# lec20 by liamei12345

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```									                          Physics I
Class 20

Electric Potential

Rev. 08-Mar-07 GB
20-1
Work Integral in
Multiple Dimensions (Review)
In multiple dimensions, the work integral looks like this:

xf
 
W   F  dx

xi
What does it mean to “dot” the force with the variable of integration?

F

dx
                                              
xi               path of integration           xf

20-2
Does work depend on the path?
Conservative Forces (Review)

For general forces, the work does depend on the path that we take.
However, there are some forces for which work does not depend on
the path taken between the beginning and ending points.
These are called conservative forces.
A mathematically equivalent way to put this is that the work done by
a conservative force along any closed path is exactly zero.
        
 Fcons  dx  0
(The funny integral symbol means a path that closes back on itself.)

20-3
Electrostatic Force
A Conservative Force

Both experimentally and mathematically, a static electric field (not
changing with time) has been found to obey the following equation,
which you will study in Physics 2:
 
 E  dx  0        
If q is a charge acted on by E , then
                     
0  q  E  dx   q E  dx   F  dx
Recall that this is one of the ways to define a conservative force.

20-4
Potential Energy
(Review)

W(A  B)  U(A)  U(B)   U
Point B:
 
B                                             A
 
U(B)    F  dx                      Point A: U(A)    F  dx
0                                              0

Point 0: U = 0 (defined)

F can be any conservative force.
Today, it will be the electrostatic force.
20-5
Electric Potential Energy

To find the electric potential energy of charge q at point A, we
define point 0 where potential energy is zero, then we calculate
 
A         A
          A  
U(A)    F  dx    q E  dx  q    E  dx 
            
0            0               0          
If we have a different charge – q´ – at point A, then we get the
following expression:
 
A           A
           A  
U(A)    F  dx    q E  dx  q    E  dx 
            
0             0                 0          

20-6
Electric Potential

Just as we noticed when we defined the electric field, we see that
there is a certain quantity that is independent of the charge
experiencing the electrostatic force. This is called the electric
potential and it is normally represented by the symbol V:
 
A
V(A)    E  dx
0

The relationship between U and V is
U  qV
V exists everywhere that an electrostatic field exists. U exists only
where there are charges in that field.

20-7
The Unit of Electric Potential

Volta invented the electric battery in 1800,
the first practical source of continuous
electrical current.
Alessandro Volta
1745-1826

The unit of electric potential is called the volt (V) in honor of Volta.

Electric field can be expressed as N/C or V/m.

20-8
Electric Potential from a
Point Source Charge
In analogy with gravitational potential energy…
The electric potential of a point source charge (q) is defined to be zero at
infinity. To find the electric potential due to q, we integrate:
 
r            r
1        q              1 q
V(r )    E  d r                           dr 
                
4   0 ( r ) 2
4  0 r
For a collection of source charges,
1 qi
V                                Positive charges make positive potentials.
4   0 ri              Negative charges make negative potentials.

Note that V is a scalar not a vector.
The electric potential is much easier to calculate than the electric field!

20-9
Calculating Electric Potential
from Point Source Charges

1 qi       1      qi
V                   r 
4   0 ri 4   0   i

9   1106  2 106 
9 10                      
     1         2 
 3.7 103 Volts                         1m

 2C            1C

1m
20-10
Relationship of Electric Field
and Electric Potential

Once we know the electric field in a region and we define a zero
point, the electric potential, V, can be calculated at any point:
 
A
V(A)    E  dx
0

Conversely, if we know V, we can find the electric field:
V                 V                V
Ex               Ey              Ez  
x                 y                z

(Note: This only works when the electric field is static.)

20-11
Equipotential Lines
and Field Lines
Equipotential Lines – Lines connecting points with equal electric
potential. (In 3D, these are equipotential surfaces.)
Electric Field Lines – Lines formed by taking tiny electric field
vector arrows and connecting them “tail to head.”
 Electric field lines are always perpendicular to equipotential lines.
 Electric field strength is higher where electric field lines are closer
together.
 Electric field strength is higher where equipotential lines are closer
together (assuming equal increments of potential per line).
 Electric field lines point in the direction of decreasing potential.

