Electric potential

Electric Potential
 We have been studying the electric field
 Next topic: the electric potential
 Note the similarity between the gravitational force and the
electric force
 Gravitation can be described in terms of a gravitational
potential and we will show that the electric potential is
analogous
 We will see how the electric potential is related to energy
and work
 We will see how we can calculate the electric potential from
the electric field and vice versa

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

University Physics, Chapter 23
February 1, 2013 1

Electric Potential Energy (1)
 The electric force, like the gravitational force, is a
conservative force
• For a conservative force, the work is path-independent
 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 − U i = −We
U i is the initial electric potential energy
U f is the final electric potential energy

February 1, 2013 2

Electric Potential Energy (2)
 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

February 1, 2013 3

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
r r
W = F ×d
 For a constant electric field the
force is r
r
F = qE
 … so the work done by the electric field on the charge is
r r
W = qE ×d = qEd cos θ r
r
Note: θ = angle between E and d
February 1, 2013 University Physics, Chapter 23 4

Constant Electric Field - Special Cases
 If the displacement is in the same
direction as the electric field
W = qEd so ∆U = −qEd
W = qEd so ∆U = −qEd
• A positive charge loses potential energy
when it moves in the direction of the
electric field.
 If the 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.


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
U “potential energy per unit charge of a
V= test particle”
q
 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
• Units: Volt, symbol V
1V = 1J/C

February 1, 2013 6

Electric Potential V
 The electric potential, V, is defined as the electric
potential energy, U, per unit charge
U
V=
q
 The electric potential is a characteristic of the
electric field, regardless of whether a charged
object has been placed in that field
(because U ∝ q)
 The electric potential is a scalar
 The electric potential is defined everywhere in
space as a value, but has no direction

February 1, 2013 7

Electric Potential Difference, ∆ V (1)
 The electric potential difference between an initial point i
and final point f can be expressed in terms of the electric
potential energy of q at each point
Uf U i ∆U
∆V = V f − Vi = − =
q q q

 Hence we can relate the change in electric potential to the
work done by the electric field on the charge
We
∆V = −
q

February 1, 2013 8

Electric Potential Difference, ∆ V (2)
 Taking the electric potential energy to be zero
at infinity we have
We ,∞ Explanation: i = ∞ , f = x,
V =−
q so that ∆V = V(x) − 0

where We,∞ is the work done by the electric field on the charge
as it is brought in from infinity

 The electric potential can be positive, negative, or zero, but
it has no direction (i.e., scalar not vector)

 The SI unit for electric potential is joules/coulomb, i.e.,
volt.
February 1, 2013 9

The Volt
 The commonly encountered unit joules/coulomb is
called the volt, abbreviated V, after the Italian physicist
Alessandro Volta (1745-1827)
1J
1V=
1C
 With this definition of the volt, we can express the units of
the electric field as
[ F ] N J/m V
[E] = = = =
[q] C C m

 For the remainder of our studies, we will use the unit V/m
for the electric field

February 1, 2013 10

Example: Energy Gain of a Proton (1)
 A proton is placed between two parallel -
+
conducting plates in a vacuum as shown.
The potential difference between the two
plates is 450 V. The proton is released
from rest close to the positive plate.
Question: What is the kinetic energy of the
proton when it reaches the negative
plate?
Answer:

The potential difference between the two plates is 450 V.

The change in potential energy of the proton is
∆U, and ∆V = ∆U / q (by definition of V), so
∆U = q ∆V = e[V(−)−V(+)] = −450 eV

February 1, 2013 11

Example: Energy Gain of a Proton (2)
Conservation of initial final
energy
∆K = − ∆U = + 450
eV
Because the proton started at
rest,
K = 1.6x10-19 C x 450 V =
7.2x10-17 J
 Because the acceleration of a charged particle across a potential difference is
often used in nuclear and high energy physics, the energy unit electron-volt (eV)
is common
 An eV is the energy gained by a charge e that accelerates across an electric
potential of 1 volt
1 eV = 1.6022 × −19 J
10
 The proton in this example would gain kinetic energy of 450 eV = 0.450 keV.

