Electric potential

University Physics, Chapter 23April 12, 2023 1

Electric PotentialElectric PotentialElectric PotentialElectric 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.


Electric Potential Energy (1)Electric Potential Energy (1)Electric Potential Energy (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

is the initial electric potential energy

is the final electric potential energy

f i e

i

f

U U U W

U

U


Electric Potential Energy (2)Electric Potential Energy (2)Electric Potential Energy (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,

The negative sign on the work:• If E does positive work then U < 0• If E does negative work then U > 0

0fU U U W


Constant Electric FieldConstant Electric FieldConstant Electric FieldConstant 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

For a constant electric field theforce is

… so the work done by the electric field on the charge is

W F d

cosW qE d qEd

Note: angle between and E d

F qE


Constant Electric Field - Special CasesConstant Electric Field - Special CasesConstant Electric Field - Special CasesConstant Electric Field - Special Cases

If the displacement is in the samedirection as the electric field

• 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

• A positive charge gains potential energy when it moves in the direction opposite to the electric field.

qEdUqEdW so so W qEd U qEd

so W qEd U qEd


Definition of the Electric PotentialDefinition of the Electric PotentialDefinition of the Electric PotentialDefinition 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

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

UV

q “potential energy per unit charge of a

test particle”


Electric Potential Electric Potential VVElectric Potential Electric Potential VV The electric potential, V, is defined as the

electric potential energy, U, per unit charge

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

UV

q


Electric Potential Difference, Electric Potential Difference, V (1)V (1)Electric Potential Difference, Electric Potential Difference, V (1)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

Hence we can relate the change in electric potential to the work done by the electric field on the charge

f if i

U U UV V V

q q q

eWV

q


Electric Potential Difference, Electric Potential Difference, V (2)V (2)Electric Potential Difference, Electric Potential Difference, V (2)V (2)

Taking the electric potential energy to be zero at infinity we have

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.

,eWV

q Explanation: i =

, f = x, so that V = V(x) 0


The VoltThe VoltThe VoltThe Volt

The commonly encountered unit joules/coulomb is called the volt, abbreviated V, after the Italian physicist Alessandro Volta (1745-1827)

With this definition of the volt, we can express the units of the electric field as

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

1 J1 V =

1 C

[ ] N J/m V[ ]

[ ] C C m

FE

q


Example: Energy Gain of a Proton (1)Example: Energy Gain of a Proton (1)Example: Energy Gain of a Proton (1)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 isU, and V = U / q (by definition of V), soU = q V = e[V()V(+)] = 450 eV


Example: Energy Gain of a Proton (2)Example: Energy Gain of a Proton (2)Example: Energy Gain of a Proton (2)Example: Energy Gain of a Proton (2)

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

The proton in this example would gain kinetic energy of 450 eV = 0.450 keV.

191 eV 1.6022 10 J

initial finalConservation of energyK = U = + 450 eVBecause the proton started at rest,K = 1.6x10-19 C x 450 V = 7.2x10-17 J


The Van de Graaff Generator (1)The Van de Graaff Generator (1)The Van de Graaff Generator (1)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


The Van de Graaff Generator (2)The Van de Graaff Generator (2)The Van de Graaff Generator (2)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


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

The Tandem Van de Graaff AcceleratorThe Tandem Van de Graaff AcceleratorThe Tandem Van de Graaff AcceleratorThe Tandem Van de Graaff Accelerator


Example: Energy of Tandem Accelerator Example: Energy of Tandem Accelerator (1)(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 energyaccelerating toward the terminal

• The stripped carbon ion with a +6e charge gainsenergy accelerating away from the terminal15 MV Tandem Van de Graaff at Brookhaven




12 -26

2

117

-26

The mass of a C nucleus is 1.99 10 kg

1

2

2 2 1.12 10 J3.36 10 m/s

1.99 10 kg

11% of the speed of light

K mv

Kv

m

v

1 2

1 2

-1911

1 and 6

7 10 MV 70 MeV

1.602 10 J70 MeV 1.12 10 J

1 eV

K U q V q V

q e q e

K e

K

University Physics, Chapter 23

Equipotential surface from eight point chargesfixed at the corners of a cube

April 12, 2023 18

Equipotential Surfaces and LinesEquipotential Surfaces and LinesEquipotential Surfaces and LinesEquipotential 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 threedimensions.

