Transcript CHAPTER 17 Electrical Energy and Current

CHAPTER 17
Electrical Energy and Current
Conservative Forces:
Work done on an object depends
only on its initial and final position.
The path from the initial to the
final position is not important
Example (Gravitational Force)
Consider the work done on an object against gravity.
W = Fg  d = mgh
Work is converted to gravitational potential energy
The Electrostatic Force (Fe) is also a conservative force.
W = Fe  d = qEd
(qE = k q21 q = Fe)
r
Work done by gravity (falling object) = -mgh
Work done by electrostatic force = -qEd
W = q E d
Potential Difference:
The change in potential energy
(Electric Potential)
of a charge, q1 divided by q1.
Potential Difference  Change in potential energy
Potential Difference (V)
(Electric Potential)
V = PE
q
• PE occurs in a uniform electric field
• q is a charge that changes position in the uniform field
• V is a scalar quantity
• V units = Joules/Coulomb
1 Volt (V) = 1 Joule/Coulomb
Example Problem (moving a positive charge against an electric field)
15cm
q
E = 250 N/C
q = +400C
d = 15cm
Work done on the charge = qEd
= (400x10-6C)(250N/C)(15x10-2m)
= 1.5 x 10-2 Joules
Work done increases the potential energy of the charge.
V = PE = qEd = Ed
q
q
V = (250N/C)(15x10-2m)
V = 38 Nm/C
V = 38 J/C
NOTE: [E] = V/m = N/C
V = 38V
Summary
Direction of Movement
Relative to E-Field
+
–
Opposite (against)
Opposite (against)
Sign of PE
+
–
+
–
Same direction (with)
Same direction (with)
–
+
Sign of Charge
Electric Field Between Parallel Plates
What is the Electric Potential Differenced in the above
diagram if E=25x102N/C and d=15cm?
Electric Potential Difference = V = VB – VA = -Ed
(Potential Difference)
= V = -(25x102 N/C)(15x10-2 m)
= V = -375 Nm/C
= V = -375 J/C
V = -375V
If the charge consisted of a
proton:
m = 1.67x10-27kg
q=1.60x10-19C
V= 38J/C
What would be its change in
potential energy and with what
velocity would it be moving at
“B” if it was at rest at point “A”?
Strategy
Calculate change in potential energy and convert potential
energy to kinetic energy. Solve for v.
PE = qV = (1.60x10-19C)(38J/C)
PE = 6.1x10-19J
PE = KE = KEB (KEA = 0)
6.1x10-18J = (1.67x10-27kg) v2
v = 6.0x104 m/s
2
Electric Potential Associated with Point Charges
Between parallel plates E is uniform.
E associated with a point charge is not uniform
E= kq = kq
r=d
2
2
r
d
V = Ed for only small d values
Calculus to the rescue!
dv = Edd
dv = k 2q dd
d
d= r
kq
= d2
d= 
 dv 
v=
dd
d=r
-1 k q
d
d=
v= k q
r
v=
kq
r
Scalar Quantity
Electric Potential Caused by Point Change
Electric Potential Caused by 2 Point Charges
When Analyzing Multiple Point Charges:
• The principle of superposition applies. (Just like with
calculating E-field due to multiple charges.)
• However “v” is a scalar quantity (J/C) and “E” was a
vector quantity (N/C)
• Scalars are much easier to add than vectors because
with scalars… we have no direction.
Example Problem (Electric Potential : linear)
Two point charges 20cm apart each with a charge of +50C
are established. What is the electric potential 10.cm from
each (midpoint)? What is the electric field at this point?
Electric Potential (V)
kq
v=
= k q1 + k q 2

r
r1
r2
v = (9.0x109Nm2/C2)(+50x10-6C)
(10x10-2m)
v = 4.5x106 Nm/C
6 Volts
6
v
=
4.5x10
v = 4.5x10 J/C
Electric Field (F)
ET =  E
Vectors
9Nm2/C2)(50x10-6C)
(9.0x10
k
q
1 =
E1 =
(10x10-2m)2
r2
E1 = 4.5x107 N/C
Similarly
E2 = 4.5x107 N/C
directed away from q1
directed away from q2
E2
E1
q1
10cm
ET = 0 N/C
q2
10cm
Example Problem (Electric Potential : 2 Dimensions)
EA2
A

