Transcript Document

25.1 Potential Difference and Electric Potential
25.2 Potential Differences in a Uniform Electric Field
25.3 Electric Potential and Potential Energy Due to Point charges
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Norah Ali Almoneef
25.1 Potential Difference and Electric Potential
When a test charge q0 is placed in an electric field E created by some source
charge distribution, the electric force acting on the test charge is q0 E.
When the test charge is moved in the field by some external agent, the work
done by the field on the charge is equal to the negative of the work done by
the external agent causing the displacement
This is analogous to the situation of lifting an object with mass in a gravitational
field—the work done by the external agent is mgh and the work done by the
gravitational force is -mgh.
the work done by the electric field on the charge is
work =F. ds =q0E“.ds
ds is the displacement of a charge
The potential energy of the charge–field system is changed by an amount dU =
q0E“.ds
For a finite displacement of the charge from point A to point B, the change in
potential energy of the system ∆U =UB - UA is
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Dividing the potential energy by the test charge gives a physical quantity that
depends only on the source charge distribution. The potential energy per unit
charge U/q0 is
independent of the value of q0 and has a value at every point in an electric field.
This quantity U/q0 is called the electric potential (or simply the potential) V.
The potential difference ∆V =VB -VA between two points A and B in an electric field
is defined as the change in potential energy of the system when a test charge is
moved between the points divided by the test charge q0:
The potential difference between A and B depends only on the source charge
distribution (consider points A and B without the presence of the test charge)
the work done by an external agent in moving a charge q through an electric
field at constant velocity is
W= q ∆V
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Definition of the Electric Potential
 The electric potential energy of a charged particle in an electric
field depends not only on the electric field but on the charge of
the particle.
 We want to define a quantity to probe the electric field that is
independent of the charge of the probe.
 We define the electric potential as
“potential energy per unit charge
U
V
of a test particle”
q
• Unlike the electric field, which is a vector, the electric potential is
a scalar.
Units: [V] = J / C, by definition, volt
 The electric potential has a value everywhere in space but has no
direction.
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Definition of the Electric Potential
 The electric potential energy of a charged particle in an electric
field depends not only on the electric field but on the charge of
the particle.
 We want to define a quantity to probe the electric field that is
independent of the charge of the probe.
 We define the electric potential as
“potential energy per unit charge
U
V
of a test particle”
q
• Unlike the electric field, which is a vector, the electric potential is
a scalar.
Units: [V] = J / C, by definition, volt
 The electric potential has a value everywhere in space but has no
direction.
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Electric Potential Energy
 The electric force, like the gravitational force, is a
conservative force.
 When an electrostatic force acts between two or more
charges within a system, we can define an electric potential
energy, U, in terms of the work done by the electric field,
We, when the system changes its configuration from some
initial configuration to some final configuration.
Change in electric potential energy = -Work done by electric field
U  U f  Ui  We
U i is the initial electric potential energy
U f is the final electric potential energy
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 Like gravitational or mechanical potential energy, we must define
a reference point from which to define the electric potential energy.
 We define the electric potential energy to be zero when all
charges are infinitely far apart.
 We can then write a simpler definition of the electric potential
taking the initial potential energy to be zero,
U  U f  0  U  W
 The negative sign on the work:
 If E does positive work then U < 0
 If E does negative work then U > 0
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Constant Electric Field - Special Cases
 Displacement is in the same direction as the electric field
W  qEd
so
U  qEd
 A positive charge loses potential energy
 when it moves in the direction of the electric field.
 Displacement is in the direction opposite to the electric
field
W  qEd
so
U  qEd
 A positive charge gains potential energy when it moves in the
direction opposite to the electric field.
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Potential Difference and Electric Potential
The potential difference between points A and B, VB
–VA, is defined as the change in potential energy
(final value minus initial value) of a charge q moved
from A to B divided by the size of the charge.
V  VB –VA = U
q
Final point
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Initial point
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Work
WAB = -U = - q(VB – VA)
Initial point
Final point
Final point
Initial point
Equation is true if the only force is the
conservative electrostatic force. That is, there are
no non conservative forces acting on the system.
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Change in Potential Energy
• As the electric field accelerates the charge, the charge gains kinetic
energy.
• As the charged particle gains kinetic energy, it loses an equal amount of
potential energy.
K = - U
• By definition, the work done by a conservative force equals the negative
change in potential energy, U.
U = -WAB = - qEd
• This equation is valid only for a uniform electric field.
Potential Energy
U = Ub – Ua = qVba
Final point
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Initial point
25-2 Potential Differences in a Uniform Electric Field (Constant Electric Field )
 Let’s look at the electric potential energy when we move a
charge q by a distance d in a constant electric field.
 The definition of work is
 
