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Quiz on the math needed today
B
What is the result of this integral:
B
1
1
A x 2  dx   x
B
A
1 1
 
A B
1
A x 2  dx
If y ( x) is known and y ( x)   z ( x)dx, what is z ( x)?
z ( x) 
dy ( x )
dx


If y (r ) is known and y (r ) 

 
z (r )  y (r ),
  
 
 z(r)ds , what is z(r)?
line integral
In a Cartesian coordinate system  
ˆ ˆ  ˆ
i
j k
x
x
x
Chapter 25
Electric Potential
A review of gravitational
potential
A
When object of mass m is on ground level B, we define that
it has zero gravitational potential energy. When we let go of
this object, if will stay in place.
When this object is moved to elevation A, we say that it has
gravitational potential energy mgh. h is the distance from B
to A. When we let go of this object, if will fall back to level B.
h
mg
B
When this object is at elevation A, it has gravitational
potential energy UA-UB= mgh. UA is the potential energy at
point A with reference to point B. When the object falls from
level A to level B, the potential energy change: ΔU =UB-UA
The gravitational force does work and causes the potential
energy change: W = mgh = UA-UB= -ΔU
We also know that the gravitational force is conservative:
the work it does to the object only depends on the two
levels A and B, not the path the object moves.
Introduction of the electric potential, a
special case: the electric field is a constant.
When a charge q0 is placed inside an electric field, it
experiences a force from
 thefield:
F  q0 E
When the charge is released, the field moves it from A
to B, doing work:
 
 
W  F  d  q0 E  d  q0 Ed
If we define the electric potential energy of the
charge at point A UA and B UB, then:
W  q0 Ed  U A  U B  U
If we define UB=0, then UA= q0Ed is the electric
potential energy the charge has at point A. We can
also say that the electric field has an electric potential
at point A. When a charge is placed there, the charge
acquires an electric potential energy that is the charge
times this potential.
Electric Potential Energy, the
general case
When a charge is moved from point A to point B in an
electric field, the charge’s electric potential energy inside
this field is changed from UA to UB: U  U B  U A
When the motion is caused by the electric field force on
the charge, the work this force does to the charge cause
the change of its electric potential energy, so:W  U


The force on the charge is: F  q0 E
So we have this final formula for electric potential energy
and the work the field force does to the charge:
 
 U  U A  U B  W (of the field force)   q 0 E  d s
B
A
Electric Potential Energy, final
discussion

Electric force is conservative. The line
integral does not depend on the path from A
to B; it only depends on the locations of A and
B.
 
 U  U A  U B   q 0 E  d s
B
A
Line integral paths
A
B
The electric potential energy of a
charge q0 in the field of a charge Q?
Reference point:
We usually define the electric potential of a
point charge to be zero (reference) at a point
that is infinitely far away from the point charge.
Applying this formula:
 
 U  U A  U B   q 0 E  d s
B
q0
R
Q
A
Where point A is where the charge q0 is, point B is
infinitely far away.



Q
E  k e 2 ˆr, d s  dr
r
   
Q
so E  d s  E  dr  k e 2 dr
r

Q
Q
and  k e 2 dr  k e
R
r
R
So the final answer is
qQ
U ( R)  k e 0
R
And the result is a scalar!
Electric Potential, the
definition

The potential energy per unit charge, U/qo, is the
electric potential



The potential is a characteristic of the field only
 The potential energy is a characteristic of the charge-field
system
The potential is independent of the value of qo
The potential has a value at every point in an electric field

The electric potential is

As in the potential energy case, electric potential also
needs a reference. So it is the potential difference ΔV
that matters, not the potential itself, unless a
reference is specified (then it is again ΔV).
U
V
qo
Electric Potential,
the formula

The potential is a scalar quantity


Since energy is a scalar
As a charged particle moves in an electric
field, it will experience a change in potential
 
 V  V A  VB (often the reference)   E  d s
B
A
Potential Difference in a
Uniform Field
The equations for electric potential can be
simplified if the electric field is uniform:
  B   
 V  V A  VB   E  d s  E  d s  E  d
B
When:
A
A
 


E  d  0, i.e., E and d the same direction,
 V  V A  VB  0, or V A  VB
This is to say that electric field lines always
point in the direction of decreasing electric
potential
Electric Potential, final
discussion



