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 :
θcos|||| rFrFWv
v
v
v
∆=∆⋅=
∆r
F
θ
Gravitational Potential EnergyLift a book by hand (Fext) at constant velocity.
Fext
mg
initial
final
h
Fext = mg
Wext = Fext h = mgh
Wgrav = -mgh
Define ∆U = +Wext = -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.
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.
0=∆+∆ UEKin
UEE KinMech +=
Electric Potential EnergyCharge in a constant field
- - - - - - - - - - - - - -
+
E
FField=qE
FExt=-qE
+
∆r
Initial position
Final position
∆Uelec = change in U when moving +q from initial to final position.
fieldextif WWUUU −=+=−=∆
rEqU
rFWU fieldfield
v
v
v
v
∆⋅−=∆→
∆⋅−=−=∆
General case
What if the E-field is not constant?
rEqUv
v
∆⋅−=∆
∫ ⋅−=∆f
i
rdEqUv
v Integral over the path
from initial (i) position to final (f)
position.
Electric Potential EnergySince Coulomb forces are conservative, it means that the change
in potential energy is path independent.
∫ ⋅−=∆f
i
rdEqUv
v
Electric Potential EnergyPositive charge in a constant field
Electric Potential EnergyNegative 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
r+Q1 +Q
2
Imagine doing work to move the objects from infinitely far apart
(initial) to the configuration drawn above (final).
11
fieldextif WWUUU −=+=−=∆
∫ ⋅−=−=∆f
i
fieldif rdFUUUv
v
∫ ⋅−=−=∆f
i
if rdEqUUUv
v
* Note that force is not
constant over the path!
r+Q1
Consider Q1
fixed and move Q2
from infinity to r.
Potential Energy between two point charges
r+Q1 +Q
2
∫ ⋅−=−=∆f
i
if rdEqUUUv
v E-field generated by Q1.
Moving Q2
through the field.
∫∞
⎟⎟⎠
⎞⎜⎜⎝
⎛−=−=∆
r
if dxx
QQUUU
21
02 4
1
πε
r
QQUUU if
21
04
1
πε=−=∆
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.
r
QQUUU if
21
04
1
πε=−=∆
r
QQU f
21
04
1
πε=
Ui=U(∞)= 0 by our convention.
Potential Energy between two
electric charges.
Potential Energy between two point charges
E. Potential energy vs. distance
E. Potential Energy of a charge
distribution
∑=⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
i i
i
r
r
q
r
q
r
qqU
0
0
3
3
2
2
1
1
0
0
4...
4 πεπε
For a continuous charge distribution, replace the sum by
an integral
Potential energy associated to the
field produced by charges qi
r1 r2
r3
q1
q2
q3
q0
E. Potential Energy of a charge
distribution
∑<
=ji ij
jiTotal r
qqU
04
1
πε
For a continuous charge distribution, replace the sum by
an integral
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
r1 r2
r3
q1
q2
q3
q0
17
r
U(r)
r
U(r)
r
QkQrU 21)( =
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
CPS Question
Just like a compressed
spring stores potential
energy.
Earlier we found that not only using forces, but also electric fields
was very useful.
rr
QQF ˆ
4
12
21
012 πε=
v
rr
QE ˆ
4
121
01 πε=
v
rr
QQEqF ˆ
4
121
02112 πε
==vv
r+Q1 +Q
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 Field
We find a similarly useful thing with electric potential energy.
r
QQU 21
012 4
1
πε=
r
QV 1
01 4
1
πε=
r
QQqVU 1
02112 4
1
πε==
r+Q1 +Q
2
•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
+Q1
Electric Field is a vector associated with
a source charge Q1.
Units are [Newtons/Coulomb].
Electric potential is a scalar associated with
a source charge Q1.
Units are [Joules/Coulomb] or [Volts].
rr
QE ˆ
4
121
01 πε=
v
r
QV 1
01 4
1
πε=
Electric Potential
V = Voltage = Electric Potential
Units are [Volts] = [Joules/Coulomb]
U = Electric Potential Energy
Units are [Joules]
Electric Field Units=[Volts]/[meter]=[Newton]/[Coulomb]
Electric Potential of a point charge
r
qV
04
1
πε=
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
∑=⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
i i
i
r
r
q
r
q
r
qqU
0
0
3
3
2
2
1
1
0
0
4...
4 πεπε
For a continuous charge distribution, replace the sum by
an integral
Potential energy associated to the
field produced by the charges qi
Electric potential due to charges qi
∑=i i
i
r
qV
04
1
πε
r1 r2
r3
q1
q2
q3
q0
“test charge”
∫ ⋅−=∆=∆f
i
rdEq
UV
v
v
Electric Potential from Electric Field
∫∫ ⋅−=⋅−==∆f
i
f
i
Ext rdEqrdFWUv
v
v
r
.
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
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).
∫ ⋅−=−=∆f
i
if rdEVVVv
v
∫∞
⋅−=∞−r
rdEVrVv
v
)()(
∫∞
⋅−==∆r
rdErVVv
v
)(
Given the Electric field vector
as a function of position, we can
compute Voltage.
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
Vi=V(∞)= 0 by our convention.
dz
dVE
dy
dVE
dx
dVEdzEdyEdxEdV zyxzyx −=−=−=⇒++=− ;;
∫∫ −=⋅−=∆b
a
b
a
dVrdEVv
v
The Electric field vector is the rate of change
of Voltage in a given direction.z
z
Vy
y
Vx
x
VE ˆˆˆ
∂∂−
∂∂−
∂∂−=
v
E. Field and E. potential
VE ∇−=rv The Electric field vector “potential gradient”
http://www.falstad.com/vector2de/
30
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) ZeroB) Non-Zero
+Q +QPoint X
E=0V=?
x
V
+Q +Q
xx
VE ˆ
∂∂−=
v
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?
0V 10V 20V 30VPoint A
Point B
A) |EA| > |EB|B) |EA| < |EB|C)|EA| = |EB|D)Not enough informationgiven.E) None of the above
32
CPS Question
Two charges, +Q and -Q, are fixed in space. The electric field at the point X midway between the charges is:A) ZeroB) Non-Zero
+Q -QPoint X
V=0E=?
x
V
+Q -Q
xx
VE ˆ
∂∂−=
v
33
CPS Question
+Q +Q
-Q-Q
Point P
The magnitude of the electric field at point P is:A) ZeroB) Non-Zero
The magnitude of the voltage at point P is:A) ZeroB) Non-Zero
All points on conductor must be at the same
electrical potential.
∫ ⋅−=∆b
a
rdEVv
v
Imagine point a at Voltage Va.
Since E=0 everywhere inside the conductor (no steepness), integral to
point b is always 0. ∆Vab=0.
E. Potential inside conductors
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