W  F d
 For a constant electric field the
force is F = qE
 the work done by the electric field on the charge is
 
W  qE  d  qEd cos
 = angle between E and d.
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Work and Potential Energy
 There is a uniform field
between the two plates
 As the positive charge moves
from A to B, work is done
 WAB=F d=q E d
 ΔPE =-W AB=-q E d
 only for a uniform field
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The positive side
is always the
“high” potential
side, regardless
of the sign of the
charge.
Important!
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The definition of
the “high” side is
done for a positive
test charge.
Usually take “low”
side as V = 0.
Conservative Forces
 A force is conservative if the work it does on a particle moving
between any two points is independent of the path taken by the
particle.
 The work done by a conservative force exerted on a particle
moving through any closed path is zero.
 Gravitational (Newton’s law of gravity) and Electrical
(Coulomb’s law of electrical force) are both conservative
forces.
 Since electrostatic force is conservative, electrostatic phenomena
can be described in terms of electrical potential energy.
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Potential Diff. In Uniform E field
Charged particle moves from A to B in uniform
E field.
B
E . ds = =  E . d
V =

A
U = qo V = - qo E . d =  qo E d cos 
Show that the potential diff. between path (1) and (2) are
the same as expected for a conservative force field.
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Potential Diff. In Uniform E field (Path independence)
Show that the potential difference between path (1) and (2)
are the same as expected for a conservative force field.
path (1)
V = path (1)
B
E . ds =  Es cos 
A
path (2)

d
=0
E . ds + -
V = -
ds
B
c
path (2)
since E
A
E . ds
c
c
V = 
ds =  Ed=  E s cos 
E.
A
Conservative force: The work is path-independent.
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Same
Electric Potential
• Unlike the electric field, which is a vector, the electric potential is
a scalar.
 The electric potential has a value everywhere in space but has no
direction.
 The SI units of electric potential are joules per coulomb.
 The unit of potential is the volt. Units: [V] = J / C, by definition, volt
 A VOLT is defined:
1 V = 1 J/C
The Electron Volt
The electron volt is defined as the energy acquired by a
particle carrying a charge equal to that on the electron (q = e) as
the result of moving through a potential difference of 1 V.
1eV = (1.6 x 10-19 C)(1.0 V) = 1.6 x 10-19 J
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Equipotential
B
V = 
E . ds = 0
c
VC = VB ( same potential)
In fact, points along
this line has the
same potential. We
have an equipotential
line.
If s is perpendicular to E (path C-B), the electric potential
does not change.
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Equipotential Surfaces
•The name equipotential surface is given to any surface
consisting of a continuous distribution of points having the same
electric potential.
•No work is done in moving a test charge between any two
points on an equipotential surface.
•The equipotential surfaces of a uniform electric field consist of a
family of planes that are all perpendicular to the field.
Equipotential surfaces are always perpendicular to electric field
lines.
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Equipotential Surface
Equipotential Surfaces (dashed blue lines) and electric field
lines (orange lines) for (a) a uniform electric field produced
by infinite sheet of charge, (b) a point charge, and (c) an
electric dipole. In all cases, the equipotential surfaces are
perpendicular to the electric field lines at every point.
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Equipotential Surfaces
•The name equipotential surface is given to any surface consisting of a continuous
distribution of points having the same electric potential.
•No work is done in moving a test charge between any two points on an equipotential
surface.
•The equipotential surfaces of a uniform electric field consist of a family of planes that are
all perpendicular to the field.
Equipotential surfaces are always perpendicular to electric field lines.
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Lines of constant V
(perpendicular to E)
Lines of constant E
Equipotential Surfaces and the Electric Field
All points on the surface of a charged conductor in electrostatic
equilibrium are at the same potential
Therefore, the electric potential is a constant everywhere on the
surface of a charged conductor in equilibrium
An ideal conductor is an equipotential surface. Therefore, if
two conductors are at the same potential, the one that is
more curved will have a larger electric field around it. This is
also true for different parts of the same conductor.
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Example
A point particle of mass m = 1.8x10-5 kg and charge q = +3.0x10-5
C is released from rest at point A and accelerates until it reaches
point B. The only force acting on the particle is the electric force
and the electric potential at A is 25 V greater than at B. What is
the speed of the particle when it reaches B?
0
qVA
qVB
KEA  EPEA  KEB  EPEB
1 2
solve for vB
mvB
2
25V
1 2
mvB  q VA  VB 
2
vB  9.13m / s
What happens if q is negative?
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Example
An ion accelerated through a potential difference of
115 V experiences an increase in kinetic energy of
7.37 x 10 –17 J. Calculate the charge on the ion.
qV= 7.37x10-17 J ,
q = 6.41x10-19 C
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V=115 V
Example
In a hydrogen atom the e- revolves around the p+ at a distance
of 5.3 x 10-11 m. Find the electric potential at the e- due to the
p+, and the electrostatic potential energy between them.
 Electric potential due to proton:
q 9 x 109 1.6 x 10 -19 
V r   k 
 27 V
-11
r
5.3 x 10
e-
r
 Electrostatic p.E. is given by:
q1q2
U12  k
r12
  e V p   1.6 x 10-19 27   4.3 x 10-18 J
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p+
Example
A proton is released from rest in a uniform E
field that has a magnitude of 8 x 104 V/m and
is directed along the positive x-axis. The
proton undergoes a displacement of 0.50 m
in the direction of E.
(a) Find the change in electric potential between
points A and B.
(b) Find the change in potential energy of the
proton for this displacement.
(a) V = Ed =  (8.0x104 V/m) (0.50m) =  4.0x104 V
(b) U = q V = (1.6 x 10-19 C) (4 .0x104 V) = 6.4 x 10-15 J
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Example
Suppose an electron is released from rest in a
uniform electric field whose magnitude is 5.90 x 103
V/m. (a) Through what potential difference will it
have passed after moving 1.00 cm? (b) How fast will
the electron be moving after it has traveled 1.00 cm?
(a) |V| = Ed = (5.90 x 103 V/m)(0.0100 m) = 59.0 V
(b) q |V| = mv2/2  v = 4.55x106 m/s
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Example
 Calculate the electrostatic potential energy between 2
protons in a Uranium nucleus separated by 2 x 10-15 m.
U r   k
q1q2
r
 9.0 x 109
~ 10 13 J
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
1.6 x 10

