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Electric Potential
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Objectives: After completing this
module, you should be able to:
• Understand an apply the concepts of electric potential
energy, electric potential, and electric potential
difference.
• Calculate the work required to move a known charge
from one point to another in an electric field created
by point charges.
• Write and apply relationships between the electric
field, potential difference, and plate separation for
parallel plates of equal and opposite charge.
Review: Work and Energy
Work isis defined
defined as
as the
the product
product of
of displacement
displacement dd
Work
and aa parallel
parallel applied
applied force
force F.
F.
and
Work == Fd;
Fd; Units:
Units: 11 JJ == 11 N
Nm
m
Work
Potential Energy
Energy U
U isis defined
defined as
as the
the ability
ability to
to do
do
Potential
work by
by virtue
virtue of
of position
position or
or condition.
condition. (Joules)
(Joules)
work
Kinetic Energy
Energy K
K isis defined
defined as
as the
the ability
ability to
to do
do work
work
Kinetic
by virtue
virtue of
of motion
motion (velocity).
(velocity). (Also
(Also in
in joules)
joules)
by
Signs for Work and Energy
Work (Fd)
(Fd) isis positive
positive ifif an
an applied
applied force
forceFF isis
Work
in the
the same
same direction
direction as
as the
the displacement
displacementd.
d.
in
The force F does positive work.
The force mg does negative work.
work
B
F
m
mg
A
d
The P.E. at B relative to A is
positive because the field can do
positive work if m is released.
P.E. at A relative to B is negative;
outside force needed to move m.
Gravitational Work and Energy
Consider work against g to move m
from A to B, a vertical height h.
Work = Fh = mgh
At level B, the potential energy U is:
U = mgh (gravitational)
B
F
m
mg
A
h
g
The external force does positive work;
the gravity g does negative work.
The
The external
external force
force FF against
against the
the g-field
g-field increases
increases the
the
potential
potential energy.
energy. IfIf released
released the
the field
field gives
gives work
work back.
back.
Electrical Work and Energy
An external force F moves +q from
A to B against the field force qE.
Work = Fd = (qE)d
At level B, the potential energy U is:
U = qEd (Electrical)
The E-field does negative work;
External force does positive work.
B ++++
Fe
+q + d
qE
E
A - - - -
The
The external
external force
force FF against
against the
the E-field
E-field increases
increasesthe
the
potential
potential energy.
energy. IfIf released
released the
the field
field gives
gives work
work back.
back.
Work and Negative Charges
Suppose a negative charge –q is
moved against E from A to B.
Work by E = qEd
At A, the potential energy U is:
U = qEd (Electrical)
No external force is required !
B ++++
qE
-q
d
E
A - - - -
The
The E-field
E-field does
does positive
positive work
work on
on –q
–q decreasing
decreasing the
the
potential
potential energy.
energy. IfIf released
released from
from BB nothing
nothing happens.
happens.
Work to Move a Charge
++
+
+
++Q++
rb
Work to move
+q from A to B.




qE
ra
Avg. Force: Favg 

F
kqQ
ra rb
kQq
(ra  rb )
Work  Fd 
ra rb
kqQ
At A: Fa  2
ra
kqQ
At B: Fb  2
rb
Distance: ra - rb
1 1
Work  kQq   
 rb ra 
Absolute Potential Energy
++
+
+
++Q++
rb
Absolute P.E. is
relative to 


ra
1 1
Work  kQq   
 rb ra 


F

It is work to bring
+q
from
infinity
to
qE
point near Q—i.e.,
from  to rb
0
 1 1  kQq
Work  kQq    
rb
 rb  
Absolute Potential
Energy:
kQq
U
r
Example 1. What is the potential energy if a
+2 nC charge moves from  to point A, 8 cm
away from a +6 C charge?
The P.E. will be positive at
point A, because the field can
do + work if q is released.
Potential
Energy:
U
A