20-12
Equipotential Lines
and Field Lines

+

equipotential
field line

20-13
Equipotential Lines –
Similarity to Topographic Maps

 Lines of equal height are
similar to equipotential
lines.
 The slope of the ground
is similar to the electric
field (pointing downhill).
 The slope is at a right
angle to the line.
 The closer the lines are
together, the steeper the
slope (the greater the
electric field magnitude).

20-14
Calculating Electric Field
from Electric Potential
Y

X

V = 100     V = 50    V=0
X=0         X=2       X=4

V    V     100
Ex                  25 V / m
x    x     4
V
Ey        0
y
20-15
Calculating Change in K.E.
from Electric Potential
e
final
 K   U  0 or  K    U
U  q V  ( e) V or  U  ( e) V
 K  (e)( V)  (1.6 1019 ) (100) 
 1.6 1017 J

V = 100
initial

V = 50         e

V=0
20-16
Calculating the Potential Energy of
a Configuration of Point Charges
+1.0 x 10–9 C
N 1 N
1 qi q j
U config   
0.6 m              0.6 m           56.                   i 1 ji 1 4   0 ri j
0.6 m                   Where did this come from? All unique pairs.
–2.0 x 10–9 C           –2.0 x 10–9 C
Equivalently:
N
U config   V[ j  1] q j
The order of numbering
doesn’t matter.
We assume the potential              56'.                   j1
energy of charges infinitely
far from each other is zero.
Perhaps surprisingly, the            Build up the configuration one charge at a time.
potential energy of this             V[j–1] is the electric potential at the location of point
configuration is also zero.          j due to all the charges already placed in the
configuration before j. V[0] = 0 by definition.
20-17
Class #20
Take-Away Concepts
Physicist General’s Health Warning:
Confusing electric potential (V) with electric potential energy
(U) will be hazardous to your score on the next exam.
1.    Electric potential (V) and electric potential energy (U):
U  qV
2.    Electric potential from point source charges:
1 qi
V
4   0 ri
3.    Relationship of electric field and electric potential.
4.    Electric field lines and equipotential lines.
5.    Calculating change in kinetic energy using electric potential.
6.    Calculating the potential energy of a configuration of charges.

20-18
Class #20
Problems of the Day
____1. The figure below shows equipotential lines in a certain
region of space. What is the direction of the electric field
in this region?
Y

X
A)   –X
B)   +X
V=0        V = 50     V = 100
C)   –Y             X = 0 cm   X = 2 cm   X = 4 cm
D)   +Y
E)   Cannot be determined.

20-19
Class #20
Problems of the Day
2. An electric field of 100 N/C is created by adjusting an electric
potential difference between two parallel plates that are 5.0 cm apart
(d). (The plates are equipotential surfaces.) What is the magnitude
of the potential difference needed to create this electric field?

20-20
Activity #20
Electric Potential

(Pencil and Paper Activity)
Objective of the Activity:

1.   Think about the relationship of electric potential and electric
potential energy.
2.   Calculate electric potential, potential energy, and kinetic
energy for particles in the electric field of point charges.
3.   Calculate the electric field for given equipotential lines.
4.   Calculate equipotential lines for a given electric field.

20-21
Class #20 Optional Material
Electron Volt Energy Units
e
final
 K   U  0 or  K    U
U  q V  ( e) V or  U  ( e) V
 K  (e)( V)  (1.6 1019 ) (100) 
 1.6 1017 J
This type of calculation for an electron or a
V = 100                         positive ion is so common, we have a special
unit of energy that makes it easier, the
initial   “electron-volt”:

V = 50         e
1eV  1.6022 1019 J
V=0
20-22
Calculating Change in K.E.
Using eV Energy Units
e
final
 K   U  0 or  K    U
U  q V  ( e) V or  U  ( e) V
 K  (1)( V )  (1) (100) 
 100 eV

When using eV energy units, charges are in
V = 100                         units of the charge of a proton.
initial

V = 50         e

V=0
20-23

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