-19
2 x 10-15

2
Example
Forces on Charged Particles
 In a CRT an electron moves 0.2 m in a straight line (from rest)
driven by an electric field of 8 x 103 V/m. Find:
(a) The force on the electron.
(b) The work done on it by the E-field.
(c) Its potential difference from start to finish.
(d) Its change in potential energy.
(e) Its final speed.
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Example
(a) Force is in opposite direction to the E-field, magnitude:
F  qE   1.6 x 10-19 8 x 103   1.3 x 10-15 N
(b) Work done by force:
Work  Fs   1.3 x 10 -15  0.2   2.6 x 10 -16 J
(c) Potential difference is defined as work/unit charge:
W
 2.6 x 10-16
3
V 


1.6
x
10
V
q
 1.6 x 10-19
Alternatively (e- opposite to p+):
b
d
a
0
V    E  d s    E dx   Ed 
 8 x 103 0.2   1.6 x 103 V
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(d) Change in potential energy:
b
U  q0  E  d s
a
 q0 V
 - 1.6 x 10 -19 1.6 x 10 3 
 - 2.6 x 10 -16 J
  work done 
(e) Loss of PE = gain in KE = ½mv2
v

2KE 
m
22.6 x 10 -16 
9.1 x 10 -31 
 2.4 x 10 7 ms -1
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Example
 A proton is accelerated across a potential difference of 600 V. Find its
change in K.E. and its final velocity.
 By definition, 1 eV = 1.6 x 10-19 J.
 Acceleration across 600 V
 Proton gains 600 eV.
K.E. = 600(1.6 x 10-19) = 9.6 x 10-17 J
 Final velocity is:
• If it started from rest
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29.6 x 10-17 
v
1.7 x 10-27 
 3.4 x 105 ms -1
Example
 Two parallel metal plates have an area A = 225 cm2 and are d
=0.5 cm apart, with a p.d. of 0.25 V between them. Calculate
the electric field.
ds
V  V  V
0V
0.25V
left
b
  E  d s
a
d

 E ds
0
d
0.1V
 E  ds
0.2V
0
 Ed
x =0
E 
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x =0.5m
V 0.25

