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Electric Potential
Symmetry, Equipotentials and Energy
What is an expert: three possibilities?
• Knows answer without waking up
• Has worked through most problems
during a previous life
• Is armed with a) logic b) intuition and
c) ability to use math
You are all well on the way to being experts
Enjoy and learn from the challenges
Today…
• Equipotentials and conductors
• Electric Potential for other symmetries
A problem of infinities
• We got V from E, now how can we get E from V
• Calculate electric field of dipole from potential
• Electric Potential Energy
– of charge in external electric field
– stored in the electric field itself (next time)
• Appendix:
– Calculate electric field of dipole from potential
Preflight 6:
Two spherical conductors are separated by a large distance.
They each carry the same positive charge Q. Conductor A has a
larger radius than conductor B.
A
B
2) Compare the potential at the surface of conductor A
with the potential at the surface of conductor B.
a) VA > VB
b) VA = VB
c) VA < VB
Potential from a charged sphere
Last time…
(where V ()  0 )
Er
Equipotential
•
•
•
•
The electric field of the charged sphere has spherical symmetry.
The potential depends only on the distance from the center of the
sphere, as is expected from spherical symmetry.
Therefore, the potential is constant along a sphere which is
concentric with the point charge. These surfaces are called
equipotentials.
Notice that the electric field is perpendicular to the equipotential
surface at all points.
Equipotentials
Defined as: The locus of points with the same potential.
•
Example: for a point charge, the equipotentials are spheres centered on
the charge.
The electric field is always perpendicular
to an equipotential surface!
Why??
Along an equipotential surface, there is NO change in V . Pick A and B on it
Then,
 
  E  dl  V  0
A
 
We can conclude then, that E  dl is zero.
B
If the dot product of the field vector and the displacement vector is zero,
then these two vectors are perpendicular, or the electric field is always
perpendicular to the equipotential surface.
Conductors
+
+
+
+
+
+
+
+
+
+
+
+
+
+
• Claim
The surface of a conductor is always an equipotential
surface (in fact, the entire conductor is an equipotential).
• Why??
If surface were not equipotential, there would be an electric
field component parallel to the surface and the charges
would move!!
Conductors
+
+
+
+
+
+
+
+
+
• Claim
+
+
+
+
+
E-field lines are always locally perpendicular to a conducting
surface
• Why??
If the E-field had a component parallel to the surface,
charges would move until the surface charge distribution
acquired a pattern that had no component of E parallel to
the surface. Remember the charge on the surface,
together with all other charges, generates E-field.
Conductors
+
+
+
+
+
+
+
+
+
• Claim
+
+
+
+
+
The local E-field magnitude at a point on the surface of a
conductor is related to the local surface charge density by
E=s/e0. (this is E right at the surface)
• Why??
Just zoom in.
Preflight 6:
B
A
3) The two conductors are now connected by a wire. How do the
potentials at the conductor surfaces compare now ?
a) VA > VB
b) VA = VB
c) VA < VB
4) What happens to the charge on conductor A after it is
connected to conductor B ?
a) QA increases
b) QA decreases
c) QA doesn’t change
Followup Question:
Two charged balls are each at the same potential V. Ball 2 is twice as
large as ball 1.
r2
r1
Ball 1
Ball 2
As V is increased, which ball will induce dielectric (air) breakdown first?
(a) Ball 1
(b) Ball 2
(c) Same Time
Smaller r  higher E  closer to breakdown
Esurface
Q
k 2
r
Q
V k
r
Esurface
V