February 1, 2013 12

The Van de Graaff Generator (1)
 A Van de Graaff generator is a device that
creates high electric potential
 The Van de Graaff generator was invented
by Robert J. Van de Graaff, an American
physicist (1901-1967)
 Van de Graaff generators can produce
electric potentials up to many 10s of
millions of volts
 Van de Graaff generators are used in
particle accelerators
 We have been using a Van de Graaff
generator in lecture demonstrations and we
will continue to use it

February 1, 2013 13

The Van de Graaff Generator (2)
 The Van de Graaff generator
works by applying a positive
charge to a non-conducting
moving belt using a corona
discharge
 The moving belt driven by an
electric motor carries the
charge up into a hollow metal
sphere where the charge is
taken from the belt by a
pointed contact connected to
the metal sphere
 The charge that builds up on
the metal sphere distributes
itself uniformly around the
outside of the sphere
 For this particular Van de
Graaff generator, a voltage
limiter is used to keep the Van
de Graaff generator from
producing sparks larger than
desired

February 1, 2013 14

The Tandem Van de Graaff Accelerator
 One use of a Van de Graaff generator is to accelerate particles
for condensed matter and nuclear physics studies
 A clever design is the tandem Van de Graaff accelerator

 A large positive electric potential is created by a huge Van de
Graaff generator
 Negatively charged C ions get accelerated toward the +10 MV
terminal (they gain kinetic energy)
 Electrons are stripped from the C and the now positively charged C
ions are repelled by the positively charged terminal and gain more
kinetic energy

February 1, 2013 15

Example: Energy of Tandem Accelerator (1)
 Suppose we have a tandem Van de Graaff accelerator
that has a terminal voltage of 10 MV (10 million volts). We
want to accelerate 12C nuclei using this accelerator.
Questions:
What is the highest energy we can attain for carbon nuclei?
What is the highest speed we can attain for carbon nuclei?
Answers:
 There are two stages to the acceleration
• The carbon ion with a -1e charge gains energy
accelerating toward the terminal
• The stripped carbon ion with a +6e charge gains
energy accelerating away from the terminal

15 MV Tandem Van de Graaff at Brookhaven

February 1, 2013 16

Example: Energy of Tandem Accelerator (2)
K = −∆U = −  q1V + q2 ( −V ) 
 
q1 = −1 e and q2 = +6 e
K = 7 e ×10 MV = 70 MeV
1.602 ×10-19 J
K = 70 MeV × = 1.12 ×10−11 J
1 eV

The mass of a 12C nucleus is 1.99 × -26 kg
10
1 2
K = mv
2
2K 2×1.12 × −11 J
10
v= = = 3.36 × 7 m/s
10
m 1.99 × -26
10 kg
v = 11% of the speed of light

February 1, 2013 17

Equipotential Surfaces and Lines
 When an electric field is present, the electric potential has a
given value everywhere in space V(x) = potential function
 Points close together that have the same electric potential form
an equipotential surface i.e., V(x) = constant value
 If a charged particle moves on an equipotential surface, no work
is done
 Equipotential surfaces exist in three
dimensions.
 We will often take advantage
of symmetries in the electric potential
and represent the equipotential surfaces
as equipotential lines in a plane
Equipotential surface from eight point charges
fixed at the corners of a cube

February 1, 2013 18

General Considerations
 If a charged particle moves perpendicular to electric
field lines, no work is done
r r r r
W = qE ×d = 0 if d ⊥ E

 If the work done by the electric field is zero, then the
electric potential must be constant
We
∆V = − = 0 ⇒ V is constant
q

 Thus equipotential surfaces and lines must always be
perpendicular to the electric field lines

February 1, 2013 19

Constant Electric Field
 Electric field lines: straight lines parallel to E
 Equipotential surfaces (3D):
planes perpendicular to E
 Equipotential lines (2D):
straight lines perpendicular to E

February 1, 2013 20

Electric Potential from a Single Point
Charge
 Electric field lines: radial lines emanating from the point
charge
 Equipotential surfaces (3D): concentric spheres
 Equipotential lines (2D): concentric circles