 50 Vm -1
d
0.5
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right
Electric Potential and Potential Energy due to point charges
Consider isolated positive point charge q.
(i.e. E directed radially outward from the
charge)
To find electric potential at a point located at
a distance r from the charge, start with the
general expression for potential difference:
B
VB VA =
E . ds
A
Where A and B are two arbitrary points as
shown.
E = kq/r2 r, where r is a unit vector directed
from the charge toward the field point.
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Electric Potential and Potential Energy due to point charges
We can express E . ds as
E . ds =
kq/r2 r . ds
The magnitude of r is 1, dot product r . ds = ds cos ,
where  is the angle between r and ds .
ds cos  is the projection of ds onto r , thus
ds cos  = dr.
B
VB-VA = -
E . ds
A
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B
=-
kq/r2 dr
A
Electric Potential and Potential Energy due to point charges
rB
B
VB-VA = -
=-
Er dr
rA
A
rB
kq
r
VB-VA =
VB-VA = kq
1
rB
Depends only on the coordinates and not on the path.
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kq/r2 dr
rA
1
rA
Electric Potential and Potential Energy due to point charges
rA = infinity (and VA = 0), we have electric
potential created by a point charge at a
distance r from the charge given by
V=
kq
r
Points at same distance r from q
have the same potential V, i.e. the
equipotential surfaces are spherical and
centered on the charge.
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Potential due to two or more charges: Superposition
q1
V=
k
S
i
qi
ri
r1
q5
q2
r2
r5
P
r4
r3
q3
q4
where potential is taken to be zero at infinity and ri is the
distance from the point P to the charge qi.
Note that this is a scalar sum rather than a vector sum.
The potential is positive if the charge is positive and negative if
the charge is negative
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Example
Finding the Electric Potential at Point P
6
5
.
0

10
C
9
2
2
V1  (8.99  10 Nm / C )
 1.12  10 4 V,
4.0m
6
(

2
.
0

10
C)
9
2
2
V2  (8.99  10 Nm / C )
 3.60  10 3 V
(3.0m) 2  (4.0m) 2
Superposition: Vp=V1+V2
Vp=1.12104 V+(-3.60103 V)=7.6103 V
5.0 mC
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-2.0 mC
Example:
How many electrons should be removed from an
initially uncharged spherical conductor of radius
0.300 m to produce a potential of 7.5 kV at the
surface?
ke q
V
r
= 7.50 x 103 V
(8.99x109 Nm2 /C 2 )q
V 
(0.300m)
q  2.50  10 7 C
=
N = 1.56 x 1012 electrons
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Example
A charge +q is at the origin. A charge –2q is at x =
2.00 m on the x axis. For what finite value(s) of x is
(a) the electric field zero ? (b) the electric potential
zero ?
q

2q
0
E  k  2 
2 
( x  2.00) 
x
x = - 4.83 m
(other root is not physically valid)
q

2q
  0
V  

 x (2.00  x) 
x = 0.667 m and x= -2.00 m
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x2 + 4.00x – 4.00 = 0
(x+4.83)(x0.83)=0
Potential Energy of a system of two charges
P
V1 = potential at a point P due to
q1, external agent must do work to
bring a second charge q2 from
infinity to P and this work = q2V1.
Definition: This work done is
equal to the potential energy
U of the two-particle system.
If two point charges are separated by a distance r12, the potential
energy of the pair of charges is given by
q1 q2
U12 = k
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r12
1
k=
4
peo
Potential Energy
Three point charges are fixed
at the positions shown. The
potential energy of this
system of charges is given by
q1 q2
U=k
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r12
q1 q3
+
r13
q2 q3
+
r23
Notes About Electric Potential Energy of Two Charges
 If the charges have the same sign, PE is positive
 Positive work must be done to force the two charges near one
another
 The like charges would repel
 If the charges have opposite signs, PE is negative
 The force would be attractive
 Work must be done to hold back the unlike charges from
accelerating as they are brought close together
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Example
What is the amount of work required to assemble four
identical point charges of magnitude Q at the corners of
a square of side s?
U = 0 + U12 + (U13 + U23) + (U14 + U24 + U34)
2
keQ 2 keQ 2  1
k
Q
1
 e 

