction
(to the right).
E = 2 K q / r2
What is the electric potential midway between the charges?
Both potentials will have the same magnitude but opposite signs; they
cancel out!
V=0
It takes work to move a charge from one potential ( Va ) to
another ( Vb ). For a charge, this is the same as having a
change in potential energy.
W = DU = q Va – q Vb = q Vab
A change in electrical potential is called a
POTENTIAL DIFFERENCE.
A potential difference of 1 volt will take or add 1 Joule of energy
to 1 Coulomb of charge.
When a potential difference (Vab) is placed across two parallel
plates (usually from a battery), a UNIFORM field is created
between the plates. This means that the electric field has the
same value at any point between the two plates.
+++++++++++++++
d = the distance
between the plates
Vab
- - - - - - - - - - - - - - -
One plate will be positively charged and one will be negatively
charged.
The field lines will go from the top ( + ) towards the bottom ( - ).
Vab = Ed
How much energy would an electron gain when moving across
a potential difference of 12.0 V between two parallel plates?
V = U/q
DU = q Vab = e Vab
DU = 1.602 x 10-19 C (12.0 V) = 1.92 x 10-18 J
The answer could also be: 12.0 eV!
What is the electric field between the two plates if they are
separated by 0.0600 m? V = Ed
12.0 V = E (0.0600 m)
ab
E = 12.0 V / (0.0600 m) = 200 V/m = 200 N/C
What is the force on the electron between the two plates?
F = qE
F = (1.602 x 10-19 C) (200 N/C)
F = 3.204 x 10-17 N = 3.20 x 10-17 N
How much work is done to move the electron 0.0600 m against
that force?
W = Fd
W = (3.204 x 10-17 N) (0.0600 m)
W = 1.92 x 10-18 J
ELECTRICITY: Circuits
When there is something to cause a force on charges and those
charges are free to move, they will begin to flow.
The movement of charge is called current.
Current ( I ) is measured in Coulombs per second.
One Coulomb per second is called an Ampere or Amp.
The unit of the Ampere is (A)
The “force” that causes current is called the electromotive
force which is a potential difference (voltage) and not really a
force.
Batteries are our primary source of voltage in circuits.
Symbol for battery:
I
+
-
As current travels it nearly always encounters some electrical
friction called RESISTANCE ( R ).
The unit of resistance is the Ohm ( W )
Resistance means that energy is being lost (usually in the form
of heat). This is electrical potential energy (not the kinetic
energy of the charges).
Resistance can be used to direct current since it will have a
greater value through paths with less resistance.
A device that is intentionally used to direct current through its
resistance is called a resistor.
Symbols for resistors:
When a current crosses a resistor, the potential of the current
drops in the direction of the current.
The following equation shows the relationship between these
three values:
V=IR
As current ( I ) encounters RESISTANCE ( R ) it loses potential ( or
voltage, V ).
If the resistance is constant, this is OHM’s LAW
If a path is followed around a circuit (of wires and resistors) the
total potential lost will equal the total created by the battery.
There are two devices used to measure current and potential
difference:
The ammeter
A
measures current in a wire.
The voltmeter
V
measures the potential
difference between two points in a wire.
The following is a simple circuit with a battery and a resistor.
An ammeter to measure the current and a voltmeter to measure
the potential difference across the resistor are included.
A
R
The battery will drive a current
counterclockwise around the
circuit.
V
The voltage from the battery will be lost through the resistor:
V – I R = 0 The voltage of the battery will be
entirely used by the circuit.
If there is more than one resistor, then you can use the concept
of equivalent resistance.
We will look at two common
arrangements. When the equivalent resistance is determined,
the single resistance (RE) replaces the entire arrangement.
Resistors in series are arranged one right after another along a
single wire of the circuit.
R1
R2
R3
RE = R1 + R2 + R3 + …
Resistors in parallel are arranged on separate wires from the
main wire of the circuit.
R1
R2
1 / R E = 1 / R 1 + 1 / R2 + 1 / R 3 + …
R3
OR: RE = [ 1 / R1 + 1 / R2 + 1 / R3 + … ]-1
What is the equivalent resistance of a 25 W, 33 W, and 15 W
resistor if they are in parallel?
1 / R E = 1 / R 1 + 1 / R 2 + 1 / R3
RE = [ 1 / 25 W + 1 / 33 W + 1 / 15 W ] -1 = 7.3 W
What is the potential difference across each one of the resistors if
hooked to a 12 V battery?
Each will have 12 V across it.
What is the current through
each resistor?
25 W
V=IR
33 W
I=V/R
I1 = 12 V / 25 W = 0.48 A
I2 = 12 V / 33 W = 0.36 A
15 W
I3 = 12 V / 15 W =
0.80 A
What would be the current through the single equivalent resistor?
IE = 12 V / 7.3 W =
1.6 A
0.48 A + 0.36 A + 0.80 A = 1.6 A
The following three resistors are hooked up in series to a 8.0 V
battery: 4.0 W, 10.0 W, and 6.0 W
What is the current through each resistor?
4.0 W
10.0 W
RE = R1 + R2 + R3
6.0 W
RE = 4.0 W + 10.0 W + 6.0 W
RE = 20.0 W
V=IR
I=V/R
IE = 8.0 V / 20.0 W = 0.40 A
What is the potential difference across each resistor?