We will often take advantageof symmetries in the electric potentialand represent the equipotential surfacesas equipotential lines in a plane


If a charged particle moves perpendicular to electric field lines, no work is done

If the work done by the electric field is zero, then the electric potential must be constant

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

General ConsiderationsGeneral ConsiderationsGeneral ConsiderationsGeneral Considerations

V

We

q0 V is constant

0 if W qE d d E


Constant Electric FieldConstant Electric FieldConstant Electric FieldConstant 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


Electric Potential from a Single Point Electric Potential from a Single Point ChargeCharge

Electric Potential from a Single Point Electric Potential from a Single Point ChargeCharge

Electric field lines: radial lines emanating from the point charge

Equipotential surfaces (3D): concentric spheres Equipotential lines (2D): concentric circles

Positive charge Negative charge


Electric Potential from Two Oppositely Electric Potential from Two Oppositely Charged Point ChargesCharged Point Charges

Electric Potential from Two Oppositely Electric Potential from Two Oppositely Charged Point ChargesCharged 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 positiveelectric potential

The blue lines represent negativeelectric potential

Close to each charge, theequipotential lines resemblethose from a pointcharge


Electric Potential from Two Identical Point Electric Potential from Two Identical Point ChargesCharges

Electric Potential from Two Identical Point Electric Potential from Two Identical Point ChargesCharges

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 equipotentiallines are alwaysperpendicular tothe electric field lines

There are only positivepotentials

Close to each charge, theequipotential linesresemble those froma point charge


Calculating the Potential from the FieldCalculating the Potential from the FieldCalculating the Potential from the FieldCalculating the Potential from the Field

Work dW done on a particle with charge q by a force F over a displacement 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

Potential difference:

Potential:

dW F ds qE ds

f

iW qE ds

fe

f i i

WV V V E ds

q

(Convention: i = , f = x)( )

xV x E ds


Example: Charge Moves in E field (1)Example: Charge Moves in E field (1)Example: Charge Moves in E field (1)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.

Idea: Integrate 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.) c f

f i f c c i i cV V V V V V E ds E ds

E ds


Example: Charge Moves in E field (2)Example: Charge Moves in E field (2)Example: Charge Moves in E field (2)Example: Charge Moves in E field (2)

12

0 (ds perpendicular to E)

cos(45 ) distance

c f

f i i c

c

i

f f

c c

V V E ds E ds

E ds

E ds E ds E

distance = sqrt(2) d by Pythagoras

f iV V Ed


Electric Potential for a Point Charge (1)Electric Potential for a Point Charge (1)Electric Potential for a Point Charge (1)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

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:

2ˆ( )

kqE r r

r

2R RR

kq kq kqV E ds dr

r r R


Electric Potential for a Point Charge (2)Electric Potential for a Point Charge (2)Electric Potential for a Point Charge (2)Electric Potential for a Point Charge (2)

The electric potential V from a point charge q at a distance r is then

Positive point charge

Negative point charge

( )kq

V rr


Electric Potential from a System of Electric Potential from a System of ChargesCharges

Electric Potential from a System of Electric Potential from a System of ChargesCharges

We calculate the electric potential from a system of n point charges by adding the potential functions from each charge

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

1 1

n ni

ii i i

kqV V

r


Example: Superposition of Electric Example: Superposition of Electric Potential (1)Potential (1)


Assume we have a system of three point charges:q1 = +1.50 Cq2 = +2.50 Cq3 = -3.50 C

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)?