 0


1

1


1




s
s  2 
s 
2 
s
keQ 2 
2 
U
4 

s 
2
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s
s
s
Example
A proton is placed in an electric field of E=105 V/m and released.
After going 10 cm, what is its speed?
Use conservation of energy.
a
b
E = 105 V/m
d = 10 cm
v
+
v
V = Vb –Va = -Ed
U = q V
U + K = 0
K = (1/2)mv2
K = -U
(1/2)mv2 = -q V = +qEd
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2qEd
m
2 1.6 10 19 C 105 Vm 1m
1.67 10  23 kg
v  1.4 106
m
s
Example
 In a CRT an electron moves 0.2 m in a straight line (from
rest) driven by an electric field of 8 x 103 V/m. Find:
(a) The force on the electron.
(b) The work done on it by the E-field.
(c) Its potential difference from start to finish.
(d) Its change in potential energy.
(e) Its final speed.
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(a) Force is in opposite direction to the E-field, magnitude:
F  qE   1.6 x 10-19 8 x 103   1.3 x 10-15 N
(b) Work done by force:
Work  Fs   1.3 x 10 -15  0.2   2.6 x 10 -16 J
(c) Potential difference is defined as work/unit charge:
W
 2.6 x 10 -16
V 

 1.6 x 103 V
-19
q
 1.6 x 10
Alternatively (e- opposite to p+):
b
d
a
0
V    E  d s    E dx   Ed 
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
8 x 10
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Almoneef
3
0.2   1.6 x 10
3
V
(d) Change in potential energy:
b
U  q0  E  d s
a
 q0 V
 - 1.6 x 10 -19 1.6 x 10 3 
 - 2.6 x 10 -16 J
  work done 
(e) Loss of PE = gain in KE = ½mv2
v 

2KE
m

22.6 x 10 -16 
9.1 x 10 -31 
 2.4 x 10 7 ms -1
51
Norah Ali Almoneef
Example
 Calculate the electrostatic potential energy between 2
protons in a Uranium nucleus separated by 2 x 10-15 m.
U r   k
q1q2
r
 9.0 x 109
~ 10 13 J
52
Norah Ali Almoneef

1.6 x 10

-19
2 x 10-15

2
Example
The charge distribution as shown is referred to as a linear
quadrupole. (a) What is the potential at a point on the axis where
x > a? (b) What happens when x >> a?
 1 2 1 
 =
V = ke Q 
 xa x xa
2keQa 2
V=
x3
53
Norah Ali Almoneef
As x >> a
2k eQa 2
x 3  xa 2
Example
When a negative charge moves in the
direction of the electric field,
1. the field does positive work on it and the
potential energy increases
2. the field does positive work on it and the
potential energy decreases
3. the field does negative work on it and the
potential energy increases
4. the field does negative work on it and the
potential energy decreases
54
Norah Ali Almoneef
Example
The electric potential energy of two point charges
approaches zero as the two point charges move
farther away from each other.
If the three point charges shown here lie at the
vertices of an equilateral triangle, the electric
potential energy of the system of three charges is
1. positive
2. negative
3. zero
4. not enough information given to decide
55
Norah Ali Almoneef
Example
The electric potential due to a point charge approaches
zero as you move farther away from the charge.
If the three point charges shown here lie at the vertices
of an equilateral triangle, the electric potential at the
center of the triangle is
1. positive
2. negative
3. zero
4. not enough information given to decide
56
Norah Ali Almoneef
E-field between two parallel plates
Assume uniform field, and 3 mm plate separation
E = |VB – VA| / d = 12 / 3.0x10-3 = 4000 V/m
E directed from A (+ve) to B (ve)
A(+ve) plate is at higher potential than –ve plate.
Potential difference between plates = potential difference
between battery terminals because all points on a conductor
in equilibrium are at the same electric potential; no potential
difference exists between a terminal and any portion of the
plate to which it is connected.
57
Norah Ali Almoneef
Example
If a positive charge be moved against the electric field, then
what will happen to the energy of the system?
If a positive charge be moved against the electric field, then
energy will be used from an outside source.
58
Norah Ali Almoneef
Example
If 80 J of work is required to transfer 4 C charge from infinity to a point,
find the potential at that point
W =80 J,
V=
59
q = 4 C,
W 80
=
= 20 V
Q
4
Norah Ali Almoneef
V =?
Example
Electric Field and Electric Potential
4. Which of the following figures have V=0 and E=0 at red
point?
q
q
q
A
60
B
q
q
q
-q
q
q
-q
q
C
Norah Ali Almoneef
D
-q
q
-q
E
Example
What is the electric potential at a distance of 0.529 A from
the proton?
V = kq/r = 9x109 N m2//C2 x1.6x10-19 C/0.529 x10-10m
V = 27. 2 J/C = 27. 2 Volts
61
Norah Ali Almoneef
Example
Two test charges are brought separately to the vicinity of a
positive charge Q
r
Q
q
Charge +q is brought to pt A, a distance r
A
from Q
Q
2r
Charge +2q is brought to pt B, a distance
2r from Q
I) Compare the potential energy of q (UA) to that of 2q (UB)
(b) UA = UB
(c) UA > UB
(a) UA < UB
The potential energy of q is proportional to Qq/r
The potential energy of 2q is proportional to Q(2q)/(2r) = Qq/r
Therefore, the potential energies UA and UB are EQUAL!!!
62
Norah Ali Almoneef
B
2q
Example
II) Suppose charge 2q has mass m and is released from rest
from the above position (a distance 2r from Q). What is its
velocity vf as it approaches r = ∞ ?
(a)
vf 
1 Qq
4pe 0 mr
(b)
vf 
1 Qq
2pe 0 mr
(c)
vf 0
The principle at work here is CONSERVATION OF ENERGY.
Initially:
The charge has no kinetic energy since it is at rest.
The charge does have potential energy (electric) = UB.
Finally:
The charge has no potential energy (U  1/R)
The charge does have kinetic energy = KE
U B  KE
63
Norah Ali Almoneef
1 Q (2q) 1
 mv 2f
4pe 0 2r
2
v 2f
1 Qq

