Transcript Presentation 04

W03D2
Work, Potential Energy
and Electric Potential
Today’s Reading Assignment: Course
Notes: Sections 4.1-4.3
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Announcements
W03D3 Reading Assignment Course
Notes: Sections 4.7-4.10
Exam One Thursday Feb 28 7:30-9:30
pm Room Assignments (See Stellar
webpage announcements)
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Outline
Electrical Work
Electric Potential Energy
Electric Potential Difference
Calculating Electric Potential Difference
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Electrical Work
Electrical force on object 1 due to interaction
between charged objects 1 and 2:
F12 = ke
q1q2
r
2
12
r̂12
Work done by electrical force moving object 1
from A to B:
B
W1 = ò F12 × d s12
A
PATH
INTEGRAL
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Concept Question: Sign of W
Suppose a fixed positively
charged object (charge qs > 0) is
at the origin and we move a
negatively charged object (charge
q1 < 0) from A to B with rA < rB ,
where r is the distance from the
origin.
1.
2.
3.
4.
Work done by the electrostatic force is positive and we do
a positive amount of work
Work done by the electrostatic force is positive and we do
a negative amount of work
Work done by the electrostatic force is negative and we do
a positive amount of work
Work done by the electrostatic force is negative and we do
a negative amount of work
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Concept Question Ans.: Sign of W
Answer 3: W is negative and we do a positive amount
of work
W is the work done by the electrical force. This is the
opposite of the work that we must do in order to move
a charged object in an electric field due to source.
The electrical force is attractive and we are moving
the positively charged object away from the source
(opposite the direction of the electric field).
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Group Problem: Work Done by
Electrical Force
A point-like charged source object
(charge qs) is held fixed. A second
point-like charged object (charge
q1)is initially at a distance rA from the
fixed source and moves to a final
distance rB from the fixed source.
What is the work done by the
electrical force on the moving object?
Hint: What coordinate system is best
suited for this problem?
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Sign of W: Negative Work
Suppose a fixed positively
charged source (charge qs >
0) is at the origin and a
positively charged object
(charge q1 > 0) moves from A
to B with rA > rB , where r is
the distance from the origin,
then W < 0.
æ 1 1ö
1 1
rA > rB Þ - > 0 and qs q1 > 0 Þ W = -ke qs q1 ç - ÷ < 0
rB rA
è rB rA ø
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Work and
Change in Kinetic Energy
W = DK
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Group Problem: Work-Kinetic
Energy In a Uniform Electric field
Consider two thin oppositely uniform charged thin plates
separated by a distance d. The surface charge densities on
the plates are uniform and equal in magnitude. An electron
with charge –e and mass m is released from rest at the
negative plate and moves to the positive plate. What is the
speed of the electron when it reaches the positive plate?
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Potential Energy Difference
Suppose charged object s is fixed
and located at the origin and
charge object 1 moves from an
initial position A, a distance rA
from the origin to a final position
B, a distance rB from the origin.
The potential energy difference
due to the interaction is defined to
be the negative of the work
done by the field in moving
object 1 in from A to B. This is the
same as the work you do in
moving object 1 from A to B.
æ 1 1ö
DU º U B - U A = - ò F12 × d s = -W = ke qs q1 ç - ÷
A
è rB rA ø
B
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Potential Energy: Zero Point
Choose the zero point for the potential energy at infinity.
U(¥) º 0
Then set rA = ∞ and rB = r .
The potential energy difference between ∞ and any point on a
circle of radius r is
keqs q1
U (r) - U (¥) = U (r) =
r
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Concept Question: Motion of Charged Objects
Two oppositely charged are released from rest in an electric
field.
1.
2.
3.
4.
Both charged objects will move from lower to higher
potential energy.
Both charged objects will move from higher to lower
potential energy.
The positively charged object will move from higher
to lower potential energy; the negatively charged
object will move from lower to higher potential
energy.
The positively charged object will move from lower to
higher potential energy; the negatively charged
object will move from higher to lower potential
energy.
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Concept Q. Ans.: Motion of Charged Objects
2. Both charged objects will move from higher to lower
potential energy so that
DU < 0
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Configuration Energy
What is the potential energy stored in a configuration of
charged objects? Start with all the charged objects at infinity.
Choose
U(¥) º 0
(1) Bring in the first charged object.
DU1 = 0
(2) Bring in the second charged object
DU 2 = U12 = keq1q2 / r12
(3) Bring in the third charged object
DU 3 = U 23 + U13 = keq2 q3 / r23 + keq1q3 / r13
(4) Configuration energy
DU = U12 + U 23 + U13 = keq1q2 / r12 + keq2 q3 / r23 + keq1q3 / r13
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Group Problem: Build It
How much energy
does it take you to
assemble the
charges into the
configuration at left,
assuming they all
started out an infinite
distance apart?