Answer: The electric potential at point P

is given by the sum of the electric potential from the three charges

V kq

i

rii1

3

kq

1

r1

q

2

r2

q

3

r3

k

q1

b

q2

a2 b2

q3

a

V 8.99109 N/C 1.5010 6 C

6.00 m

2.5010 6 C

8.00 m 2 6.00 m 2

3.5010 6 C

8.00 m

V 562 V

r1

r2 r3


Calculating the Field from the Potential Calculating the Field from the Potential (1)(1)


We can calculate the electric field from the electric potential starting with

Which allows us to write

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

We,

q

S

VE

s

dW qE ds

qdV qE ds E ds dV


Math Reminder - Partial DerivativesMath Reminder - Partial DerivativesMath Reminder - Partial DerivativesMath Reminder - Partial Derivatives

Given a function V(x,y,z), the partial derivatives are

Example: V(x,y,z) = 2xy2 + z3

they act on x, y, and z independently

Meaning: partial derivatives give the slope along the respective direction

V

x

V

y

V

x

V

x2y2

V

y4xy

V

x3z2




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

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

; ; x y z

V V VE E E

x y z

also written as E V E V


Example: Graphical Extraction of the Field Example: Graphical Extraction of the Field from the Potentialfrom the Potential

Example: Graphical Extraction of the Field Example: Graphical Extraction of the Field from the Potentialfrom the Potential

Assume a system of three point chargesq1 6.00 C 2 3.00 Cq 3 9.00 Cq 1 1, 1.5 cm,9.0 cmx y x2 , y2 6.0 cm,8.0 cm 3 3, 5.3 cm,2.0 cmx y


Example: Graphical Extraction of the Example: Graphical Extraction of the Field from the Potential (2)Field from the Potential (2)

Example: Graphical Extraction of the Example: Graphical Extraction of the Field from the Potential (2)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

The direction of the electric field points from the positive equipotential line to the negative potential line

52000 V 0 V1.3 10 V/m

1.5 cmS

VE

s


Electric Potential Energy for aElectric Potential Energy for aSystem of ParticlesSystem of Particles

Electric Potential Energy for aElectric Potential Energy for aSystem of ParticlesSystem 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


Electric Potential Energy for aElectric Potential Energy for aPair of Particles (1)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)


So, the electric potential energy of this two charge system is

where

Hence the electric potential of the two charge system is

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)

2 1( )U q V r 11( )

kqV r

r

1 2kq qU

r




Example: Electric Potential Energy (1)Example: Electric Potential Energy (1)Example: Electric Potential Energy (1)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?



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


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:



Example: Potential Due to Charges (1)Example: Potential Due to Charges (1)Example: Potential Due to Charges (1)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?

Answer:• Draw the picture• 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

q1=+q, q2=-2q, q3=+q

C


Example: Potential Due to Charges (2)Example: Potential Due to Charges (2)Example: Potential Due to Charges (2)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?

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

C

31 2( ) , where 2cos(30 )

qq q dV C R

R R R

2( ) 0

q q qV C

R R R

q1=+q, q2=-2q, q3=+q


Example: Four Charges (1)Example: Four Charges (1)Example: Four Charges (1)Example: Four Charges (1)

Consider a system of four point charges as shown. The four point charges have the values q1 =+1.0 C, q2 = +2.0 C, q3 = -3.0 C, and q4 = +4.0 C. 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?



Start: Bring in q1 from infinity• No work required

Bring in q2 from infinity



q1

q2

q3

q4

a

1 2q qU k

a

1 3 2 31 2

2 2

q q q qq qU k k k

a b a b

b

1 3 2 3 3 41 2 1 4 2 4

2 2 2 2

q q q q q qq q q q q qU k k k k k k

a b b aa b a b



1 31 2 1 4

2 2

2 3 3 42 4

2 2

q qq q q qU k

a b a b

q q q qq q

b aa b

Energy of the complete assembly:

= sum of pairs

Answer: 1.2 · 10-3 J


Example: 12 Electrons on a CircleExample: 12 Electrons on a CircleExample: 12 Electrons on a CircleExample: 12 Electrons on a CircleQuestion: 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 CircleExample: 12 Electrons on a CircleExample: 12 Electrons on a CircleExample: 12 Electrons on a Circle

Part I: Superposition principle:

Part II: pair of electrons on opposite ends of the circle produce fields at the center that cancel each other.

12

1

12( )

i

e eV C k k

R R

( ) 0E C

Symmetry dictates that the field be 0, otherwise, which direction would the field vector point?

Electric potential

Documents

Transcript of Electric potential