2pe 0 mr
Example
: The electron in the Bohr model of the atom can exist at only certain orbits. The
smallest has a radius of .0529nm, and the next level has a radius of .212m.
a) What is the potential difference between the two levels?
b) Which level has a higher potential?
q
V k
r
e
V1  k
r1
r1
19
1
.
6

10
V1  (9 109 )
 27.2V
9
.0529 10
19
1
.
6

10
V2  (9 109 )
 6.79V
9
.0212 10
r2
r1 is at a higher potential.
potential diff  V  27.2  6.79  20.4V
64
Norah Ali Almoneef
+e
Example
65
Norah Ali Almoneef
Example
What is the electric potential difference between A and B?
VA  VB  240V
66
Norah Ali Almoneef
Example
Ex: Given that q1 = +2.40 nC and q2 = -6.50 nC.
(a) what is the electric potential at points A and
B?
The potential is a scalar (not vector) sum of the electric potentials
produced by the individual charges:

 

Electric Potential
k  2.40 10 9 C k  6.5 109 C
VA 
 737V

at A:
.050m
.050m
Electric Potential
k  2.40 10 9 C k  6.5 109 C 
 704V

VB 
at B:
.060m
.080m
67

Norah Ali Almoneef

Example
Ex: Given that q1 = +2.40 nC and q2 = -6.50 nC.
(b) what is the work done by the electric field on a
point charge of 2.50 nC that travels from A to
B?
V = EPE/q0
WAB = EPEA – EPEB
WAB  q0VA  q0VB
 2.50nC  737V   2.50nC  704V 
 82.5nJ
68
Norah Ali Almoneef
Example
What is the total energy of the electron in a hydrogen atom? In
the ground state the electron orbit about the proton has a radius
equal to one Bohr radius rB = 5.29x10-11 m.
qe = -qp = -1.60x10-19 C, me = 9.11x10-31 kg and ve = 2.2x106 m/s
Total Energy  E  KE  PE
EPE  qe V
kqp 
 kqp
18
EPE  qe 
J
 r   
  4.35  10
 B

