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Electric Potential Energy
and Potential
q0
A
E
B
d = 14.6 •10-15m
235
Unucleus + n
(92 protons)
Ba Kr
(56 p) (36 p)
Today…
• Conservative Forces and Energy Conservation
– Total energy is constant and is sum of kinetic and potential
• Electric Potential Energy
• Introduce Concept of Electric Potential
– A property of the space and sources as is the Electric Field
– Potential differences drive all biological & chemical reactions,
as well as all electric circuits
• Path-independence of Potential
• Calculating Electric Potentials: N point charges
• Appendices:
– A. Proof of path-independence
– B. Electric dipole potential
Conservation of Energy of a particle from Phys 211
• Kinetic Energy (K) K  1 mv 2
– non-relativistic
2
• Potential Energy (U) U ( x , y , z )
– determined by force law
• for Conservative Forces: K+U is constant
– total energy is always constant
• examples of conservative forces
– gravity; gravitational potential energy
– springs; coiled spring energy (Hooke’s Law): U(x)=kx2
– electric; electric potential energy (today!)
• examples of non-conservative forces (heat)
– friction
– viscous damping (terminal velocity)
Example: Gravitational Force is conservative
(and attractive)
• Consider a comet in a highly elliptical orbit
U(r)
pt 2
pt 1
0
U(r1)
GMm
U (r )  
r
• At point 1, particle has a lot of potential
energy, but little kinetic energy
• At point 2, particle has little potential
energy, but a lot of kinetic energy
More
potential
energy
U(r2)
Less
potential
energy
Total energy = K + U
is constant!
Electric forces are conservative, too
• Consider a charged particle traveling
through a region of static electric field:
+
• A negative charge is attracted to
the fixed positive charge
• negative charge has more
potential energy and less kinetic
energy far from the fixed positive
charge, and…
• more kinetic energy and less
potential energy near the fixed
positive charge.
• But, the total energy is conserved
• We will now discuss electric
potential energy and the
electrostatic potential….
Electric potential energy
• Imagine two positive charges, one with charge Q1,
the other with charge Q2:
Q2
Q1
• Initially the charges are very far apart, so we say
that the initial energy Ui of interaction is zero
(we are free to define the energy zero somewhere).
• If we want to push the particles together, this will
require work (since they want to repel).
Q1
Q2
 the final energy Uf of the system will increase by
the same amount:
DU = Uf – Ui = DW
Electric potential energy, cont.
• Pretend Q1 is fixed at the origin.
• What is the work required to move Q2, initially at
infinity, a distance r away? Q1
Q2
r
• Remember – work is force times distance:
• What if Q2 were negative (but Q1 still positive)?
• Then the work “required” by us would be negative
 the charges would like to come together.
• In this case the final energy is negative!
Particles will move to minimize
the final potential energy.
Electric Potential Energy
•
In addition to discussing the energy of a “test” charge in a Coulomb
field, we can speak of the electric potential energy of the field itself!
•
Reasons?
– Work had to be done to assemble the charges (from infinity)
into their final positions.
– This work is the potential energy of the field.
– The potential energy of a system of N charges is defined to be
the algebraic sum of the potential energy for every pair of
charges.
•
This theme continues in the course with E in capacitors and B in
inductors – these devices store electric and magnetic energy.
•
We will start with a couple of example calculations…
Electric Potential Energy
• Example 1: Nuclear Fission
Unucleus + n
235
(92 protons)
d = 14.6 •10-15m
Ba Kr
(The electrons
are all
“far” away)
(56 p) (36 p)
What is the potential energy of the two nuclei?
1.6  1019 C
(9  10 )(56e)(36e)
8
kq1q2
e

e(2

10
V) = 200 MeV
U

15
d
14.6  10 m
Compare this to the typical energy released in a chemical reaction, ~10eV.
By allowing the two fragments to fly apart, this potential energy 
kinetic energy  heat  drives a turbine  generate electricity.
9
(What holds the protons together in the nucleus to begin with? Gluons!
The “strong force”  very short range, very very strong!)
Electric Potential Energy
-q
• Example 2: What is the potential energy of
this collection of charges?
2d
d
+2q
d
-q
Step 1: Bring in +2q from infinity. This costs nothing.
Step 2: Bring in one -q charge. The force is
attractive! The work required is negative:
U
(2q)(q)
4 0 d
Step 3: Bring in 2nd -q charge. It is attracted to the +2q, but repelled
from the other -q charge. The total work (all 3 charges) is
 (2q)(q) (2q)(q) (q)(q) 
q2 
1 
U 