Positive charge Negative charge
February 1, 2013 21

Electric Potential from Two Oppositely
Charged Point Charges
 The electric field lines from two oppositely charge point charges are
a little more complicated
 The electric field lines originate on the positive charge and
terminate on the negative charge
 The equipotential lines are always perpendicular to the electric field
lines
 The red lines represent positive
electric potential
 The blue lines represent negative
electric potential
 Close to each charge, the
equipotential lines resemble
those from a point
charge

February 1, 2013 22

Electric Potential from Two Identical Point
Charges
 The electric field lines from two identical point charges are
also complicated
 The electric field lines originate on the positive charge and
terminate at infinity
 Again, the equipotential
lines are always
perpendicular to
the electric field lines
 There are only positive
potentials
 Close to each charge, the
equipotential lines
resemble those from
a point charge

February 1, 2013 23

Calculating the Potential from the Field
 Work dW done on a particle with charge q by a force F over
a displacement ds:
r r r r
dW = F ×ds = qE ×ds
 Work done by the electric force on the particle as it moves
in the electric field from some initial point i to some final
point f
f r r
W = ∫ qE ×ds
i

 Potential difference:
We f r r
∆V = V f − Vi = − = − ∫ E ×ds
q i

 Potential:
r ∞ r r (Convention: i = ∞, f = x)
V ( x ) = ∫ E ×ds
x

February 1, 2013 24

Example: Charge Moves in E field (1)
 Given the uniform electric
field E, find the potential
difference Vf-Vi by moving a
test charge q0 along the path
icf, where cf forms a 45º
angle with the field.
r r
 Idea: Integrate E ×ds along
the path connecting i and c,
then c and f. (Imagine that we
move a test charge q0 from i
to c and then from c to f.)
V f − Vi = ( V f − Vc ) + ( Vc − Vi ) = − ∫ ( i
c r r
) ( f r r
E ×ds + − ∫ E ×ds
c )

Example: Charge Moves in E field (2)

c r r f r r
V f − Vi = − ∫ E ×ds − ∫ E ×ds
i c
c r r
∫ i
E ×ds = 0 (ds perpendicular to E)
f r r f
∫ c
E ×ds = ∫ E ds cos(45°) = E × distance ×
c
1
2

distance = sqrt(2) d by Pythagoras

V f − Vi = − Ed

February 1, 2013 26

Electric Potential for a Point Charge (1)
 We’ll derive the electric potential for a point source q, as a
function of distance R from the source
• That is, V(R)
 Remember that the electric field from a point charge q at
a distance r is given by
r r kq
E (r ) = 2 r
ˆ
r
 The direction of the electric field from a point charge is
always radial
• V is a scalar
 We integrate from distance R (distance from the point
charge) along a radial to infinity:
∞
∞ r r ∞ kq  kq  kq
V = ∫ E g = ∫ 2 dr = −   =
ds
R R r
 r R R
February 1, 2013 27

Electric Potential for a Point Charge (2)
 The electric potential V from a point charge q at a distance r is
then
kq
V (r ) = Negative point charge
r

Positive point charge

February 1, 2013 28

Electric Potential from a System of
Charges
 We calculate the electric potential from a system of n point
charges by adding the potential functions from each charge
n
kqi
n
V = ∑ Vi = ∑
i =1 i =1 ri
 This summation produces an electric potential at all points in
space – a scalar function
 Calculating the electric potential from a group of point
charges is usually much simpler than calculating the electric
field
• It’s a scalar

February 1, 2013 29

Example: Superposition of Electric
Potential (1)
 Assume we have a system of
three point charges:
q1 = +1.50 µC
q2 = +2.50 µC
q3 = -3.50 µC
 q1 is located at (0,a)
q2 is located at (0,0)
q3 is located at (b,0)
a = 8.00 m and b = 6.00 m
Question: What is the electric
potential at point P located at
(b,a)?

