Download Lecture Notes: Y F Chapter 23

Review: Work, Energy, Potential Energy, Conservation of Energy
v r
∆W = F ⋅ ∆x
Unit are Newtons · Meters = Joules
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an
st
Di
e
rc
Fo
e
on
kd
or
W
Wa →b = ∫
b
a
v r
b
F ⋅ dl = ∫ F cos θdl
a
Forces like Gravity and the Electric Force are Conservative Forces [i.e. No Friction]
Work done by conservative forces can be described as a change in Potential Energy
Wa →b = (U a − U b ) = −(U b − U a ) = − ∆U
∆U can be positive or negative
∆U positive – Work is “done” in moving from point a to point b
∆U negative – Energy is “released” in moving from point a to point b
To avoid confusion it is often helpful to think about a ball rolling up or down a hill –
∆U positive – Work is “done” in rolling the ball UP the hill – higher up means higher EP
∆U negative – Energy is “released” when the ball rolls down - lower down means lower EP
Electrostatic Potential Energy in a Uniform Electric Field
+
b
yb
r
E
a
ya
+ q
y
-
Move a positive charge in a vertical direction
W =∫
b
a
r
E = − Eyˆ
constant
r
dl = yˆ dl
v r
b
b
F ⋅ dl = ∫ − qEyˆ ⋅ yˆ dl = ∫ − qEdl = − qE ( yb − ya ) = − qE∆y
a
a
W = − ∆U
Looks like U gravity = mg∆h
∆U a →b = qE∆y
b
+
c
Now move from b → c
+
r
E
a
-
W =∫
y
c
v r
F ⋅ dl
b
r
r
But F is perpendicular to dl
So no work is done and there is
no change in U
∆U a →c = qE∆y
Now, instead, go directly from a → c and compute ∆U
c
b
∆y
a
+
φ
L
+
r
E
-
W =∫
c
a
y
v r
c
F ⋅ dl = ∫ F cos(ϕ )dl
a
c
W = − qE cos(ϕ ) ∫ dl
a
W = − qE cos(ϕ ) L
Path length is given by:
y = L cos(ϕ ) → L =
∆y
cos(ϕ )
Therefore: W = − qE∆y
Work done on a charge in an electric Field is INDEPENDENT OF PATH
Since Wa →b = (U a − U b ) = −(U b − U a ) = − ∆U is independent of path,
all that matters are the values of the potential energy at the start and
At the finish
Work and Potential Energy – Keeping the signs straight
Wa →b = ∫
b
a
r v b
F ⋅ dl = ∫ F cos φdl
a
Work done by a Force
Work and Potential Energy
Wa →b = ∫
b
a
r v b
F ⋅ dl = ∫ F cos φdl
Work done by a Force
a
The sign is critically important
r
r
F = qE
r
F" ME"
Direction of Motion
The Work in the equations above refers to the
Work DONE BY THE ELECTRIC FIELD
If I slowly move the charge,* I will always be
applying a force which is equal and opposite to the
Electrostatic Force. The work DONE BY ME will
be equal and opposite to the work done by the
E field.
* don’t change in kinetic energy
The Path of Integration Does NOT matter for a
“conservative” force such as the Electrostatic Force
Wa →b = ∫
b
a
r v b
F ⋅ dl = ∫ F cos φdl
a
Electrostatic Potential Energy of Two Point Charges
Fr =
b
b
a
a
Wa →b = ∫ Fr dr = ∫
qq0
qq0
=
dr
4πε 0 r 2
4πε 0
Wa →b = U a − U b = − ∆U =
∫
b
a
qq0
4πε 0 r 2
qq0
1
=
dr
2
r
4πε 0
0
qq0
4πε 0
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ ra rb ⎠
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ ra rb ⎠
The Path of Integration Does NOT matter for a
“conservative” force such as the Electrostatic Force
Wa →b = ∫
b
a
Wa →b
r v b
F ⋅ dl = ∫ F cos φdl
a
qq0
= U a − U b = − ∆U =
4πε 0
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ ra rb ⎠
There is NO ABSOLUTE ZERO for Potential Energy
It is CHANGE in Potential Energy That is Important
We can Pick any Distance to be the “Baseline” for Potential Energy
For “Convenience” we pick the zero of potential energy to refer the
Energy of the charges when they are VERY far away from each other
We pick ZERO for potential energy to be “Infinite Separation”
qq0
− ∆U =
4πε 0
qq0
− ∆U =
4πε 0
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ ra rb ⎠
⎛
1⎞
⎜⎜ 0 − ⎟⎟
rb ⎠
⎝
Let ra = ∞
Define
U (r ) =
qq0 1
4πε 0 r
The Potential Energy:
qq0 1
U (r ) =
4πε 0 r
is related to the amount of work that must be done to bring
two charges together from far away.