4




4


d
4


d
4


d
4  0 2d 
2

0
0
0 
1
A negative amount of work was required to bring these charges
from infinity to where they are now (i.e., the attractive forces
between the charges are larger than the repulsive ones).
Lecture 5, ACT 1
• Consider the 3 collections of point charges shown
below.
– Which collection has the smallest potential energy?
d
-Q
-Q
d
d
d
-Q
+Q
-Q
d
d
d
-Q
+Q
(a)
(b)
+Q
d
+Q
(c)
Lecture 5, ACT 1
• Consider the 3 collections of point charges shown
below.
– Which collection has the smallest potential energy?
d
-Q
-Q
+Q
d
d
d
d
d
d
-Q
-Q
-Q
+Q
(a)
(b)
+Q
d
+Q
(c)
• We have to do positive work to assemble the charges in (a) since
they all have the same charge and will naturally repel each other. In
(b) and (c), it’s not clear whether we have to do positive or negative
work since there are 2 attractive pairs and one repulsive pair.
(a)
Q2
U  3
40 d
1
(b) (c)
(b)
Q2
U 
40 d
1
1
(c) U   4
0
Q2
2d
U
0
(a)
Preflight 5:
A
Two charges which are equal in
magnitude, but opposite in sign are
placed at equal distances from
point A.
2) If a third charge is added to the system and placed at point A,
how does the electric potential energy of the charge collection
change ?
a) increases
b) decreases
c) doesn’t change
Electric potential
• Consider that we have three charges fixed in space.
• The potential energy of an added test
charge q0 at point P is just
U of
q0 at P
 Q1
Q3 
Q2
 q0  k  k
k

 r
r2 p
r3 p 
 1p
Q1
Q2
r1p
r2p
• Note that this factors: q0 x (the effects of all other charges)
Q3
r3p
q0
• Just as we previously defined the electric field as the force/charge,
we now define the electric potential as the potential energy/charge:
V(x,y,z) = U(x,y,z)/q0
(U = qV)
• U depends on what qo is, but V is independent of qo (can be + or -)
•
Units of electric potential are volts: 1 V = 1 J/C
• V(x,y,z) is a scalar field, defined everywhere in space.
Electric potential difference, in terms of E
• Suppose charge q0 is moved from
pt A to pt B through a region of
space described by electric field E.
Fwe supply = -Felec
Felec
2
q0
A
E
B
• Force on the charge due to E  work WAB≡WAB will have to be
done to accomplish this task:

• To get a positive test charge from lower potential to higher
potential you need to invest energy - you need to do work
• The overall sign of this: A positive charge would “fall” from
a higher potential to a lower one
• If a positive charge moves from high to low potential, it can
do work on you; you do “negative work” on the charge
Preflight 5:
A
E
C
B
4) Points A, B, and C lie in a uniform electric field. What is the
potential difference between points A and B?
ΔVAB = VB - VA
a) ΔVAB > 0
b) ΔVAB = 0
c) ΔVAB < 0
5) Point C is at a higher potential than point A.
True
False
Preflight 5:
9) A positive charge is released from rest in a region of
electric field. The charge moves:
a) towards a region of smaller electric potential
b) along a path of constant electric potential
c) towards a region of greater electric potential
Preflight 5:
13) If you want to move in a region of electric field without
changing your electric potential energy. How would you choose
the direction ?
You would have to move
perpendicular to the field
if you wish to move without changing
electric potential.
Lecture 5, ACT 2
• A single charge ( Q = -1mC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
– What is the sign of the potential difference
between A and B? (Δ VAB  VB - VA )
(a) ΔVAB < 0 (b) Δ VAB = 0
B

-1mC
(c) Δ VAB > 0
A

x
Lecture 5, ACT 2
• A single charge ( Q = -1mC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
B

– What is the sign of the potential difference
between A and B? (Δ VAB  VB - VA )
(a) ΔVAB < 0 (b) Δ VAB = 0
-1mC
A