The difference in potential is the
meaningful quantity
We often take the value of the potential to
be zero at some convenient point in the
field
Electric potential is a scalar characteristic
of an electric field, independent of any
charges that may be placed in the field
Electric Potential, electric
potential energy and Work
When there is electric field, there is electric potential V.
When a charge q0 is in an electric field, this charge
has an electric potential energy U in this electric field:
U = q0 V.
When this charge q0 is move by the electric field force,
the work this field force does to this charge equals the
electric potential energy change -ΔU:
W = -ΔU = -q0 ΔV.
Units


The unit for electric potential energy is the unit for energy joule
(J).
The unit for electric potential is volt (V):
1 V = 1 J/C
 This unit comes from U = q0 V (here U is electric potential energy,
V is electric potential, not the unit volt)
 It takes one joule of work to move a 1-coulomb charge through a
potential difference of 1 volt
 
But from  V  E  d s

B

A
We also have the unit for electric potential as 1 V = 1 (N/C)m

So we have that 1 N/C (the unit of E) = 1 V/m
 This indicates that we can interpret the electric field as a
measure of the rate of change with position of the electric
potential
Electron-Volts, another unit often used in
nuclear and particle physics


Another unit of energy that is commonly used in
atomic and nuclear physics is the electron-volt
One electron-volt is defined as the energy a
charge-field system gains or loses when a charge of
magnitude e (an electron or a proton) is moved
through a potential difference of 1 volt

1 eV = 1.60 x 10-19 J
Direction of Electric Field,
energy conservation




As pointed out before, electric field
lines always point in the direction
of decreasing electric potential
So when the electric field is
directed downward, point B is at a
lower potential than point A
When a positive test charge moves
from A to B, the charge-field
system loses potential energy
through doing work to this charge
Where does this energy go?
It turns into the kinetic energy of the object
(with a mass) that carries the charge q0.
PLAY
ACTIVE FIGURE
2502
Equipotentials = equal potentials


Points B and C are at a
lower potential than point A
Points B and C are at the
same potential


All points in a plane
perpendicular to a uniform
electric field are at the same
electric potential
The name equipotential
surface is given to any
surface consisting of a
continuous distribution of
points having the same
electric potential
Charged Particle in a Uniform
Field, Example
Question: a positive charge (mass
m) is released from rest and moves
in the direction of the electric field.
Find its speed at point B.
Solution: The system loses potential
energy: -ΔU=UA-UB=qEd
The force and acceleration are in
the direction of the field
Use energy conservation to find its
speed:
1 2
mv  qEd
2
2qEd
v
m
Potential and Point Charges


A positive point charge
produces a field
directed radially
outward
The potential difference
between points A and B
will be
 1 1
VB  VA  keq   
 rB rA 
Potential and Point Charges,
cont.



The electric potential is independent of the
path between points A and B
It is customary to choose a reference
potential of V = 0 at rA = ∞
Then the potential at some point r is
q
V  ke
r
Electric Potential of a Point
Charge


The electric potential in
the plane around a
single point charge is
shown
The red line shows the
1/r nature of the
potential
Electric Potential with Multiple
Charges

The electric potential due to several point
charges is the sum of the potentials due to
each individual charge


This is another example of the superposition
principle
The sum is the algebraic sum

qi
V  ke 
i ri
V = 0 at r = ∞
Immediate application:
Electric Potential of a Dipole
1
1
V  V  V  k e q(      )
r - r
r - r
Work on the board to prove it.


The graph shows the
potential (y-axis) of an
electric dipole
The steep slope between
the charges represents the
strong electric field in this
region
Potential Energy of Multiple
Charges


Consider two charged
particles
The potential energy of
the system is
q1q2
U  ke
r12

Use the active figure to
move the charge and see
the effect on the potential
energy of the system
2509
PLAY
ACTIVE FIGURE
More About U of Multiple
Charges


If the two charges are the same sign, U is
positive and external work (not the one from
the field force) must be done to bring the
charges together
If the two charges have opposite signs, U is
negative and external work is done to keep
the charges apart
U with Multiple Charges, take 3 as an
example


If there are more than
two charges, then find
U for each pair of
charges and add them
For three charges:
 q1q2 q1q3 q2q3 
U  ke 



r
r
r
13
23 
 12

The result is independent
of the order of the
charges
Finding E From V
This is straight forward
 




From  V   E  d s We have E  V  ( i  j  k )V
x
x
x
dV
If E is one dimensional (say along the x-axis) E x  
dx
If E is only a function of r (the point charge case):
 
dV

E (r )  
, and E(r)  E (r )r
dr
Find V for an Infinite Sheet of
Charge






a constant
2

0

We know the E 

From V   E  d s
We have V  Ed
The equipotential lines are the
dashed blue lines
The electric field lines are the
brown lines
The equipotential lines are
everywhere perpendicular to the
field lines
d
E and V for a Point Charge



The equipotential lines
are the dashed blue
lines
The electric field lines
are the brown lines
The equipotential lines
are everywhere
perpendicular to the
field lines
E and V for a Dipole



The equipotential lines
are the dashed blue
lines
The electric field lines
are the brown lines
The equipotential lines
are everywhere
perpendicular to the
field lines
When you use a computer (program) to calculate
electric Potential for a Continuous Charge
Distribution:

Consider a small
charge element dq


Treat it as a point charge
The potential at some
point due to this charge
element is
dq
dV  ke
r
V for a Continuous Charge
Distribution, cont.