But the electron in a hydrogen atom also has kinetic energy:
1
2
KE  me ve
2
KE  2.20  10 18 J
Total energy of e- in hydrogen atom is EPE + KE = -2.15x10-18 J
This is the electron’s binding energy, i.e., how much energy is
required to rip off an electron!
69
Norah Ali Almoneef
Example:
Three identical point charges (q = +2.0 μC each) are brought
from infinity and fixed to a straight line so that the spacing
between adjacent charges is d = 0.40 m. Determine the electric
potential energy of this group.
q
U=k
U=k
q1 q2
r12
+
q1 q3
r13
+
q2 q3
r23
2x10-6 x2x10-6
0.4
+
EPE = 0.23 J
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Norah Ali Almoneef
q
d
2x10-6 x2x10-6
+
0.4
q
d
2x10-6 x2x10-6
0.8
Example
71
Norah Ali Almoneef
Example
A
C
E
Points A, B, and C lie in a
B
uniform electric field.
What is the potential difference between points A and B?
ΔVAB =VB -VA
a) ΔVAB > 0
b) ΔVAB = 0
c) ΔVAB < 0
The electric field, E, points in the direction of decreasing
potential
Since points A and B are in the same relative horizontal
location in the electric field there is on potential difference
between them
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Norah Ali Almoneef
Example
A
Points A, B, and C lie in a E
B
uniform electric field.
Point C is at a higher potential than point A.
True
C
False
As stated previously the electric field points in the direction of
decreasing potential
Since point C is further to the right in the electric field and
the electric field is pointing to the right, point C is at a lower
potential
The statement is therefore false
73
Norah Ali Almoneef
Example
A
C
Points A, B, and C lie in a E
B
uniform electric field.
If a negative charge is moved from point A to point B, its
electric potential energy
a) Increases.
b) decreases.
c) doesn’t change.
The potential energy of a charge at a location in an electric field
is given by the product of the charge and the potential at the
location
As shown in Example, the potential at points A and B are the
same
Therefore the electric potential energy also doesn’t change
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Norah Ali Almoneef
Example
A
C
Points A, B, and C lie in a E
B
uniform electric field.
Compare the potential differences between points A and C
and points B and C.
a)VAC > VBC
b) VAC = VBC
c) VAC < VBC
In Example 4 we showed that the the potential at points A and
B were the same
Therefore the potential difference between A and C and the
potential difference between points B and C are the same
Also remember that potential and potential energy are scalars
and directions do not come into play
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Norah Ali Almoneef
Example
A positive charge is released from rest in a region of
electric field. The charge moves:
a) towards a region of smaller electric potential
b) along a path of constant electric potential
c) towards a region of greater electric potential
A positive charge placed in an electric field will experience a
force given by F  q E
But E is also given by E   dV
Therefore
F  q E  q
V
d
Since q is positive, the force F points in the direction opposite to
increasing potential or in the direction of decreasing potential
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Norah Ali Almoneef
Example
If you want to move in a region of electric field without
changing your electric potential energy.You would move
a) Parallel to the electric field
b) Perpendicular to the electric field
The work done by the electric field when a charge moves
from one point to another is given by
b
Wa b
  b  
  F  dl   q0 E  dl
a
a
The way no work is done by the electric field is if the
integration path is perpendicular to the electric field giving a
zero for the dot product
77
Norah Ali Almoneef
Example
2x10-10 C from point
A to point B and does 5x10-6 J of work.
 What is the difference in potential energies of A and B
(EPEA – EPEB)?
 Electric force moves a charge of
EPEA – EPEB = 5x10-6 J
 What is the potential difference between A and B (VA – VB)?
V = 25000 V Point A is higher potential
WAB = - (EPEB - EPEA )
∆PE: = EPEA - EPEB = 5x10-6 J
∆V: VA –VB = (EPEA – EPEB ) / q= 5x10-6J/2x10-10C = 25000 V
78
Norah Ali Almoneef
Example
What is the electric potential at the center of the square?
r 2  r 2  0.10 
2
2r 2  .01
r  .071m
45º
45º
r
r
6


q
5

10
  6.34 105V
V  k  9 109 
r
 .071 
r
r
V  k
i
6
q
9   10  10
V  k  9 10 
r
.071

qi
r
Vtotal  1.27 106  (1.27 106 )  6.34 105  6.34 105
79
Vtotal  1.27 106
Norah Ali Almoneef
J
C