I must do POSITIVE work to
bring two like charges together
I must do NEGATIVE work to
bring two like charges together
Electrostatic Potential Energy for Several Points Charges
What is the electrostatic potential
energy of q0 due to q1, q2, and q3 ?
Electrostatic Potential Energy is a SCALAR and adds algebraically
This is different from the Electric Field; a VECTOR that adds vectorially
What is the potential energy of charge q2 in the field of q1 and q3 ?
The Electrostatic Potential Energy of a point charge q0
and a collection of Charges qi :
Sample Problem #3
q1
q2
q3
What is the Total potential energy of this charge distribution?
∞
Start by removing ALL charges to
.
This corresponds to the “zero” of Potential Energy
∞
Start by removing ALL charges to
.
This corresponds to the “zero” of Potential Energy:
Step 1: All Charges at
∞
U =0
Step 2: Bring Charge q3 in from infinity
q1
U =0
q2
q3
This requires no work
because there are no
other charges present.
Step 3: Bring Charge q2 in from infinity ,
q1
q2
q3
1 ⎛ q 2 q3 ⎞
⎜⎜
⎟⎟
U=
4πε 0 ⎝ r23 ⎠
Step 4: Bring Charge q1 in from infinity ,
q1
q2
1 ⎛ q2 q3 q1q2 q1q3 ⎞
⎜⎜
⎟⎟
U=
+
+
4πε 0 ⎝ r23
r12
r13 ⎠
q3
The Electrostatic Potential Energy of
a Collection of Point Charges:
U=
1
4πε 0
∑
i< j
qi q j
rij
Note: BE CAREFUL NOT TO DOUBLE COUNT
Electrostatic Force → Electrostatic Field
r
1 qi
F (r ) = q ∑
r̂
2 i
i 4πε 0 ri
v
1 qi
(
)
E r =∑
r̂
2 i
i 4πε 0 ri
We defined the concept of the Electric Field as
a generalization of the Electric Field.
The Electric Field assigns a VECTOR to each point in Space
Electrostatic Potential Energy of a charge q0
and a collection of charges qi
⎞
q0 ⎛ q1 q2 q3
q0
⎜⎜ + + + ...⎟⎟ =
U=
4πε 0 ⎝ r1 r2 r3
⎠ 4πε 0
qi
∑i r
i
If the charges qi remain fixed and I move q0, I can express
Potential Energy as a Function of the position of q0
q0
qi
U (r ) =
∑
4πε 0 i ri
Electrostatic Potential Energy → Electrostatic Potential
q0
qi
U (r ) =
∑
4πε 0 i ri
qi
V (r ) =
∑
4πε 0 i ri
1
We defined the concept of the Electric Potential as
a generalization of the Electric Potential Energy.
The Electric Field assigns a Scalar to each point in Space
Electrostatic Potential
U (r )
1 q
V (r ) =
=
q0
4πε 0 r
q0 q
U (r ) =
4πε 0 r
Electrostatic Potential of
set of POINT CHARGES
Electrostatic Potential of
POINT CHARGE
IMPORTANT: Like Work -Electrostatic Potential is
a “relative” quantity.
Wa →b = ∫
b
Va − Vb = ∫
b
a
a
r r b r r
F ⋅ dl = ∫ q0 E ⋅ dl
a
r r b
E ⋅ dl = ∫ E cos φdl
a
UNITS – Volt = Joule/Coulomb
∆V = 1.5 volts
Example 23.9
Find the potential at any height y between two oppositely charged parallel plates
Va − Vb = ∫
b
a
V (y) =
r r b
E ⋅ dl = ∫ E cos φdl
a
U ( y ) q0 Ey
=
= Ey
q0
q0
We have chosen V=0 at point b
Example 23.9
Conducting Sphere Revisited
Important Note – A conducting surface is always an equipotential surface
Equipotential Surfaces and Electric Field Lines
Equipotential Surfaces
are always perpendicular
to Electric Field Lines
End of Chapter 23
You are responsible for the material covered in T&F Sections 23.1-23.4
You are expected to:
•
Understand the following terms:
Work, Potential Energy, Electrostatic Potential, Potential Difference, Equipotential Surface
•
Understand the relationship between work and potential energy. In particular, be able to distinguish
between work done “by the field” and work done by the force moving a charge in a field.
•
Be able to calculate the electrostatic potential energy of simple charge geometries using either the
sum rule for discreet geometries or the integral rule for continuous geometries
•
CLEARLY DISTINGUISH BETWEEN POTENTIAL ENERGY AND ELECTROSTATIC POTENTIAL
involved.
•
Be able to calculate the electrostatic potential at a point in space due to simple charge geometries
using either the sum rule for discreet geometries or the integral rule for continuous geometries
•
Understand the convention of setting the zero of potential energy at “infinity”
Recommended Y&F Exercises chapter 23:
1,6,9,14,21,23,35