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Electric Potential
Equipotentials and Energy
Last time: Potential energy and potential
• 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
• Electrostatic potential at any point in space
– sources give rise to V(r):
allows us to calculate the potential V
everywhere if we know the electric field
E (define VA = 0 somewhere)
Potential from charged spherical shell
(See Appendix A for a more complex example)
• E-field (from Gauss' Law)
•
r < a: Er = 0
•
r >a:
• Potential
• r > a:
• r < a:
V
Q
4pe0 a
1
Q
Er =
4pe 0r 2
Q
4pe0 r
a
a
a
r
What does the result mean?
• This is the plot of the radial component of the electric field
of a charged spherical shell:
Er
Notice that inside the shell, the electric
field is zero. Outside the shell, the electric
field falls off as 1/r2.
The potential for r > a is given by the
integral of Er. This integral is simply the
area underneath the Er curve.
a R
r
V
Q
4pe0 a
Q
4pe0 r
a
a R
r
a
1
1
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
the inner surface of the spherical shell?
(a)
(b)
(c)
a
Q
b
Eout
1
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
the inner surface of the spherical shell?
(a)
(b)
a
Q
b
(c)
• How to start?? The only thing we know about the potential is
its definition:
• To calculate Va, we need to know the electric field E
• Outside the spherical shell:
• Apply Gauss’ Law to sphere:
• Inside the spherical shell:
E=0
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 the surface, there is NO change in V (it’s an equipotential!)
Therefore,
 
  E  dl  V  0
B
A
 
We can conclude then, that E  dl is zero.
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.
Electric Dipole 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
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!!
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
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.
2
An uncharged spherical conductor has
2A 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
2B
(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
2C
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
An uncharged spherical conductor has
2A 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).
2B
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.
2C
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.
Sparks
• High electric fields can ionize nonconducting
materials (“dielectrics”)
Dielectric
Insulator
Conductor
Breakdown
• Breakdown can occur when the field is greater than
the “dielectric strength” of the material.
– E.g., in air, Emax  3  106 N/C  3  106 V/m  30 kV/cm
What is ΔV?
d  2mm
Ex.
V  Emax  d
Vdoorknob
Vfinger
Arc discharge equalizes
the potential
 30 kV/cm • 0.2 cm
 6 kV
Note: High humidity can also bleed the charge off  reduce ΔV.
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 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!
Conservation of 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
Preflight 6:
A
E
C
B
6) If a negative charge is moved from point A to point B, its
electric potential energy
a) increases
b) decreases
c) doesn’t change
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)
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
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
• If we know the electric field everywhere, we can
calculate the potential, e.g.,
• The place where V=0 is “arbitrary” (often at infinity)
• Physically, V is what counts
• Equipotential surfaces are surfaces where the
potential is constant
• Conductors are equipotentials
• “Breakdown” can occur if the electric field exceeds
the “dielectric strength”
• Next time  capacitors
Appendix A: Electrical potential examples
Calculate the potential V(r)
at the point shown (r<a)
I
II
uncharged
conductor
III
IV
r a
solid sphere
with total
charge Q
c
b
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r < a)
• Where do we know the potential,
and where do we need to know it?
V=0 at r =  ...
III
IV
uncharged
conductor
r a
we need r < a ...
• Determine E(r) for all regions
in between these two points
I
II
c
b
solid sphere
with total
charge Q
• Determine V for each region by integration
... and so on ...
• Check the sign of each potential difference V
V > 0 means we went “uphill”
V < 0 means we went “downhill”
(from the point of view
of a positive charge)
Calculating Electric Potentials
Calculate the potential V(r)
at the point shown (r < a)
• Look at first term:
III
IV

dl
• Line integral from infinity to c has to be
positive, pushing against a force:
Line integral is going “in”
which is just the opposite of
what usually is done
- controlled by limits
• What’s left?
I
II
r a

E
c
1  Q
1 Q



4pe o  r    4pe o c
c
b
Calculating Electric Potentials
I
II
Calculate the potential V(r)
at the point shown (r < a)
• Look at third term:

dl
• Line integral from b to a, again has to be
positive, pushing against a force:
Line integral is going “in”
which is just the opposite
of what usually is done
a
1  Q
1 Q(b  a )



4pe o  r   b 4pe o
ab
- controlled by limits
• What’s left?
Previous slide
we have calculated this already
III
IV

E
r a
c
b
Calculating Electric Potentials
I
II
Calculate the potential V(r)
at the point shown (r < a)
• Look at last term:

dl
III
IV

E
r a
c
b
• Line integral from a to r, again has to be
positive, pushing against a force.
• But this time the force doesn’t vary the
same way, since “r ’ ” determines the
amount of source charge
Q
r3
a
3
Q
This is the charge
that is inside “r”
and sources field
• What’s left to do?
• ADD THEM ALL UP!
• Sum the potentials
r
1  Q r 
1 Q  r2 


 3
1  2 

4pe o  a 2  a 4pe o 2a  a 
2
Calculating Electric Potentials
I
II
Calculate the potential V(r)
at the point shown (r < a)
• Add up the terms from I, III and IV:
I
III

dl
III
IV

E
r a
c
b
IV

dl
Potential increase
from moving into
the sphere
The potential difference
from infinity to a if the
conducting shell
weren’t there
An adjustment to
account for the fact
that the conductor
is an equipotential,
V= 0 from c → b
Calculating Electric Potentials
Summary
The potential as a function of r for all 4
regions is:
I
r > c:
II
b < r < c:
III
a < r < b:
IV
r < a:
I
II
III
IV
r a
c
b
Let’s try some numbers
Q = 6m C
a = 5cm
b = 8cm
c = 10cm
I
II
III
IV
r a
I
r > c: V(r = 12cm) = 449.5 kV
II
b < r < c: V(r = 9cm) = 539.4 kV
III
a < r < b: V(r = 7cm) = 635.7 kV
IV
r < a: V(r = 3cm) = 961.2 kV
c
b
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



E normal ( 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
What does grounding do?
1. Acts as an “infinite” source or sink of
charge.
2. The charges arrange themselves in such a
way as to minimize the global energy (e.g.,
E0 at infinity, V0 at infinity).
3. Typically we assign V = 0 to ground.