8 cm
+Q
kQq
U
r
(9 x 10
9 Nm 2
C2
U
U == 1.35
1.35 mJ
mJ
+2 nC
+6 C
)(6 x 10 C)(+2 x 10 C)
-6
-9
(0.08 m)
Positive potential energy
Signs for Potential Energy
Consider Points A, B, and C.
For +2 nC at A: U = +1.35 mJ
Questions:
If +2 nC moves from A to B,
does field E do + or – work?
Does P.E. increase or decrease?
A

B
12 cm
8 cm
C
+Q
4 cm
+6 C
Moving
positive q
+2 nC
The field E does positive work, the P.E. decreases.
If +2 nC moves from A to C (closer to +Q), the
field E does negative work and P.E. increases.
Example 2. What is the change in potential
energy if a +2 nC charge moves from  to B?
Potential
Energy:
kQq
U
r
A

UB 
(9 x 19
12 cm
8 cm
From Ex-1: UA = + 1.35 mJ
9 Nm 2
C2
B
+Q
+6 C
)(6 x 10-6 C)(+2 x 10-9 C)
(0.12 m)
U = UB – UA = 0.9 mJ – 1.35 mJ
 0.900 mJ
U
U == -0.450
-0.450 mJ
mJ
Note that P.E. has decreased as work is done by E.
Moving a Negative Charge
Consider Points A, B, and C.
Suppose a negative -q is moved.
Questions:
If -q moves from A to B, does
field E do + or – work? Does
P.E. increase or decrease?
A

B
12 cm
8 cm
C
+Q
4 cm
+6 C
Moving
negative q
-
The field E does negative work, the P.E. increases.
What happens if we move a –2 nC charge from A to B
instead of a +2 nC charge. This example follows . . .
Example 3. What is the change in potential
energy if a -2 nC charge moves from  to B?
Potential
Energy:
kQq
U
r
From Ex-1: UA = -1.35 mJ
(Negative due to – charge)
UB 
(9 x 19
9 Nm 2
C2
A

UB – UA = -0.9 mJ – (-1.35 mJ)
12 cm
8 cm
+Q
+6 C
)(6 x 10-6 C)(-2 x 10-9 C)
(0.12 m)
B
 0.900 mJ
U
U == +0.450
+0.450 mJ
mJ
A
from aa ++ charge
charge gains
gains P.E.
P.E.
A –– charge
charge moved
moved away
away from
Properties of Space
Electric Field
.
E
r
+ ++
+
++Q++
E is a Vector
An electric field is a property of
space allowing prediction of the
force on a charge at that point.
F
E ;
q
F  qE
The field E exist independently of
the charge q and is found from:
kQ
Electric Field : E  2
r
Electric Potential
Electric potential is another property
of space allowing us to predict the
P.E. of any charge q at a point.
U
Electric
V ;
Potential:
q
U  qV
The units are: joules per coulomb (J/C)
U
P. V 
q
r
+ ++
+
++Q++
Potential
For example, if the potential is 400 J/C at point P,
a –2 nC charge at that point would have P.E. :
U = qV = (-2 x 10-9C)(400 J/C);
U
U == -800
-800 nJ
nJ
The SI Unit of Potential (Volt)
From the definition of electric potential as P.E.
per unit charge, we see that the unit must be
J/C. We redefine this unit as the volt (V).
U
V ;
q
1 joule 

1 volt =

1 coulomb 

A
A potential
potential of
of one
one volt
volt at
at aa given
given point
point means
means that
that
aa charge
charge of
of one
one coulomb
coulomb placed
placed at
at that
that point
point will
will
experience
experience aa potential
potential energy
energy of
of one
one joule.
joule.
Calculating Electric Potential
Electric Potential Energy and Potential:
kQq
U
U
; V
r
q
Substituting for
U, we find V: V 
kQ
V
r