February 1, 2013 30

Example: Superposition of Electric
Potential (2)
r1
Answer:
 The electric potential at point P is
given by the sum of the electric r2 r3
potential from the three charges
kqi3  q1 q2 q3   q1 q2 q3 
V =∑ = k + +  = k + + 
i=1 ri  r1 r2 r3  b a +b a
2 2

 
1.50 ⋅10−6 C −6 −6

(
V = 8.99 ⋅109 N/C )
 6.00 m
+
2.50 ⋅10 C
+
−3.50 ⋅10 C 
8.00 m 
(8.00 m ) + (6.00 m )
2 2

 


V = 562 V

February 1, 2013 31

Calculating the Field from the Potential (1)
 We can calculate the electric field from the electric
potential starting with
We,∞ r r
V =− dW = qE gds
q
 Which allows us to write
r r r r
− qdV = qE gds ⇒ E g = −dV
ds

 If we look at the component of the electric field along the
direction of ds, we can write the magnitude of the electric
field as the partial derivative along the direction s
∂V
ES = −
∂s

February 1, 2013 32

Math Reminder - Partial Derivatives
 Given a function V(x,y,z), the partial derivatives are
∂V ∂V ∂V
they act on x, y, and z independently
∂x ∂y ∂x

 Example: V(x,y,z) = 2xy2 + z3

∂V
= 2y 2
∂x
Meaning: partial derivatives
∂V give the slope along the
= 4 xy
∂y respective direction
∂V
= 3z 2
∂x

February 1, 2013 33

Calculating the Field from the Potential (2)
 We can calculate any component of the electric field by
taking the partial derivative of the potential along the
direction of that component
 We can write the components of the electric field in terms
of partial derivatives of the potential as
∂V ∂V ∂V
Ex = − ; Ey = − ; Ez = −
∂x ∂y ∂z
 In terms of graphical representations of the electric
potential, we can get an approximate value for the electric
field by measuring the gradient of the potential
perpendicular to an equipotential line
r r r
E = −∇V also written as E = −∇V

February 1, 2013 34

Example: Graphical Extraction of the Field
from the Potential
 Assume a system of three point charges
q1 = −6.00 µC q2 = −3.00 µ C q3 = +9.00 µ C
( x1 , y1 ) = ( 1.5 cm,9.0 cm ) (x2 , y2 ) = (6.0 cm, 8.0 cm ) ( x3 , y3 ) = ( 5.3 cm, 2.0 cm )

February 1, 2013 35

Example: Graphical Extraction of the Field
from the Potential (2)
 Calculate the magnitude of the
electric field at point P
 To perform this task, we draw a
line through point P perpendic-
ular to the equipotential line
reaching from the equipotential
line of +1000 V to the line of –
1000 V
 The length of this line is 1.5 cm.
So the magnitude of the
electric field can be
approximated as
ES = −
∆V
=
( +2000 V ) − ( 0 V ) = 1.3 × 5 V/m
10
∆s 1.5 cm
 The direction of the electric
field points from the positive
equipotential line to the
negative potential line

February 1, 2013 36

Electric Potential Energy for a
System of Particles
 So far, we have discussed the electric potential energy of a
point charge in a fixed electric field
 Now we introduce the concept of the electric potential
energy of a system of point charges
 In the case of a fixed electric field, the point charge itself
did not affect the electric field that did work on the
charge
 Now we consider a system of point charges that produce
the electric potential themselves
 We begin with a system of charges that are infinitely far
apart
• This is the reference state, U = 0
 To bring these charges into proximity with each other, we
must do work on the charges, which changes the electric
potential energy of the system
February 1, 2013 37

Pair of Particles (1)
 To illustrate the concept of the electric potential energy of
a system of particles we calculate the electric potential
energy of a system of two point charges, q1 and q2 .
 We start our calculation with the two charges at infinity
 We then bring in point charge q1
• Because there is no electric field and no corresponding electric
force, this action requires no work to be done on the charge
 Keeping this charge (q1) stationary, we bring the second
point charge (q2) in from infinity to a distance r from q1
• That requires work q2V1(r)

February 1, 2013 38

Pair of Particles (2)
 So, the electric potential energy of this two charge system is
kq1
U = q2V1 (r ) where V1 (r ) =
r