  1.27 106 V

Example A proton is moved from the negative plate to the positive
plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the
electric field is uniform with a magnitude of 1500N/C.
a) How much work would be required to move a proton from the negative
to the positive plate?
b) What is the potential difference between the plates?
c) If the proton is released from rest at the positive plate, what speed will it
have just before it hits the negative plate?
W  Fx cos
1
FE  qE
W  qEx
N
W  (1.6 10 C )(1500 )(.015m)
C
W  3.6 10 18 J
19
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Norah Ali Almoneef
Example
A proton is moved from the negative plate to the positive plate of
a parallel-plate arrangement. The plates are 1.5cm apart, and
the electric field is uniform with a magnitude of 1500N/C.
b) What is the potential difference between the plates?
V  Ed
N
V  (1500 )(.015m)
C
J
V  22.5
C
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Norah Ali Almoneef
Example
A proton is moved from the negative plate to the positive plate of
a parallel-plate arrangement. The plates are 1.5cm apart, and
the electric field is uniform with a magnitude of 1500N/C.
c) If the proton is released from rest at the positive plate, what
speed will it have just before it hits the negative plate?
UE  K
qV 
v
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Norah Ali Almoneef
1 2
mv
2
2qV
m
2(1.6  10 19 )( 22.5)
v
1.67  10  27
m
v  6.57  10 4
s
Example
Calculate the electric potential energy between each pair of
charges and add them together.
6
6
q1q2
9 ( 4 10 )( 4 10 )
U12  k
 (9 10 )
 0.72 J
r
.2
6
6
q2 q3
9 (4 10 )( 4 10 )
U 23  k
 (9 10 )
 0.72 J
r
.2
6
6
q1q3
9 (4 10 )( 4 10 )
U13  k
 (9 10 )
 0.72 J
r
.2
U total  .72 J  (.72 J )  (.72 J )
83
U total  .72 J
Norah Ali Almoneef
EXAMPLE
Charges +Q and –Q are arranged at the corners of a
square as shown. When the magnitude of the electric
field E and the electric potential V are determined at P,
the center of the square, we find that
A. E ≠ 0 and V > 0.
B. E = 0 and V = 0.
C. E = 0 and V > 0.
D. E ≠ 0 and V < 0.
E. None of these is correct.
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Norah Ali Almoneef
EXAMPLE
Two equal positive charges are placed in an external
electric field. The equipotential lines shown are at 100 V
intervals. The potential for line c is
A.100 V.
200V
00V
a
b
c
B.100 V.
C.200 V.
D.200 V.
E.zero
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Norah Ali Almoneef
Q
Q
EXAMPLE
Two equal positive charges are placed in an external
electric field. The equipotential lines shown are at 100 V
intervals. The work required to move a third charge, q =
e, from the 100 V line to b is
A.100 eV.
B.100 eV.
200V
00V
a
b
c
C.200 eV.
D.200 eV.
E.zero
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Norah Ali Almoneef
Q
Q
EXAMPLE
The potential at a point due to a unit positive point
charge is found to be V. If the distance between the
charge and the point is tripled, the potential becomes
A. V/3.
B. 3V.
C. V/9.
D. 9V.
E. 1/V 2 .
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Norah Ali Almoneef
EXAMPLE
A proton is released from rest in a uniform electric field that has a
magnitude of 8.0 104 V/m and is directed along the positive x axis.
The proton undergoes a displacement of 0.50 m in the direction of E.
(a) Find the change in electric potential between points A and B.
(b) Find the change in potential energy of the proton for
this displacement.
(a)
V   Ed   (8.0 *104 V / m)(0.50 m)
 4.0 *104 V
U  q0 V  eV
(b)
 (1.6 *10 19 C )( 4.0 *10 4 V )
 6.4 *10 15 J
The negative sign means the potential energy of the proton
decreases as it moves in the direction of the electric field. As
the proton accelerates in the direction of the field, it gains
kinetic energy and at the same time loses electric potential
energy (because energy is conserved).
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Norah Ali Almoneef
EXAMPLE
A proton is released from rest at point B, where
the potential is 0 V. Afterward, the proton
1. moves toward A with an increasing speed.
2. moves toward A with a steady speed.
3. remains at rest at B.
4. moves toward C with a steady speed.
5. moves toward C with an increasing speed.
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Norah Ali Almoneef
EXAMPLE
90
Norah Ali Almoneef
EXAMPLE
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Norah Ali Almoneef
Summary
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.
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Norah Ali Almoneef
Summary
Parallel Plates
Consider Two parallel plates of equal
VA + + + +
and opposite charge, a distance d apart.
+q
Constant E field: F = qE
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.
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Norah Ali Almoneef
E
Summary of Formulas
Electric Potential
Energy and Potential
Electric Potential Near
Multiple charges:
WorkAB = q(VA – VB)
Oppositely Charged
Parallel Plates:
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Norah Ali Almoneef
kQq
U
U
; V
r
q
kQ
V 
r
Work BY E-field
V  Ed ;
V
E
d
Summary
 Electric potential difference
B
V =
E . ds = =  E . d
A
 Since only differences in potential matter,
the location of the
zero of the potential can be chosen as we wish
 Electric potential of a point charge
 If we chose the zero to be infinitely far from a point charge,
we can write its potential as
kq
V 
 Equipotentials
r
These are surfaces of equal potential difference. The surface of a
conductor in equilibrium is an equipotential.
95