V=IR
V1 = 0.40 A ( 4.0 W ) =
1.6 V
V2 = 0.40 A ( 10.0 W ) =
4.0 V
V3 = 0.40 A ( 6.0 W ) =
2.4 V
Resistors in series will all have the same current through them
but the potential difference for each depends on the resistance
of each.
Resistors in parallel will all have the same potential difference
across them but the current for each depends on the resistance
of each.
Resistors
Current
Potential
Difference
Series
Same for all
Resistors
Depends on
Resistors
Parallel
Depends on
Resistors
Same for all
Resistors
Resistors take the energy from the circuit and expel it as heat.
The rate of this energy transfer is the power ( P ) of the resistor.
P = I V = V2 / R = I 2 R
A light bulb is rated as:
75 W / 120 V
What is the resistance of the bulb?
P = V2 / R
R = V2 / P
R = ( 120 V )2 / 75 W = 192 W = 190 W
What is the power output of the above bulb if connected to a
40.0 V potential difference.
P = V2 / R = ( 40 V )2 / 192 W
P = 8.333 W = 8.33 W
Kirchhoff’s rules for circuits:
Loop Rule: The sum of all potential differences around any
closed path in a circuit must equal zero (like we had with our
single loop earlier).
Resistors create ( - ) Vab ( I R ) in the direction of current.
Batteries create ( + ) Vab in the direction of current.
The loop rule is a conservation of energy rule. The net
change in energy should be zero if you are back to the
starting point.
Point Rule: The sum of all currents into and out of a given
point must equal zero.
S Iin = S Iout
The point rule is a conservation of charge rule. charge
cannot be created or destroyed at any point in a circuit.
Electricity Practice Problems
What is the current drawn from the battery in the circuit below?
Vab = 12.0 V
R1 = 2.00 W
R2 = 3.00 W
R3 = 4.00 W
Vab
R1
R2
Vab = I RE
R3
RE = 2.00 W + 3.00 W + 4.00 W = 9.00 W
12.0 V = ( I ) 9.00 W
I = 1.33 A
V
A voltmeter is attached as shown. What should the voltmeter
read?
V = I R = ( 1.3333 A ) 3.00 W = 4.00 V
What is the total power output of the entire circuit?
P = I V = 1.3333 A ( 12.0 V ) =
16.0 W
P = V2 / RE = ( 12.0 V )2 / 9.00 W =
16.0 W
P = I2 RE = ( 1.3333 A )2 ( 9.00 W ) =
16.0 W
The ammeter in the following circuit reads 1.45 A. Determine
the potential difference (VOLTAGE) of the battery.
R1 = 1.00 W
R2 = 2.00 W
R3 = 4.00 W
R1
R2
R3
V1 = ( 1.45 A ) 1.00 W = 1.45 V
V2 = ( 1.45 A ) 2.00 W =
2.90 V
V3 = ( 1.45 A ) 4.00 W =
5.80 V
Vab = 1.45 V + 2.90 V + 5.80 V = 10.15 V
Vab = ?
What is the current drawn from the battery?
Vab = 10.0 V
R1 = 1.00 W
R2 = 2.00 W
R3 = 2.00 W
Vab
R1
R2
R4 = 3.00 W
RE = ?
1 / RE = 1 / 2.00 W + 1 / 2.00 W + 1 / 3.00 W
1 / RE = 1.3333 W-1
RE = 0.750 W
1.00 W + 0.750 W = 1.750 W
R3
I1 = 10.0 V / 1.750 W = 5.71 A
R4
Two charges are arranged 60.0 cm apart as shown. The charge on
the left is + 3.00 mC, and the charge on the right is – 3.00 mC. What
is the electric field midway between the two points?
E =
60.0 cm
E =
kq
S
2 (8.987 x 109 Nm2/C2) 3 x 10-6 C
r2
=
2kq
r2
= 5.99 x 105 N/C
(0.30 m)2
What is the electric potential at the same point?
V =
S
kq
r
= 0
What is the force on an electron at that point?
F = q E = 1.602 x 10-19 C ( 5.99 x 105 N/C ) = 9.60 x 10-14 N
Three charges of equal magnitude but different sign are set up at
the corners of an isosceles triangle as shown. What is the
direction of the force on the charge at the top. The signs are
indicated.
The charge will be repelled
by the lower left charge.
+
The charge will be attracted
to the lower right charge.
+
-
The distances and magnitudes are the same for each pair meaning
they will produce the same forces; the upward component of one will
be equal and opposite to the downward component of the other.
The resultant will be to the right.
What would be the kinetic energy of the alpha particle if it moved
1.00 cm from one plate to another plate if there is a potential
difference of 24.0 V between the plates?
V = U/q
DK = DU = q Vab = 2e Vab
48.0 eV
DK = 2 ( 1.602 x 10-19 C ) (24.0 V) = 7.69 x 10-18 J
What is the averge speed of an electron if it is orbiting a proton at a
k q 1 q2
distance of 1.50 x 10-10 m?
F =
r2
(8.987 x 109 Nm2/C2) (1.602 x 10-19 C)(1.602 x 10-19 C)
F =
(1.50 x 10-10 m)2
F = 1.025 x 10-8 N = mv2 / r
v2 = ( 1.50 x 10-10 m ) 1.025 x 10-8 N / ( 9.109 x 10-31 kg )
v = 1.30 x 106 m/s