kQq
r
q
  kQ
kQ
P. V 
r
r
+ ++
+
++Q++
Potential
r
The
The potential
potential due
due to
to aa positive
positive charge
charge isis
positive;
positive; The
The potential
potential due
due to
to aa negative
negative
charge
charge isis positive.
positive. (Use
(Use sign
sign of
of charge.)
charge.)
Example 4: Find the potential at a distance
of 6 cm from a –5 nC charge.
P. q = –4 C
r 6 cm
-- Q -- Q = -5 nC
kQ
V

r

9 x 10
Negative V at
Point P :
9 Nm 2
C
2

(5 x 10-9 C)
(0.06 m)
VVPP == -750
-750 V
V
What would be the P.E. of a –4 C
charge placed at this point P?
U = qV = (-4 x 10-6 C)(-750 V);
U
U == 3.00
3.00 mJ
mJ
Since
released.
Since P.E.
P.E. isis positive,
positive, EE will
will do
do ++ work
work ifif qq isis released.
Potential For Multiple Charges
The Electric
Electric Potential
Potential VV in
in the
the vicinity
vicinity of
of aa number
number
The
of charges
charges isis equal
equal to
to the
the algebraic
algebraic sum
sum of
of the
the
of
potentials due
due to
to each
each charge.
charge.
potentials
Q1 - r1
r3
Q3 -
A
r2
+
Q2
kQ1 kQ2 kQ3
VA 


r1
r2
r3
kQ
V 
r
Potential isis ++ or
or –– based
based on
on sign
sign of
of the
the charges
charges Q.
Q.
Potential
Example 5: Two charges Q1= +3 nC and
Q2 = -5 nC are separated by 8 cm. Calculate
the electric potential at point A.

B
kQ1 kQ2
VA 

r1
r2

2 cm
9 x 10
(3 x 10 C)
kQ1
C2

 450 V
r1
(0.06 m)
kQ2

r2

9 x 10
9 Nm 2
9 Nm 2
C
2

-9
(5 x 10-9 C)
(0.02 m)
VA = 450 V – 2250 V;
 2250 V
Q1 + +3 nC
6 cm
A 
2 cm
V
-1800 V
V
VAA == -1800
Q2 = -5 nC
Example 5 (Cont.): Calculate the electric potential
at point B for same charges.
kQ1

r1

kQ2

r2
kQ1 kQ2
VB 

r1
r2
9 x 109 Nm
2
C2

(3 x 10-9 C)
(0.02 m)