r
Ex. V  100 kV
100 103 V
r
 0.03m  3cm
6
3 10 V/m
High Voltage
Terminals must
be big!
Other symmetries
• What about cylindrical and planar symmtries?
• E-fields are easily obtained using techniques from last
two lectures – Gauss’ Law, etc
• To get potential, integrate -E from reference point to
any point (x,y,z). Where to start?
• Charge distribution extends to infinity, so we can not
pick rinfinity for “place” where V0.
• Must pick some other point
• This point and other points form an equipotential
surface where V=0
• Discuss planar symmetry, then cylindrical symmetry
planar symmerty
s
• Let’s assume s is positive
x
• Pick x=0 plane to be V=0
• Integrate –E from this plane to
get V(x)
x=0
• Potential drops on both sides
since E // dl on both sides
V(x)
x
x
s
V(x) = 
x
2e 0
s
slope = 
2e 0
x=0
V ( x)    dxE ( x )
0
x
   dx( )
0
s
2e 0
+ for x>0
- for x<0
s
planar symmerty
s
V(x)
x
x=0
V(x)
x
x
x=d
x=0
• With more planes, just use superposition
•
V (d )  
s
d
e0
s at x=0, -s at x=d
x=0
• Still pick x=0 plane to be V=0
• Second plane has offset wedge and opposite charge
• Add potential graphs from both source planes to get final result
cylindrical symmetry
• Consider an infinite line charge on the z-axis
• Again, we can’t set V at infinity = 0, since line charge
extends to infinity
• Equipotential surfaces will be cylinders, V has cyl sym
• Pick one of these cylinders (r=a) to be V=0
Segment of
infinite
line charge
l C/m
End on view
a
V=0 here
cylindrical symmetry
End on view
r
V (r )    drE (r )
a
l r dr
l
r


ln  

2e 0 a r
2e 0  a 
a
V=0 here
V(r)
Curve shows V(r) for l pos
a
r
What happens when l neg?
Still have zero at a, and
curve flips through that
point on x-axis…
cylindrical symmetry
• More complicated cylindrically symmetrical charge
distributions can also be solved by superposition
Vtotal(r) = V1(r)+ V2(r)+ V3(r)+ V4(r)+….
• But…
Must make sure you keep V(a)=0 for all of the
potential functions you add up.
Charge on Conductors?
• How is charge distributed on the surface of a conductor?
– KEY: Must produce E=0 inside the conductor and E normal to the
surface .
Spherical example (with little off-center charge):
+ + +
+
- -- +
- +
+ -+q - +
+ - +
+ - +
+
+ + +
E=0 inside conducting shell.
charge density induced on
inner surface non-uniform.
charge density induced on
outer surface uniform
E outside has spherical
symmetry centered on spherical
conducting shell.
1
Lecture 6, ACT 1
An uncharged spherical conductor has
1A a weirdly shaped cavity carved out of
it. Inside the cavity is a charge -q.
How much charge is on the cavity wall?
(a) Less than< q
1B
(b) Exactly q
(c) More than q
How is the charge distributed on the cavity wall?
(a) Uniformly
(b) More charge closer to –q
(c) Less charge closer to -q
1C
How is the charge distributed on the outside of the sphere?
(a) Uniformly
(b) More charge near the cavity
(c) Less charge near the cavity
-q
Lecture 6, ACT 1
An uncharged spherical conductor has
1A a weirdly shaped cavity carved out of
it. Inside the cavity is a charge -q.
-q
How much charge is on the cavity wall?
(a) Less than< q
(b) Exactly q
(c) More than q
By Gauss’ Law, since E=0 inside the conductor, the
total charge on the inner wall must be q
(and therefore -q must be on the outside surface
of the conductor, since it has no net charge).
Lecture 6, ACT 1
1B
How is the charge distributed on the cavity wall?
(a) Uniformly
(b) More charge closer to -q
(c) Less charge closer to -q
-q
The induced charge will distribute itself nonuniformly to
exactly cancel E everywhere in the conductor. The surface
charge density will be higher near the -q charge.
Lecture 6, ACT 1
1C
How is the charge distributed on the outside of the
sphere?
(a) Uniformly
(b) More charge near the cavity
(c) Less charge near the cavity
-q
As in the previous example, the charge will be uniformly
distributed (because the outer surface is symmetric).
Outside the conductor the E field always points directly to
the center of the sphere, regardless of the cavity or
charge.
Note: this is why your radio, cell phone, etc. won’t
work inside a metal building!
Conductors
versus
Charges move to
cancel electric field
in the conductor
Insulators
E=0  equipotential
surface
Charge distribution
on insulator
unaffected by
external fields
Charge can sit “inside”
All charge on surface
Charges cannot
move at all
(Appendix B describes method of “images” to find
the surface charge distribution on a conductor
[only for your reading pleasure!])
Charge on Conductor Demo
• How is the charge distributed on a nonspherical conductor?? Claim largest charge
density at smallest radius of curvature.
• 2 spheres, connected by a wire, “far” apart
• Both at same potential
r
rS
But:

L
Smaller sphere
has the larger
surface charge
density !
Equipotential Example
• Field lines more closely
spaced near end with most
curvature – higher E-field
• Field lines ^ to surface
near the surface (since
surface is equipotential).
• Near the surface,
equipotentials have similar
shape as surface.
• Equipotentials will look
more circular (spherical) at
large r.
Electric Dipole Equipotentials
•First, let’s take a look at the equipotentials:
Electric Fish
Some fish have the ability to
produce & detect electric fields
• Navigation, object detection,
communication with other electric
fish
• “Strongly electric fish” (eels) can
stun their prey
Dipole-like equipotentials
More info: Prof. Mark Nelson,
Beckman Institute, UIUC
Black ghost knife fish
-Electric current flows down the voltage gradient
-An object brought close to the fish alters the
pattern of current flow
E from V?
• We can obtain the electric field E from the
potential V by inverting our previous relation
between E and V:

r

r  xˆ dx
V
V+dV

dV   E  xˆ dx   E x dx
• Expressed as a vector, E is the negative gradient of V
• Cartesian coordinates:
• Spherical coordinates:
Preflight 6:
This graph shows the electric
potential at various points
along the x-axis.
8) At which point(s) is the electric field zero?
A
B
C
D
E from V: an Example
• Consider the following electric potential:
• What electric field does this describe?
... expressing this as a
vector:
• Something for you to try:
Can you use the dipole potential to obtain the
dipole field? Try it in spherical coordinates ...
you should get (see Appendix):
2
Lecture 6, ACT 2
2
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
(c)
Ex < 0
Lecture 6, ACT 2
2
The electric potential in a region of space is given by
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
We know V(x) “everywhere”
To obtain Ex “everywhere”, use
(c)
Ex < 0
The Bottom Line
If we know the electric field E everywhere,