To find the total potential, you need to
integrate to include the contributions from all
the elements
dq
V  ke 
r

This value for V uses the reference of V = 0 when
P is infinitely far away from the charge
distributions
V for a Uniformly Charged
Ring

P is located on the
perpendicular central
axis of the uniformly
charged ring

The ring has a radius a
and a total charge Q
keQ
dq
V  ke 

r
a2  x 2
V for a Uniformly Charged Disk


The ring has a radius R
and surface charge
density of σ
P is along the
perpendicular central
axis of the disk

V  2πkeσ  R 2  x 2



1
2

 x

V for a Finite Line of Charge

A rod of line ℓ has a
total charge of Q and a
linear charge density of
λ
V
keQ
  a2 
ln 

a

2




Prove that V is everywhere the same on a
charged conductor in equilibrium





Inside the conductor, because
is E 0, E  ds  0 , so ΔV=0
On the surface, consider two
points on the surface of the
charged conductor as shown
E is always perpendicular to
the displacement ds
Therefore, E  ds  0
Therefore, the potential
difference between A and B is
also zero
Summarize on potential V of a
charged conductor in equilibrium

V is constant everywhere on the surface of a
charged conductor in equilibrium



ΔV = 0 between any two points on the surface
The surface of any charged conductor in
electrostatic equilibrium is an equipotential surface
Because the electric field is zero inside the
conductor, we conclude that the electric potential is
constant everywhere inside the conductor and equal
to the value at the surface
E Compared to V



The electric potential is a
function of r
The electric field is a
function of r2
The effect of a charge on
the space surrounding it:
 The charge sets up a
vector electric field which
is related to the force
 The charge sets up a
scalar potential which is
related to the energy
Cavity in a Conductor



Assume an irregularly
shaped cavity is inside
a conductor
No charges are inside
the cavity
The electric field inside
the conductor must be
zero (can you prove
that?)
Cavity in a Conductor, cont


The electric field inside does not depend on the
charge distribution on the outside surface of the
conductor
For all paths between A and B,
B
VB  VA   E  ds  0
A

A cavity surrounded by conducting walls is a fieldfree region as long as no charges are inside the
cavity
Millikan Oil-Drop Experiment –
Experimental Set-Up
PLAY
ACTIVE FIGURE
Millikan Oil-Drop Experiment



Robert Millikan measured e, the magnitude of
the elementary charge on the electron
He also demonstrated the quantized nature
of this charge
Oil droplets pass through a small hole and
are illuminated by a light
Oil-Drop Experiment, 2


With no electric field
between the plates, the
gravitational force and
the drag force (viscous)
act on the electron
The drop reaches
terminal velocity with
FD  mg
Oil-Drop Experiment, 3

When an electric field is
set up between the
plates


The upper plate has a
higher potential
The drop reaches a
new terminal velocity
when the electrical
force equals the sum of
the drag force and
gravity
Oil-Drop Experiment, final


The drop can be raised and allowed to fall
numerous times by turning the electric field on and
off
After many experiments, Millikan determined:




q = ne where n = 0, -1, -2, -3, …
e = 1.60 x 10-19 C
This yields conclusive evidence that charge is
quantized
Use the active figure to conduct a version of the
experiment
Van de Graaff
Generator




Charge is delivered continuously to a
high-potential electrode by means of a
moving belt of insulating material
The high-voltage electrode is a hollow
metal dome mounted on an insulated
column
Large potentials can be developed by
repeated trips of the belt
Protons accelerated through such large
potentials receive enough energy to
initiate nuclear reactions
Electrostatic Precipitator





An application of electrical discharge in gases
is the electrostatic precipitator
It removes particulate matter from
combustible gases
The air to be cleaned enters the duct and
moves near the wire
As the electrons and negative ions created by
the discharge are accelerated toward the
outer wall by the electric field, the dirt
particles become charged
Most of the dirt particles are negatively
charged and are drawn to the walls by the
electric field