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ELECTRIC POTENTIALELECTRIC POTENTIAL
Summer , 2008Summer , 2008
Chapter 24 Electric PotentialIn this chapter we will define the electric potential ( symbol V ) associated with the electric force and accomplish the following tasks:
Calculate V if we know the corresponding electric field Calculate the electric field if we know the corresponding potential V Determine the potential V generated by a point charge Determine the potential V generated by a continuous charge distribution Determine the electric potential energy U of a system of charges Define the notion of an equipotential surface Explore the geometric relationship between equipotential surfaces and electric field lines Explore the potential of a charged isolated conductor
(24 - 1)
Picture a Region ofspace Where there is an Electric
Field
• Imagine there is a particle of charge q at some location.
• Imagine that the particle must be moved to another spot within the field.
• Work must be done in order to accomplish this.
So, when we move a charge in an Electric Field ..
Move the charge at constant velocity so it is in mechanical equilibrium all the time.
Ignore the acceleration at the beginning because you have to do the same amount of negative work to stop it when you get there.
When an object is moved from one point to another in an Electric Field, It takes energy (work) to move it. This work can be done by an external force (you).
You can also think of this as the FIELDFIELD
doing the negativenegative of this amount of work on the particle.
The net work done by a conservative (field)force on a particle moving
around a closed path is
ZERO!
IMPORTANT
The work necessary to move a charge from an initial point to a final point is INDEPENDENT OF THE PATH CHOSEN!
Electric Potential EnergyElectric Potential Energy
When an electrostatic force acts between two or more charged particles, we can assign an ELECTRIC POTENTIAL ENERGY U to the system.
The change in potential energy of a charge is the amount of work that is done by an external force in moving the charge from its initial position to its new position.
It is the negative of the work done by the FIELD in moving the particle from the initial to the final position.
Definition – Potential Energy
PE or U is the work done by an external agent in moving a charge from a REFERENCE POSITION to a different position.
A Reference ZERO is placed at the most convenient position Like the ground level in many gravitational
potential energy problems.
Example:
E
Zero Levelq
F
d
Work by External Agent
Wexternal = F d = qEd= U
Work done by the Fieldis:Wfield= -qEd = -Wexternal
AN IMPORTANT DEFINITION
Just as the ELECTRIC FIELD was defined as the FORCE per UNIT CHARGE:
We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE:
q
FE
q
UV
VECTOR
SCALAR
UNITS OF POTENTIAL
VOLTCoulomb
Joules
q
UV
Furthermore…
VqW
so
q
W
q
UV
applied
applied
If we move a particle through a potential difference of V, the work from an external
“person” necessary to do this is qV
Consider Two Plates
OOPS …
Look at the path issue
An Equipotential Surface is defined as a surface on which the potential is constant.
0VIt takes NO work to move a charged particlebetween two points at the same potential.
The locus of all possible points that require NO WORK to move the charge to is actually a surface.
A collection of points that have the same
potential is known as an equipotential
surface. Four such surfaces are shown in
the figure. The work done by as it moves
a charge be
E
q
Equipotential surfaces
tween two points that have a
potential difference is given by:
V
W q V
2 1
2 1
For path I : 0 because 0
For path II: 0 because 0
For path III:
For path IV:
When a charge is moved on an equipotential surface 0
The work don
I
II
III
IV
W V
W V
W q V q V V
W q V q V V
V
Note :
e by the electric field is zero: 0W
W q V
(24 - 10)
SA B
V
q
F
E
r
Consider the equipotential surface at
potential . A charge is moved
by an electric field from point A
to point B along a path
V q
E
The electric field E is perpendicular
to the equipotential surfaces
.
Points A and B and the path lie on S
r
Lets assume that the electric field forms an angle with the path .
The work done by the electric field is: cos cos
We also know that 0. Thus: cos 0
0, 0,
E r
W F r F r qE r
W qE r
q E r
0 Thus: cos 0
The correct picture is shown in the figure belo
9
w
0
(24 - 11)S
V
E
Field Lines are Perpendicular to the Equipotential Lines
Equipotential
)(0 ifexternal VVqWork
dF W
Keep in Mind
Force and Displacement are VECTORS!
Potential is a SCALAR.
UNITS
1 VOLT = 1 Joule/Coulomb For the electric field, the units of N/C can be
converted to: 1 (N/C) = 1 (N/C) x 1(V/(J/C)) x 1J/(1 NM) Or
1 N/C = 1 V/m So an acceptable unit for the electric field is now
Volts/meter. N/C is still correct as well.
In Atomic Physics
It is sometimes useful to define an energy in eV or electron volts.
One eV is the additional energy that an proton charge would get if it were accelerated through a potential difference of one volt.
1 eV = e x 1V = (1.6 x 10-19C) x 1(J/C) = 1.6 x 10-19 Joules.
Coulomb Stuff: A NEW REFERENCE: INFINITY
204
1
r
qE
Consider a unit charge (+) being brought from infinity to a distance r from a Charge q:
q r
To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate.
x
AB
r
r
unit
AB
rrkq
r
drkq
dr
qk
dVV
B
A
112
2sr
sE
Set the REFERENCE LEVEL OF POTENTIALat INFINITY so (1/rA)=0.