allows us to calculate the potential function V everywhere (keep
in mind, we often define VA = 0 at some convenient place)
If we know the potential function V everywhere,
allows us to calculate the electric field E everywhere
• Units for Potential! 1 Joule/Coul = 1 VOLT
Electric Potential Energy
• The Coulomb force is a CONSERVATIVE force
(i.e., the work done by it on a particle which
moves around a closed path returning to its
initial position is ZERO.)
• Therefore, a particle moving under the influence
of the Coulomb force is said to have an
electric potential energy defined by:
this “q” is the “test charge”
in other examples...
• The total energy (kinetic + electric potential) is then
conserved for a charged particle moving under the
influence of the Coulomb force.
3
Lecture 6, ACT 3
3A
Two test charges are brought
separately to the vicinity of a
positive charge Q.
Q
r
q
– charge +q is brought to pt A, a
distance r from Q.
– charge +2q is brought to pt B, a
Q
distance 2r from Q.
– Compare the potential energy of q
(UA) to that of 2q (UB):
(a) UA < UB
3B
(b) UA = UB
A
2r
2q
(c) UA > UB
• Suppose charge 2q has mass m and is released from
rest from the above position (a distance 2r from Q).
What is its velocity vf as it approaches r = ?
(a)
(b)
B
(c)
Lecture 6, ACT 3
3A
• Two test charges are brought
separately to the vicinity of positive
charge Q.
– charge +q is brought to pt A, a
distance r from Q.
– charge +2q is brought to pt B, a
distance 2r from Q.
Q
r
q
A
Q
2r
– Compare the potential energy of q (UA)
to that of 2q (UB):
(a) UA < UB
(b) UA = UB
(c) UA > UB
• The potential energy of q is proportional to Qq/r.
• The potential energy of 2q is proportional to Q(2q)/(2r).
• Therefore, the potential energies UA and UB are EQUAL!!!
2q
B
Lecture 6, ACT 3
3B
• Suppose charge 2q has mass m and is released from
rest from the above position (a distance 2r from Q).
What is its velocity vf as it approaches r = ?
(a)
(b)
(c)
• What we have here is a little combination of 111 and 112.
• The principle at work here is CONSERVATION OF ENERGY.
• Initially:
• The charge has no kinetic energy since it is at rest.
• The charge does have potential energy (electric) = UB.
• Finally:
• The charge has no potential energy (U  1/R)
• The charge does have kinetic energy = KE
Energy Units
MKS:
U = QV
1 coul-volt
for particles (e, p, ...)
1 eV
= 1 joule
= 1.6x10-19 joules
Accelerators
• Electrostatic: Van de Graaff
electrons  100 keV ( 105 eV)
• Electromagnetic: Fermilab
protons  1TeV ( 1012 eV)
Summary
• Physically, V is what counts
• The place where V=0 is “arbitrary” (at infinity for
sperically symmetric charges)
• At some specific place for charge distributions that
extend to infinity
• Conductors are equipotentials and E is always
perpendicular to the surface of a conductor
• Find E from V:
• Potential Energy


E   V
U  qV
• Next time  capacitors
Appendix A: Electric Dipole
The potential is much easier to
calculate than the field since it
is an algebraic sum of 2 scalar
terms.
z
+q
a
r1
r
q
a
r2-r1
-q
• Rewrite this for special case r>>a:

Now we can use this potential to calculate the E
field of a dipole (after drawing a picture)
(remember how messy the direct calculation was?)
r2
Appendix A: Electric Dipole
z
+q
aq
a
-q
• Calculate E in spherical coordinates:
the dipole
moment

r1
r
r2
Appendix A: Dipole Field
y=
z
+q
a q
a
Etot
r
E
q
0
Er
-q
0
/



/
x=
Sample Problem
• Consider the dipole shown at the
right.
– Fix r = r0 >> a
z
+q
– Define qmax such that the polar
component of the electric field has its
maximum value (for r = r0).
a q
a
What is qmax?
-q
(a) qmax = 0
(b) qmax = 45
r1
r
(c) qmax = 90
• The expression for the electric
field of a dipole (r >> a) is:
• The polar component of
E is maximum when sinq is maximum.
• Therefore, Eq has its maximum value when q = 90.
r2
Appendix B: FYI: Induced charge distribution on
conductor via “method of images”
• Consider a source charge
brought close to a
conductor:
+
-
+
+
+
+
• Charge distribution
- +
“induced” on conductor
by source charge:
• Induced charge distribution is “real” and sources Efield so that the total is zero inside conductor!
– resulting E-field is sum of field from source charge
+
and induced charge distribution
+
-
– E-field is locally perpendicular to surface
+
+
+
• With enough symmetry, can solve for s on conductor
– how? Gauss’ Law



Enormal ( rsurface )  E ( rsurface ) 
+
+
- +
- +
- +

s ( rsurface )
eo
Appendix B: (FYI) Induced charge distribution on
conductor via “method of images”
• Consider a source charge brought
close to a planar conductor:
-
• Charge distribution “induced” on
conductor by source charge
– conductor is equipotential
+
-
-
– E-field is normal to surface
– this is just like a dipole
• Method of Images for a charge (distribution) near a flat
conducting plane:
– reflect the point charge through the surface and put a
charge of opposite sign there
– do this for all source charges
– E-field at plane of symmetry - the conductor surface 
 
s ( rsurface )

determines s.
Enormal ( rsurface )  E ( rsurface ) 
eo