copyright © 2009 pearson education, inc. chapter 23 electric potential

Copyright © 2009 Pearson Education, Inc.

Chapter 23Electric Potential


The electrostatic force is conservative – potential energy can be defined.

Change in electric potential energy is negative of work done by electric force:

23-1 Electrostatic Potential Energy and Potential Difference


Electric potential is defined as potential energy per unit charge:

Unit of electric potential: the volt (V):

1 V = 1 J/C.



Only changes in potential can be measured, allowing free assignment of V = 0:




Conceptual Example 23-1: A negative charge.

Suppose a negative charge, such as an electron, is placed near the negative plate at point b, as shown here. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change?


Analogy between gravitational and electrical potential energy:




Electrical sources such as batteries and generators supply a constant potential difference. Here are some typical potential differences, both natural and manufactured:



Example 23-2: Electron in CRT.

Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference Vb – Va = Vba = +5000 V. (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration?


23-2 Relation between Electric Potential and Electric Field

The general relationship between a conservative force and potential energy:

Substituting the potential difference and the electric field:



The simplest case is a uniform field:



Example 23-3: Electric field obtained from voltage.

Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is 0.050 m, calculate the magnitude of the electric field in the space between the plates.



Example 23-4: Charged conducting sphere.

Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q.



The previous example gives the electric potential as a function of distance from the surface of a charged conducting sphere, which is plotted here, and compared with the electric field:



Example 23-5: Breakdown voltage.

In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize O2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3.0 × 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 cm.


23-3 Electric Potential Due to Point Charges

To find the electric potential due to a point charge, we integrate the field along a field line:



Setting the potential to zero at r = ∞ gives the general form of the potential due to a point charge:



Example 23-6: Work required to bring two positive charges close together.

What minimum work must be done by an external force to bring a charge q = 3.00 μC from a great distance away (take r = ∞) to a point 0.500 m from a charge Q = 20.0 µC?



Example 23-7: Potential above two charges.

Calculate the electric potential (a) at point A in the figure due to the two charges shown, and (b) at point B.


23-4 Potential Due to Any Charge Distribution

The potential due to an arbitrary charge distribution can be expressed as a sum or integral (if the distribution is continuous):

or



Example 23-8: Potential due to a ring of charge.

A thin circular ring of radius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center.



Example 23-9: Potential due to a charged disk.

A thin flat disk, of radius R0, has a uniformly distributed charge Q. Determine the potential at a point P on the axis of the disk, a distance x from its center.


An equipotential is a line or surface over which the potential is constant.

Electric field lines are perpendicular to equipotentials.

The surface of a conductor is an equipotential.

23-5 Equipotential Surfaces



Example 23-10: Point charge equipotential surfaces.

For a single point charge with Q = 4.0 × 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to V1 = 10 V, V2 = 20 V, and V3 = 30 V.


23-5 Equipotential SurfacesEquipotential surfaces are always perpendicular to field lines; they are always closed surfaces (unlike field lines, which begin and end on charges).



A gravitational analogy to equipotential surfaces is the topographical map – the lines connect points of equal gravitational potential (altitude).


The potential due to an electric dipole is just the sum of the potentials due to each charge, and can be calculated exactly. For distances large compared to the charge separation:

23-6 Electric Dipole Potential


23-7 E Determined from V

If we know the field, we can determine the potential by integrating. Inverting this process, if we know the potential, we can find the field by differentiating:

This is a vector differential equation; here it is in component form:


23-7 E Determined from V

Example 23-11: E for ring and disk.

Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge and (b) a uniformly charged disk.


23-8 Electrostatic Potential Energy; the Electron Volt

The potential energy of a charge in an electric potential is U = qV. To find the electric potential energy of two charges, imagine bringing each in from infinitely far away. The first one takes no work, as there is no field. To bring in the second one, we must do work due to the field of the first one; this means the potential energy of the pair is:


One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt:

1 eV = 1.6 × 10-19 J.