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Electric Potential Difference
Change in potential energy per test charge in
moving the test object (charge qt) from A to B:
B
B
DU
F
DV º
= -ò × d s = -ò E × d s
qt
q
A t
A
Units: Joules/Coulomb = Volts
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Demonstration
Van de Graaf
D29
Breakdown of dry air
33 kV/cm
Video of Tesla Coil
http://www.youtube.com/watch?v=FY-AS13fl30
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How Big is a Volt?
AA Batteries
1.5 V
High Voltage
Transmission Lines
100 kV-700
kV
Car Batteries
12 V
Van der Graaf
300 kV
US Outlet (AC)
120 V
Tesla Coil
500 kV
Distribution
Power Lines
120 V- 70
kV
Lightning
10-1000 MV
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E Field and Potential: Effects
If you put a charged particle, (charge q), in a field:
F = qE
To move a charged particle, (charge q), in a field
and the particle does not change its kinetic energy
then:
DU = qDV
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Concept Question: Motion of Charged Objects
Two oppositely charged are released from rest in an electric
field.
1.
2.
3.
4.
Both charged objects will move from lower to higher
electric potential.
Both charged objects will move from higher to lower
electric potential.
The positively charged object will move from higher
to lower electric potential; the negatively charged
object will move from lower to higher electric
potential.
The positively charged object will move from lower to
higher electric potential; the negatively charged
object will move from higher to lower electric
potential.
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Concept Q. Ans.: Motion of Charged Objects
Two oppositely charged are released from rest in an electric
field.
3. The positively charged object will move from
higher to lower electric potential; the negatively
charged object will move from lower to higher electric
potential.
For the positively charged object:
DV < 0 Þ DU = qDV < 0
For the negatively charged object:
DV > 0 Þ DU = qDV < 0
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Potential & External Work
Change in potential energy in moving the
charged object (charge q) from A to B:
DU = qDV
Conservation of Energy Law:
Wext = DK + DU
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Demonstration
Kelvin Water Drop in 26-152
Wimshurst Machine in 32-152
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Demonstration: Kelvin Water
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Potential Created by Pt Charge
B
DV = VB - VA = - ò E × d s
A
B dr
r̂
= - ò kQ 2 × d s = -kQ ò 2
A
A r
r
æ 1 1ö
= kQ ç - ÷
è rB rA ø
B
Take V = 0 at r = ∞:
kQ
VPoint Charge (r) =
r
r̂
E = kQ 2
r
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Concept Question: Two Point
Charges
The work done in moving a
positively charged object that
starts from rest at infinity and
ends at rest at the point P
midway between two charges
of magnitude +Q and –Q
1.
2.
3.
4.
is positive.
is negative.
is zero.
can not be determined – not enough info is given.
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Concept Question Answer: Two
Point Charges
3. Work from ∞ to P is
zero.
The potential at ∞ is zero.
The potential at P is zero because equal and
opposite potentials are superimposed from the
two point charges (remember: V is a scalar,
not a vector)
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Potential Landscape
Positive Charge
Negative Charge
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Continuous Charge
Distributions
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Continuous Charge Distributions
Break distribution into
infinitesimal charged elements of
charge dq. Electric Potential
difference between infinity and P
due to dq.
dq
dV º Vdq (P) - Vdq (¥) = ke
r
Superposition Principle: V (P) - V (¥) = k
e
Reference Point:
dq
ò r
source
dq
V (¥) = 0 Þ V (P) = ke ò
r
source
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Calculating Electric Potential Difference
for Continuous Distributions
1. Choose V (¥) = 0
2. Choose integration r¢
variables
ì l d s¢
ïï
3. Identify
d q¢ = í s d a¢
ï
ïî r d v ¢
4. Choose field point variables r
5. Calculate source to field
d q¢
point distance r - r ¢
V (r) - V (¥) = ke ò
source r - r ¢
6. Define limits of integral
7. Integrate
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Worked Example
Consider a uniformly
charged ring with total
charge Q. Find the electric
potential difference between
infinity and a point P along
the symmetric axis a
distance z from the center of
the ring.
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Worked Example: Charged Ring
l = Q / 2p R
dq¢ = l ds¢ = l Rdq ¢
Choose
V (¥) = 0
r = z k̂
r¢ = Rr̂
r - r¢ = R2 + z 2
dq¢
1
l Rdq ¢
dV =
=
4pe 0 r - r¢ 4pe 0 R 2 + z 2
1
V (z) = kel ò
2p
Rdq ¢
=
ke l R
R +z
R +z
ke 2p Rl
k eQ
V (z) =
=
2
2
R +z
R2 + z 2
0
2
2
2
2
ò
b
a
dq ¢
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Group Problem
A thin rod extends along
the x-axis from x = -l /2 to
x = l/2 . The rod carries a
uniformly distributed
positive charge +Q.
Calculate the electric
potential difference
between infinity and at a
point P along the x-axis.
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