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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 is defined as the product of displacement d
and a parallel applied force F.
Work = Fd; Units: 1 J = 1 N m
Potential Energy U is defined as the ability to do work
by virtue of position or condition. (Joules)
Kinetic Energy K is defined as the ability to do work
by virtue of motion (velocity). (Also in joules)
Signs for Work and Energy
Work (Fd) is positive if an applied force F is
in the same direction as the displacement d.
The force F does positive work.
The force mg does negative 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 external force F against the g-field increases the
potential energy. If released the field gives work 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 external force F against the E-field increases the
potential energy. If released the field gives work 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 E-field does positive work on –q decreasing the
potential energy. If released from B nothing happens.
Work to Move a Charge
++
+
+
++Q++
rb
Work to move
+q from A to B.
B

A

qE
ra
Avg. Force: Favg 

F+
kqQ
ra rb
kQq
Work  Fd 
(ra  rb )
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 .
B

ra
1 1
Work  kQq   
 rb ra 
A

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 mC charge?
The P.E. will be positive at
point A, because the field can
do + work if q is released.
Potential
Energy:
U
kQq
U
r
(9 x 10
9 Nm 2
C2
U = 1.35 mJ
A

+2 nC
8 cm
+Q
+6 mC
)(+6 x 10-6 C)(+2 x 10-9 C)
(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 mC
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 A 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 mC
)(+6 x 10-6C)(+2 x 10-9C)
(0.12 m)
DU = UB – UA = 0.9 mJ – 1.35 mJ
 0.900 mJ
DU = -0.450 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 mC
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 A 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 mC
)(6 x 10-6C)(-2 x 10-9C)
(0.12 m)
B
 0.900 mJ
DU = +0.450 mJ
A – charge moved away from a + charge gains P.E.
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.
Electric
Potential:
U
V ;
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 = -800 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 potential of one volt at a given point means that
a charge of one coulomb placed at that point will
experience a potential energy of one 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 potential due to a positive charge is
positive; The potential due to a negative
charge is positive. (Use sign of charge.)
Example 4: Find the potential at a distance
of 6 cm from a –5 nC charge.
P. q = –4 mC
r 6 cm
-- Q -- Q = -5 nC
kQ
V

r

9 Nm2
9 x 10
Negative V at
Point P :
C
2

(5 x 10-9C)
(0.06 m)
VP = -750 V
What would be the P.E. of a –4 mC
charge placed at this point P?
U = qV = (-4 x 10-6 mC)(-750 V);
U = 3.00 mJ
Since P.E. is positive, E will do + work if q is released.
Potential For Multiple Charges
The Electric Potential V in the vicinity of a number
of charges is equal to the algebraic sum of the
potentials due to each charge.
Q1 - r1
r3
Q3 -
A
r2
+
Q2
kQ1 kQ2 kQ3
VA 
+
+
r1
r2
r3
kQ
V 
r
Potential is + or – based on sign of the charges Q.
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 10-9C)
(0.02 m)
VA = 450 V – 2250 V;
 2250 V
Q1 + +3 nC
6 cm
A 
2 cm
VA = -1800 V
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 10-9C)
(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 10-9C)
(0.10 m)
B
 450 V
A 
2 cm
VB = +900 V
Q2 = -5 nC
Example 5 (Cont.): Discuss meaning of the potentials
just found for points A and B.
Consider Point A:
VA = -1800 V
For every coulomb of positive charge
placed at point A, the potential energy
will be –1800 J. (Negative P.E.)
B
2 cm
Q1 + +3 nC
6 cm
The field holds on to this positive
charge. An external force must do
+1800 J of work to remove each
coulomb of + charge to infinity.
A 
2 cm
Q2 = -5 nC
Example 5 (Cont.): Discuss meaning of the potentials
just found for points A and B.
Consider Point B:
VB = +900 V
B 
2 cm
For every coulomb of positive charge
placed at point B, the potential energy
will be +900 J. (Positive P.E.)
For every coulomb of positive charge,
the field E will do 900 J of positive
work in removing it to infinity.
Q1 + +3 nC
6 cm
A 
2 cm
Q2 = -5 nC
Potential Difference
The potential difference between two points A and B
is the work per unit positive charge done by electric
forces in moving a small test charge from the point of
higher potential to the point of lower potential.
Potential Difference: VAB = VA - VB
WorkAB = q(VA – VB)
Work BY E-field
The positive and negative signs of the charges may
be used mathematically to give appropriate signs.
Example 6: What is the potential difference between
points A and B. What work is done by the E-field if a
+2 mC charge is moved from A to B?
B 2 cm
VA = -1800 V
VB = +900 V
Q1 + +3 nC
VAB= VA – VB = -1800 V – 900 V
VAB = -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 = -5.40 mJ
E-field does negative work.
Thus, an external force was required to move the charge.
Example 6 (Cont.): Now suppose the +2 mC charge
is moved from back from B to A?
B 2 cm
VA = -1800 V
VB = +900 V
Q1 + +3 nC
VBA= VB – VA = 900 V – (-1800 V)
VBA = +2700 V
6 cm
A 
2 cm
This path is from
high to low potential. Q2 - -5 nC
WorkBA = q(VB – VA) = (2 x 10-6 C )(+2700 V)
Work = +5.40 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 E field: F = qE
+q
F = qE
VB - - - -
Work = Fd = (qE)d
Also, Work = q(VA – VB)
So that: qVAB = qEd
and
VAB = Ed
The potential difference between two oppositely
charged parallel plates is the product of E and d.
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 E-field expressed in volts per meter (V/m) is
known as the potential gradient and is equivalent to
the N/C. The volt per meter is the better unit for
current electricity, the N/C is better electrostatics.
Summary of Formulas
Electric Potential
Energy and Potential
Electric Potential Near
Multiple charges:
WorkAB = q(VA – VB)
Oppositely Charged
Parallel Plates:
kQq
U
U
; V
r
q
kQ
V 
r
Work BY E-field
V  Ed ;
V
E
d
CONCLUSION: Chapter 25
Electric Potential