 Hence the electric potential of the two charge system is
kq1q2
U=
r
 If the two point charges have the same sign, then we must do positive
work on the particles to bring them together from infinity (i.e., we must
put energy into the system)
 If the two charges have opposite signs, we must do negative work on the
system to bring them together from infinity (i.e., energy is released
from the system)


Example: Electric Potential Energy (1)
 Consider three point charges at
fixed positions arranged an equal
distance d from each other with
the values
• q1=+q
• q2=-4q
• q3=+2q
Question:
What is the electric potential
energy U of the assembly of these
charges?

February 1, 2013 40

Answer:
 The potential energy is equal to
the work we must do to
assemble the system, bringing in
each charge from an infinite
distance

 Let’s build the system by
bringing the charges in from
infinity, one at a time

February 1, 2013 41

 Bringing in q1 doesn’t cost any
work

 With q1 in place, bring in q2

 We then bring in q3

 The work we must do to bring q3
to its place relative to q1 and q2
is then:

February 1, 2013 42

Example: Potential Due to Charges (1)
Question: Consider the system of 3
charges as indicated in the figure.
Relative to V=0 at infinity, what is the
electric potential V at point C, the
center of the triangle?
C
Answer:
• Draw the picture q1=+q, q2=-2q, q3=+q
• The task is different from the
previous example. Rather than
assembling the system, we are to
find the potential at the center.
• Use symmetry and superposition
principle for the 3 point charges


Example: Potential Due to Charges (2)
Consider the system of 3 charges as
indicated in the figure. Relative to
V=0 at infinity, what is the electric
potential V at point C, the center of
the triangle? C
q1 q2 q3 d
V (C ) = + + , where R =
R R R 2 cos(30o ) q1=+q, q2=-2q, q3=+q

+ q −2 q + q
⇒ V (C ) = + + =0
R R R

The charges sum to 0, the distances are all
identical.


Example: Four Charges (1)
 Consider a system of four
point charges as shown. The
four point charges have the
values q1 =+1.0 µC, q2 = +2.0 µC,
q3 = -3.0 µC, and q4 = +4.0 µC.
The charges are placed such
that a = 6.0 m and b = 4.0 m.
Question:
What is the electric potential
energy of this system of four
point charges?

February 1, 2013 45

 Start: Bring in q1 from infinity
• No work required
 Bring in q2 from infinity q2 q4
qq
U =k 1 2
a

 Bring in q3 from infinity a
q1q2 q1q3 q2 q3
U =k +k +k q1 q3
a b a 2 + b2

b
 Bring in q4 from infinity
q1q2 q1q3 q2 q3 q1q4 q2 q 4 q3q4
U =k +k +k +k +k +k
a b a +b
2 2
a +b
2 2 b a

February 1, 2013 46


Energy of the complete assembly:

 q1q2 q1q3 q1q4
U =k + +
 a b a2 + b2
q2 q3 q2 q4 q3q4 
+ + + 
a +b
2 2 b a 

= sum of pairs

Answer: 1.2 · 10-3 J

February 1, 2013 47

Example: 12 Electrons on a Circle
Question: Consider a system of 12 electrons
arranged on a circle with radius R as
indicated in the figure. Relative to V=0 at
infinity, what are the electric potential V
and the electric field E at point C?

Answer:
• Draw the picture
• This is a two-part question
• Part I is about the electric potential
• Part II is about the electric field

The task is to find both at the center
of a circle.

Part I: The question did not ask about

assembling this system of electrons.
Instead, use symmetry and super-position
principle for 12 point charges.
Part II: Use symmetry and consider pairs of
charges on opposite sides

Example: 12 Electrons on a Circle
Part I: Superposition principle:

V (C ) = ∑
12
k
( −e ) =k
( −12e )
i =1 R R

Part II: pair of electrons on
opposite ends of the circle produce
fields at the center that cancel each
other. r
E (C ) = 0
Symmetry dictates that the field be 0, otherwise,
which direction would the field vector point?


Electric potential

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