Download Gravitational Potential Energy

Briefly review the concepts of potential
energy and work.
•Potential Energy = U = stored work in a system
•Work = energy put into or taken out of system by forces
•Work done by a (constant) force F :
v
v
v
F
v
W = F ⋅ ∆r =| F || ∆r | cos θ
θ
∆r
Gravitational Potential Energy
Lift a book by hand (Fext) at constant velocity.
final
Fext = mg
Wext = Fext h = mgh
h
Fext
Define ∆U = +Wext = -Wgrav= mgh
initial
mg
Wgrav = -mgh
Note that get to define U=0,
typically at the ground.
U is for potential energy, do not confuse with
“internal energy” in Thermo.
Gravitational Potential Energy (cont)
For conservative forces Mechanical Energy
is conserved.
EMech = E Kin + U
Gravity is a conservative force.
Coulomb force is also a conservative force.
Friction is not a conservative force.
If only conservative forces are acting, then
∆EMech=0.
∆EKin + ∆U = 0
Electric Potential Energy
Charge in a constant field
∆Uelec = change in U when moving +q from initial to final position.
∆U = U f − U i = +Wext = −W field
FExt=-qE
FField=qE
+
Final position
∆r
E
+
Initial position
--------------
v
v
∆U = −W field = − F field ⋅ ∆r
v
v
→ ∆U = −qE ⋅ ∆r
General case
What if the E-field is not constant?
v v
∆U = −qE ⋅ ∆r
v v
∆U = −q ∫ E ⋅ dr
f
Integral over the path
from initial (i) position to final (f)
position.
i
Electric Potential Energy
Since Coulomb forces are conservative, it means that the change
in potential energy is path independent.
v v
∆U = −q ∫ E ⋅ dr
f
i
Electric Potential Energy
Positive charge in a constant field
Electric Potential Energy
Negative charge in a constant field
Observations
• If we need to exert a force to “push” or “pull” against the
field to move the particle to the new position, then U
increases. In other words “we want to move the particle to
the new position” and “the field resists”.
• If we need to exert a force to “hold” the particle so the field
will not move the particle to the new position, then U
decreases. In other words, “the field wants to move the
particle”, and “we resist”.
• Both can be summarized in the following statement:
“If the force exerted by the field opposes the motion of the
particle, the field does negative work and U increases,
otherwise, U decreases”
Potential Energy between two point
charges
+Q1
r
+Q2
Imagine doing work to move the objects from infinitely far apart
(initial) to the configuration drawn above (final).
Potential Energy between two point
charges
∆U = U f − U i = +Wext = −W field
f
v
v
∆U = U f − U i = − ∫ F field ⋅ dr
i
f
v
* Note that force is not
constant over the path!
v
∆U = U f − U i = − ∫ qE ⋅ dr
i
Consider Q1 fixed and move Q2 from infinity to r.
+Q1
r
11
Potential Energy between two point
charges
+Q1
r
+Q2
f
v
v
∆U = U f − U i = − ∫ qE ⋅ dr
i
r
1 Q1 ⎞
⎟ dx
2 ⎟
4
πε
x
0
⎠
⎝
⎛
∆U = U f − U i = − ∫ Q2 ⎜⎜
∞
∆U = U f − U i =
E-field generated by Q1.
Moving Q2 through the field.
1 Q1Q2
4πε 0 r
Potential Energy between two point
charges
We also need to define the zero point for potential energy.
This is arbitrary, but the convention is U=0 when all charged
objects are infinitely far apart.
∆U = U f − U i =
Uf =
1 Q1Q2
4πε 0 r
1 Q1Q2
4πε 0 r
Ui =U(∞)= 0 by our convention.
Potential Energy between two
electric charges.
E. Potential energy vs. distance
E. Potential Energy of a charge
distribution
q1
Potential energy associated to the
field produced by charges qi
q2
r1
r2
q3
r3
U=
q0
4πε 0
⎛ q1
⎜
⎜r
⎝ 1
+
⎞
q2 q3
q
+ + ...⎟⎟ = 0
r2 r3
⎠ 4πε 0
qi
∑r
i
q0
For a continuous charge distribution, replace the sum by
an integral
E. Potential Energy of a charge
distribution
To calculate the TOTAL potential
energy we have to consider all the
fields produced by all the charges qi
on the other(n-1) charges qj
q1
q2
r1
r2
q3
r3
q0
U Total =
1
4πε 0
∑
i< j
qi q j
rij
For a continuous charge distribution, replace the sum by
an integral
i
CPS Question
U (r ) =
Both cases below are for two
point charges separated by a
distance r.
Which graph is correct for two
negative charges? A) Left Plot,
B) Right Plot
kQ1Q2
r
U(r)
U(r)
r
r
Just like a compressed
spring stores potential
energy.
17
Electric Field
r
+Q1
+Q2
Earlier we found that not only using forces, but also electric fields
was very useful.
v
F12 =
1 Q1Q2
rˆ
4πε 0 r 2
v
E1 =
v
v
1 Q1
1 Q1
rˆ F12 = qE1 = Q2
rˆ
2
4πε 0 r
4πε 0 r 2
•Electric Field is the force per unit of charge due to the presence of Q1
•Electric field from Q1 is there even if Q2 is not there.
All of the above are vectors!