For point charges
i i
i
r
qV
04
1
r1
r2
r3
P
q1
q3
q2
Consider the group of three point charges shown in the
figure. The potential generated by this group at any
point P is calculated using the principle of super
V
Potential due to a group of point charges
1
31 2
1 2
2 3
31 21 2 3
1 2 3
1 2 3
position
We determine the potentials , and generated
by each charge at point P.
1 1 1, ,
4 4 4
We add the thr
1
ee terms:
1 1
4 4 4
o o
o o o
o
V ,
qq qV
r r
V V
qq qV V V
r r r
V V V V
1.
2.
3r
1 2
11 2
The previous equation can be generalized for charges as
1 1 1 1...
follo
4 4 4 4
ws:n
n i
o o o n o i
q qq q
r r
n
Vr r
(24 - 5)
1 2 3V V V V
For a DISTRIBUTION of charge:
volume r
dqkV
dqO A
Potential created by a line of charge of
length L and uniform linear charge density λ at point P.
Consider the charge element at point A, a
distance from O. From triangle OAP we
dq dx
x
Example :
2 2
2 2
2 20
2 2
2
2
0
2
2
2
2
have:
Here is the distance OP
The potential created by at P is:
1 1
4 4
4
ln4
l
l
n ln4
n
o o
L
o
L
o
o
r d x d
dV dq
dq dxdV
r d x
dxV
d x
V x d x
V L L x
dxx
d
x dd x
(24 - 8)
A
B
VV+dV
The work done by the electric field is given by:
(eqs.1)
also cos cos (eqs.2)
If we compare these two equations we have:
cos cos
From triangle PAB we see that
o
o
o o
W q dV
W Fds Eq ds
dVEq ds q dV E
ds
cos is the
component of along the direction s.
Thus:
s
s
E
E E
VE
s
s
VE
s
(24 - 13)Calculating the electric field E from the potential V
A
B
VV+dV
We have proved that: s
VE
s
s
VE
s
The component of in any direction is the negative of the rate
at which the electric potential changes with distance in this direction
E
If we take s ro be the - , -, and -axes we get:
If we know the function ( , , )
we can determine the components of
and thus the vector itself
x
y
z
x y z
V x y z
E
VE
xV
Ey
VE
z
E
ˆˆ ˆV V VE i j k
x y z
(24 - 14)
Potential energy of a system of point charges
We define as the work required to assemble the
system of charges one by one, bringing each charge
from infinity to its final position
Using the above de
U
U
finition we will prove that for
a system of three point charges is given by:U
2 3 1 31 2
12 23 134 4 4o o o
q q q qq qU
r r r x
y
O
q1
q2
q3
r12
r23
r13
(24 - 15)
, 1
each pair of charges is counted only once
For a system of point charges the potential energy is given by:
Her1
e is the separation between and 4
T
i
ij
ni j
i jo iji j
i j
n q U
r q qq q
Ur
Note :
he summation condition is imposed so that, as in the
case of three point charges, each pair of charges is counted only once
i j
1
1
2
2 2
1 1 22
12 12
3
3 3
1 2
13 23
1 3 23
13
Bring in
0
(no other charges around)
Bring in
(2)
(2)4 4
Bring in
(3)
1(3)
4
1
4
o o
o
o
q
W
q
W q V
q q qV W
r r
q
W q V
q qV
r r
q q qW
r
Step2 :
Step 1 :
e
St p3 :
3
23
1 2 3
2 3 1 31 2
12 23 134 4 4o o o
q q q qq
q
r
W W W W
qW
r r r
(24 - 16)
O
x
y
q1
Step 1
x
y
O
r12q1
q2
Step 2
Step 3
x
y
O
r12
r13
r23
q1
q2
q3
path
A
B
conductor
0E
We shall prove that all the points on a conductor
(either on the surface or inside) have the same
potential
Potential of an isolated conductor
Consider two points A and B on or inside an conductor. The potential difference
between these two points is give by the equation:
We already know that the electrostatic field
B A
B
B A
A
V V
V V E d S
88888888888888
inside a conductor is zero
Thus the integral above vanishes and for any two points
on or inside the conductor.B A
E
V V
A conductor is an equipotential surface
(24 - 17)
We already know that the surface of a conductor
is an equipotential surface. We also know that
the electric field lines are perpendicular to the
equipote
Isolated conductor in an external electric field
ntial surfaces.
From these two statements it follows that the electric field vector is
perpendicular to the conductor surface, as shown in the figure.
All the charges of the conductor reside on the surface and a
E
rrange
themselves in such as way so that the net electric field inside the
conductor
The electric field just out side the conductor is:
0.
in
outo
E
E
(24 - 18)
Electric field and potential
in and around a charged
conductor. A summary
All the charges reside on the conductor surface.
The electric field inside the conductor is zero 0
The electric field just outside the conductor is:
The electric field
in
outo
E
E
1.
2.
3.
4. just outside the conductor is perpendicular
to the conductor surface
All the points on the surface and inside the conductor have the same potential
The conductor is an eequipotential surf
5.
ace
0inE
ˆouto
E n
E
E
n̂
n̂(24 - 19)
In the figure, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?
Continuing
dd
s
ddd
dds
12.152
4
5
42
222
222
s
1 2 3
45 6
d
qxV
d
qk
d
qk
d
q
d
qkV
d
qqqq
d
qqk
r
qkV
i i
i
91035.8
93.093.8812.1
108
12.1
5533
2/
22