30cm
EAl
60cm
52cm
q2 = +50 C
q1 = -50 C
Calculate the electric potential of point A

v= k(
v=
kq
r
q1
q2
r1 + r2
)
(
-6C + 50x10-6C
-50x10
v = (9.0x109Nm2/C2) 60x10-2m 30x10-2m
)
v = (9.0x109Nm2/C2)(-8.3x10-5C/m + 1.66x10-4C/m)
v = (9.0x109Nm2/C2)(8.3x10-5C/m)
v = 7.5x105Nm/C
v = 7.5x105J/C
v = 7.5x105 V
Equipotential Lines (Surfaces)
Gravitational
Gravitational
Field Lines
Lines of
Equipotential
Earth
Line of Equipotential is merely a line (surface in
3-dimensional system) where potential energy
remains constant as an object moves along the
line
Lines of Equipotential are perpendicular to force
field lines so that no work is done when the
object moves
W = F  d cos 
Equipotential Lines (Surfaces)
Electrostatic
Equipotential Lines (surfaces)
• encircle the charged particle
• perpendicular to field lines
• never cross each other because field lines never cross
each other
Electric Field Lines
• directed away from a positive charge (i.e. direction a
positive “test charge” would move)
• closer together indicates greater E-field
• exit perpendicular to the surface
• never cross each other
Test Yourself
Draw E-fields and lines (surfaces) of Equipotential for the
following situations.
a)
-q
b)
+q
+q
+q
-q
c)
Current and Resistance
(Electric) Current (I):
The rate at which charge is flowing
(through a wire).
I=
Q
t
Ampere (A):
SI unit for current
1 Ampere = 1 Amp = 1 A = 1 Coulomb/sec
Conventional Current: The flow of positive charge.
If conventional current is flowing to
the right, then in reality, electrons
are flowing to the left.
Positive charge, protons, don’t
move.
• The current in a light bulb is 0.835 A. How long does it take for a
total charge of 1.67 C to pass through the filament of the bulb?
Resistance:
Ohm ():
A measure of what must be overcome
to make charge flow
SI unit for resistance
1 ohm = 1  = 1 Volt/Amp
V
R=
I
If a large current results from a small V, then the
resistance must be small.
If a small current results from a large V, then the
resistance must be large.
Example Problem
A flow of 24 coulombs of charge passes through a wire in 2.1
seconds where a voltage of 37 volts is applied across the wire.
Calculate a) how many charges moved through the wire, b)
the current in the wire, and c) the resistance of the wire.
Strategy:
Extract the data.
Q = 24 coulombs
t = 2.1 seconds
V = 37 V
Apply Proper Formulas
a) 24 coulombs
1 electrons
20electrons
1.5x10
1.60x10-19 coulombs
Q
24 coulombs
b) I =
=
t
2.1 seconds
11 amps
I = 11 coulombs/sec
V
37 Volts
c) R =
=
I
11 amps
R = 3.4 Volts/amp 3.4 ohms
Resistance
What factors might affect the resistance of a wire?
l
A
R
l
A
R=
l
A
 = resistivity of the material
= ohms -meter
Resistivity is another physical
property of a material
Resistors and Energy Loss
It is really friction in the wire that results in resistance to
flow of charge.
Friction causes heat.
A resistor should get hot when voltage causes a current
passes through it (Ex: filament in a light bulb)
Derivation:
Voltage = Joules/coulomb
Current = Coulomb/sec
Joules x Coulomb
Voltage x Current = Coulomb
sec
V x I = Joules/sec = Watts
Power has units of Joules/sec or Watts
P = IV
P = I2R
P =?
V = IR