The electron volt is often a much more convenient unit than the joule for measuring the energy of individual particles.




Example 23-12: Disassembling a hydrogen atom.

Calculate the work needed to “disassemble” a hydrogen atom. Assume that the proton and electron are initially separated by a distance equal to the “average” radius of the hydrogen atom in its ground state, 0.529 × 10-10 m, and that they end up an infinite distance apart from each other.


A cathode ray tube contains a wire cathode that, when heated, emits electrons. A voltage source causes the electrons to travel to the anode.

23-9 Cathode Ray Tube: TV and Computer Monitors, Oscilloscope


The electrons can be steered using electric or magnetic fields.



Televisions and computer monitors (except for LCD and plasma models) have a largecathode ray tube as their display. Variations in the field steer the electrons on their way to the screen.



An oscilloscope displays an electrical signal on a screen, using it to deflect the beam vertically while it sweeps horizontally.



Review Questions

ConcepTest 23.1aConcepTest 23.1a Electric Potential Energy I Electric Potential Energy I A) proton

B) electron

C) both feel the same force

D) neither – there is no force

E) they feel the same magnitude force but opposite direction

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. Which side. Which feels the larger electric force?feels the larger electric force?

ConcepTest 23.1aConcepTest 23.1a Electric Potential Energy I Electric Potential Energy I A) proton

B) electron

C) both feel the same force

D) neither – there is no force

E) they feel the same magnitude force but opposite direction

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

Since F = qEF = qE and the proton and electron

have the same charge in magnitudesame charge in magnitude, they

both experience the same forcesame force. However,

the forces point in opposite directionsopposite directions

because the proton and electron are

oppositely chargedoppositely charged.

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. Which side. Which feels the larger electric force?feels the larger electric force?

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

A) proton

B) electron

C) both feel the same acceleration

D) neither – there is no acceleration

E) they feel the same magnitude acceleration but opposite direction

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. Which side. Which has the larger acceleration?has the larger acceleration?

ConcepTest 23.1bConcepTest 23.1b Electric Potential Energy II Electric Potential Energy II

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

A) proton

B) electron

C) both feel the same acceleration

D) neither – there is no acceleration

E) they feel the same magnitude acceleration but opposite direction

Since F = maF = ma and the electron is much less electron is much less

massivemassive than the proton, the electron electron

experiences the larger accelerationexperiences the larger acceleration.

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. Which side. Which has the larger acceleration?has the larger acceleration?

ConcepTest 23.1bConcepTest 23.1b Electric Potential Energy II Electric Potential Energy II

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

A) proton

B) electron

C) both acquire the same KE

D) neither – there is no change of KE

E) they both acquire the same KE but with opposite signs

ConcepTest 23.1cConcepTest 23.1c Electric Potential Energy III Electric Potential Energy III

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. When side. When it strikes the opposite plate, it strikes the opposite plate, which one has more KE?which one has more KE?

E

electronelectron

protonproton

E

ElectronElectron

ProtonProton++

--

A) proton

B) electron

C) both acquire the same KE

D) neither – there is no change of KE

E) they both acquire the same KE but with opposite signs

Since PE = PE = qVqV and the proton and electron

have the same charge in magnitudesame charge in magnitude, they

both have the same electric potential energysame electric potential energy

initially. Because energy is conserved, they

both must have the same kinetic energysame kinetic energy after

they reach the opposite plate.

ConcepTest 23.1cConcepTest 23.1c Electric Potential Energy III Electric Potential Energy III

A A protonproton and an and an electronelectron are in are in a constant electric field created a constant electric field created by oppositely charged plates. by oppositely charged plates. You release the You release the protonproton from the from the positivepositive side and the side and the electronelectron from the from the negativenegative side. When side. When it strikes the opposite plate, it strikes the opposite plate, which one has more KE?which one has more KE?

Which group of charges took more work to bring together from a very large initial distance apart?