Electric Potential
r
+Q1
+Q2
We find a similarly useful thing with electric potential energy.
U12 =
1 Q1Q2
4πε 0 r
V1 =
1 Q1
4πε 0 r
U12 = qV1 = Q2
1 Q1
4πε 0 r
•Electric potential is the electric potential energy per unit of charge due
to the presence of Q1
•Electric potential from Q1 is there even if Q2 is not there.
All of the above are scalars!
Electric Potential
1 Q1
E1 =
rˆ
4πε 0 r 2
Electric Field is a vector associated with
a source charge Q1.
Units are [Newtons/Coulomb].
1 Q1
V1 =
4πε 0 r
Electric potential is a scalar associated with
v
a source charge Q1.
Units are [Joules/Coulomb] or [Volts].
V = Voltage = Electric Potential
Units are [Volts] = [Joules/Coulomb]
U = Electric Potential Energy
Units are [Joules]
Electric Field Units=[Volts]/[meter]=[Newton]/[Coulomb]
+Q1
Electric Potential of a point charge
V=
1
q
4πε 0 r
Analogy: Electrical pressure or electrical "height"
Positive charges want to get away from higher voltage towards lower voltage.
Just like a gas wants to move from high to low pressure.
Electric Potential of a charge
distribution
q1
Potential energy associated to the
field produced by the charges qi
q2
r1
r2
q3
r3
q0
“test charge”
U=
q0
4πε 0
⎛ q1
⎜
⎜r
⎝ 1
+
⎞
q2 q3
q
+ + ...⎟⎟ = 0
r2 r3
⎠ 4πε 0
qi
∑r
i
Electric potential due to charges qi
1
qi
V=
∑
4πε 0 i ri
For a continuous charge distribution, replace the sum by
an integral
i
Electric Potential from Electric Field
f
r
v
f
v
v
∆U = WExt . = − ∫ F ⋅ dr = − ∫ qE ⋅ dr
i
i
f
∆V =
v v
∆U
= − ∫ E ⋅ dr
q
i
Equipotential surfaces
Elevation is a scalar.
Contour lines show paths of
constant elevation.
Gravitational potential
VG=gh
If I stand at a certain
elevation I have a
gravitational potential energy
[Joules] = mgh=mVG
Equipotential surfaces
Electric potential (Voltage) is a
scalar.
Contour lines show paths of
constant Voltage (equipotentials).
If a charge q is at a certain
Voltage, it has an electrical
potential energy [Joules] = qV
Equipotential surfaces and field lines
•Equi-potential surfaces and field lines are always mutually
perpendicular
•The Field is not necessarily constant on equipotential surfaces
•Larger density of equipotentials means larger variations of V,
and larger values of |E|
E. Field and E. potential
What is Electric Field?
Electric Field is a vector that is
analogous to a steepness
vector.
Steepness cannot have one
number at a given position, it
depends which direction you
look (vector).
E. Field and E. potential
f
v
v
v
v
∆V = V f − Vi = − ∫ E ⋅ dr
Given the Electric field vector
as a function of position, we can
compute Voltage.
i
r
V (r ) − V (∞ ) = − ∫ E ⋅ dr
V =V(∞)= 0 by our convention.
i
∞
r
v
v
∆V = V ( r ) = − ∫ E ⋅ d r
∞
If we integrate the “steepness” over a path, we
find the change in elevation.
This does not depend on our path taken.
E. Field and E. potential
b
v
v
b
∆V = −∫ E ⋅ dr = −∫ dV
a
a
− dV = E x dx + E y dy + Ez dz
⇒E
∂V
∂V
∂V
E=−
xˆ −
yˆ −
zˆ
∂x
∂y
∂z
v
r
v
x
=−
dV
dV
dV
; Ey = −
; Ez = −
dx
dy
dz
The Electric field vector is the rate of change
of Voltage in a given direction.
The Electric field vector “potential gradient”
E = −∇V
http://www.falstad.com/vector2de/
CPS question
Two identical charge, +Q and +Q, are fixed in space. The
electric potential (V) at the point X midway between the
charges is:
A) Zero
V
B) Non-Zero
v
E=−
E=0
V=?
+Q
Point X
∂V
xˆ
∂x
+Q
+Q
+Q
x
30
CPS Question
Drawn are a set of equipotential lines. Consider the electric
field at points A and B. Which of the following statements is
true?
A) |EA| > |EB|
B) |EA| < |EB|
Point B
C)|EA| = |EB|
10V 20V 30V
0V
Point A
D)Not enough information
given.
E) None of the above
CPS Question
Two charges, +Q and -Q, are fixed in space. The
electric field at the point X midway between the charges is:
A) Zero
B) Non-Zero
V
V=0
E=?
+Q
Point X
v
E=−
+Q
∂V
xˆ
∂x
-Q
-Q
x
32
CPS Question
+Q
+Q
Point P
-Q
The magnitude of the electric field at point P
is:
A) Zero
B) Non-Zero
The magnitude of the voltage at point P is:
A) Zero
B) Non-Zero
-Q
33
E. Potential inside conductors
All points on conductor must be at the same
electrical potential.
b
v
v
∆V = − ∫ E ⋅ dr
a
Imagine point a at Voltage Va.
Since E=0 everywhere inside the conductor (no steepness), integral to
point b is always 0. ∆Vab=0.