9 x 10
9 Nm 2
C
2

VB = 1350 V – 450 V;
2 cm
 1350 V
Q1 + +3 nC
6 cm
(5 x 10-9 C)
(0.10 m)
B
 450 V
A 
2 cm
V
VBB == +900
+900 V
V
Q2 = -5 nC
Example 5 (Cont.): Discuss meaning of the potentials
just found for points A and B.
Consider Point A:
V
VAA == -1800
-1800 V
V
For
For every
every coulomb
coulomb of
of positive
positive charge
charge
placed
placed at
at point
point A,
A, the
the potential
potential energy
energy
will
will be
be –1800
–1800 J.J. (Negative
(Negative P.E.)
P.E.)
The
The field
field holds
holds on
on to
to this
this positive
positive
charge.
charge. An
An external
external force
force must
must do
do
+1800
+1800 JJ of
of work
work to
to remove
remove each
each
coulomb
coulomb of
of ++ charge
charge to
to infinity.
infinity.
B
2 cm
Q1 + +3 nC
6 cm
A 
2 cm
Q2 = -5 nC
Example 5 (Cont.): Discuss meaning of the potentials
just found for points A and B.
Consider Point B:
V
+900 V
V
VBB == +900
For
For every
every coulomb
coulomb of
of positive
positive charge
charge
placed
placed at
at point
point B,
B, the
the potential
potential energy
energy
will
will be
be +900
+900 J.J. (Positive
(Positive P.E.)
P.E.)
For
For every
every coulomb
coulomb of
of positive
positive charge,
charge,
the
the field
field EE will
will do
do 900
900 JJ of
of positive
positive
work
work in
in removing
removing itit to
to infinity.
infinity.
B 
2 cm
Q1 + +3 nC
6 cm
A 
2 cm
Q2 = -5 nC
Potential Difference
The
The potential
potential difference
difference between
between two
two points
points A
A and
and BB
isis the
the work
work per
per unit
unit positive
positive charge
charge done
done by
by electric
electric
forces
forces in
in moving
moving aa small
small test
test charge
charge from
from the
the point
point of
of
higher
higher potential
potential to
to the
the point
point of
of lower
lower potential.
potential.
Potential
=V -V
Potential Difference:
Difference: VVAB
AB = VAA - VBB
Work
= q(V – V )
WorkAB
AB = q(VAA – VBB)
Work
Work BY
BY E-field
E-field
The
The positive
positive and
and negative
negative signs
signs of
of the
the charges
charges may
may
be
be used
used mathematically
mathematically to
to give
give appropriate
appropriate signs.
signs.
Example 6: What is the potential difference between
points A and B. What work is done by the E-field if a
+2 C charge is moved from A to B?
B 2 cm
V
Q1 + +3 nC
VBB == +900
VAA == -1800
-1800 V
V V
+900 V
V
VAB= VA – VB = -1800 V – 900 V
VVAB
= -2700 V
AB = -2700 V
Note point B is at
higher potential.
6 cm
A 
Q2
-
2 cm
-5 nC
WorkAB = q(VA – VB) = (2 x 10-6 C )(-2700 V)
Work
Work == -5.40
-5.40 mJ
mJ
E-field does negative work.
Thus, an external force was required to move the charge.
Example 6 (Cont.): Now suppose the +2 C charge
is moved from back from B to A?
B 2 cm
V
Q1 + +3 nC
VBB == +900
VAA == -1800
-1800 V
V V
+900 V
V
VBA= VB – VA = 900 V – (-1800 V)
6 cm
A 
2 cm
This path is from
VVBA
= +2700 V
BA = +2700 V high to low potential. Q2 - -5 nC
WorkBA = q(VB – VA) = (2 x 10-6 C )(+2700 V)
Work
Work == +5.40
+5.40 mJ
mJ
E-field does positive work.
The work is done BY the E-field this time !
Parallel Plates
Consider Two parallel plates of equal
VA + + + +
and opposite charge, a distance d apart.
Constant
Constant EE field:
field: FF == qE
qE
Work = Fd = (qE)d
+q
F = qE
VB - - - -
Also, Work = q(VA – VB)
So that: qVAB = qEd and
VVAB
= Ed
AB = Ed
The
The potential
potential difference
difference between
between two
two oppositely
oppositely
charged
charged parallel
parallel plates
plates isis the
the product
product of
ofEE and
and dd..
E
Example 7: The potential difference between
two parallel plates is 800 V. If their separation is 3 mm, what is the field E?
VA + + + +
+q
F = qE
VB - - - -
V  Ed ;
E
V
E
d
80 V
E
 26, 700 V/m
0.003 m
The
The E-field
E-field expressed
expressed in
in volts
volts per
per meter
meter (V/m)
(V/m) isis
known
known as
as the
the potential
potential gradient
gradient and
and isis equivalent
equivalent to
to
the
the N/C.
N/C. The
The volt
volt per
per meter
meter isis the
the better
better unit
unit for
for
current
current electricity,
electricity, the
the N/C
N/C isis better
better electrostatics.
electrostatics.
Summary of Formulas
Electric
Electric Potential
Potential
Energy
Energy and
and Potential
Potential
kQq
U
U
; V
r
q
Electric
Electric Potential
Potential Near
Near
Multiple
Multiple charges:
charges:
kQ
V 
r
Work
= q(V – V )
WorkAB
AB = q(VAA – VBB)
Work
Work BY
BY E-field
E-field
Oppositely
Oppositely Charged
Charged
Parallel
Parallel Plates:
Plates:
V  Ed ;
V
E
d
CONCLUSION: Chapter 25
Electric Potential