x
(c) Δ VAB > 0
•The simplest way to get the sign of the potential difference is to imagine
placing a positive charge at point A and determining which way it would move.
Remember that a positive charge will always “fall” to lower potential.
•A positive charge at A would be attracted to the -1mC charge;
therefore NEGATIVE work would be done to move the charge from A to B.
•You could also determine the sign directly from the definition:
Since
, ΔVAB <0 !!
Δ VAB is Independent of Path
q0
E
A
B
• The integral is the sum of the tangential (to the path) component
of the electric field along a path from A to B.
• This integral does not depend upon the exact path chosen to
move from A to B.
• Δ VAB is the same for any path chosen to move from A to B
(because electric forces are conservative).
Does it really work?
• Consider case of constant field:
– Direct: A - B
B
h
A
q
C
E
r
dl
• Long way round: A - C – B
3
• So here we have at least one example of a case in which the
integral is the same for BOTH paths.
• In fact, it works for all paths  see Appendix A.
Preflight 5:
A
E
C
B
7) Compare the potential differences between points A to C
and points B to C.
a) VAC > VBC
b) VAC = VBC
c) VAC < VBC
Lecture 5, ACT 3
3
A positive charge Q is moved from A to B
along the path shown by the arrow. What
is the sign of the work done to move the
charge from A to B?
(a) WAB < 0
(b) WAB = 0
(c) WAB > 0
A
B
Lecture 5, ACT 3
A positive charge Q is moved from A to B
along the path shown by the arrow. What
is the sign of the work done to move the
charge from A to B?
3
(a) WAB < 0
•
(b) WAB = 0
A
B
(c) WAB > 0
A direct calculation of the work done to move a positive
charge from point A to point B is not easy.
• Neither the magnitude nor the direction of the field is
constant along the straight line from A to B.
•
But, you DO NOT have to do the direct calculation.
•
•
Remember: potential difference is INDEPENDENT OF
THE PATH!!
Therefore we can take any path we wish.
Choose a path along the arc of a circle centered at
the charge. Along this path
at every point!!
Electric Potential: where is it zero?
• So far we have only considered potential differences.
• Define the electric potential of a point in space as the
potential difference between that point and a reference
point.
• a good reference point is infinity ... we often set V = 0
• the electric potential is then defined as:
• for a point charge at origin, integrate in from infinity along some
r
axis, e.g., the x-axis
• here “r” is distance to origin
V (r )  V ()    dl  E

Potential from a point charge q
line
integral
Potential from N charges
The potential from a collection of N charges is just
the algebraic sum of the potential due to each charge
separately (this is much easier to calculate than the
net electric field, which would be a vector sum).
Q1
Q2
r1p
r2p

Q3
r3p
Vat P
1 Q1
1 Q2
1 Q3



4o r1 p 4o r2 p 4o r3 p
P
Appendix B: Calculation of potential from a dipole
4
Preflight 5:
+5 μC
-3 μC
A
11) Two charges q1 = + 5 μC, q2 = -3μC are placed at equal distances
from the point A. What is the electric potential at point A?
a) VA < 0
b) VA = 0
c) VA > 0
Lecture 5, ACT 4
Which of the following charge distributions produces V(x) = 0
for all points on the x-axis? (we define V(x)  0 at x =  )
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
(a)
+2mC
-2mC
x
-1mC
-2mC
(b)
x
-1mC
(c)
+1mC
Lecture 5, ACT 4
Which of the following charge distributions produces V(x) = 0
for all points on the x-axis? (we define V(x)  0 at x =  )
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
(a)
+2mC
-2mC
x
-1mC
-2mC
(b)
x
-1mC
(c)
+1mC
The key here is to realize that to calculate the total potential at a point, we
must only make an ALGEBRAIC sum of the individual contributions.
Therefore, to make V(x)=0 for all x, we must have the +Q and -Q
contributions cancel, which means that any point on the x-axis must be
equidistant from +2mC and -2mC and also from +1mC and -1mC.
This condition is met only in case (a)!
Summary
•Potential energy stored in a static charge distribution
–work we do to assemble the charges
•Electric potential energy of a charge in the presence of
a set of source charges
–potential energy of the test charge equals the potential from
the sources times the test charge: U = qV
• If we know the electric field E,
allows us to calculate the potential function V
everywhere (define VA = 0 above)
• Potential due to n charges:
Appendix A: Independent of Path?
•Consider any path from A to B as
being made up of a succession of arc
plus radial parts as above. The work
along the arcs will always be 0
(since
), leaving just the
sum of the radial parts, e.g.:
B
r2
A
r1
q
In going from rA to rB, all terms with intermediate radii will
cancel, leaving just the initial and final radii:
Therefore it's general!
Appendix B: Electric Dipole
The potential is much easier to
calculate than the field since it is
an algebraic sum of 2 scalar
terms.
z
r1
+q
a
r
q
a
r2-r1
-q
• Rewrite this for special case r>>a:

In Lecture 7 you will see how to use this potential
to calculate the E field of a dipole
(remember how messy the direct calculation was?)
r2