+1

+1

+1

d d

d

+1+2d

Both took the same amount of work.

ConcepTest 23.2ConcepTest 23.2 Work and Potential EnergyWork and Potential Energy

The work needed to assemble a collection of charges is the same as the total PEtotal PE of those charges:

r

QQkPE

21

Which group of charges took more work to bring together from a very large initial distance apart?

+1

+1

+1

d d

d

+1+2d

Both took the same amount of work.

For case 1:For case 1: only only 11 pair pair

dk

dkPE

212

))((

For case 2:For case 2: there are there are 33

pairspairs

dk

dkPE

13

113

))((added over added over

all pairsall pairs

ConcepTest 23.2ConcepTest 23.2 Work and Potential EnergyWork and Potential Energy

A) V > 0

B) V = 0

C) V < 0

AA BB

What is the electric What is the electric

potential at point A?potential at point A?

ConcepTest 23.3aConcepTest 23.3a Electric Potential IElectric Potential I

Since Q2 (which is positivepositive) is closercloser

to point A than Q1 (which is negative)

and since the total potential is equal

to V1 + V2, the total potential is

positivepositive.

A) V > 0

B) V = 0

C) V < 0

AA BB


potential at point A?potential at point A?

ConcepTest 23.3aConcepTest 23.3a Electric Potential IElectric Potential I

A) V > 0

B) V = 0

C) V < 0

AA BB


potential at point B?potential at point B?

ConcepTest 23.3bConcepTest 23.3b Electric Potential IIElectric Potential II

Since Q2 and Q1 are equidistant

from point B, and since they have

equal and opposite charges, the

total potential is zerozero.

A) V > 0

B) V = 0

C) V < 0

AA BB


potential at point B?potential at point B?

ConcepTest 23.3bConcepTest 23.3b Electric Potential IIElectric Potential II

Follow-up:Follow-up: What is the potential What is the potential at the origin of the at the origin of the x yx y axes? axes?

Four point charges are Four point charges are

arranged at the corners of a arranged at the corners of a

square. Find the square. Find the electric electric

field field EE and the and the potential potential VV at at

the the center of the squarecenter of the square..

A) E = 0 V = 0

B) E = 0 V 0

C) E 0 V 0

D) E 0 V = 0

E) E = V regardless of the value

--QQ

--QQ ++QQ

++QQ

ConcepTest 23.4ConcepTest 23.4 Hollywood SquareHollywood Square

Four point charges are Four point charges are

arranged at the corners of a arranged at the corners of a

square. Find the square. Find the electric electric

field field EE and the and the potential potential VV at at

the the center of the squarecenter of the square..

A) E = 0 V = 0

B) E = 0 V 0

C) E 0 V 0

D) E 0 V = 0

E) E = V regardless of the value

--QQ

--QQ ++QQ

++QQ

The potential is zeropotential is zero: the scalar

contributions from the two positive

charges cancel the two minus charges.

However, the contributions from the

electric field add up as vectors, and

they do not cancel (so it is non-zeroit is non-zero).

ConcepTest 23.4ConcepTest 23.4 Hollywood SquareHollywood Square

Follow-up:Follow-up: What is the direction What is the direction of the electric field at the center?of the electric field at the center?

At which point At which point

does does VV = 0? = 0?

1

3

2

4

+Q –Q

E) all of them

ConcepTest 23.5aConcepTest 23.5a Equipotential Surfaces IEquipotential Surfaces I

At which point At which point

does does VV = 0? = 0?

1

3

2

4

+Q –Q

E) all of them

All of the points are equidistant from both chargesAll of the points are equidistant from both charges. Since

the charges are equal and opposite, their contributions to

the potential cancel outcancel out everywhereeverywhere along the mid-plane

between the charges.

ConcepTest 23.5aConcepTest 23.5a Equipotential Surfaces IEquipotential Surfaces I

Follow-up:Follow-up: What is the direction of the electric field at all 4 points? What is the direction of the electric field at all 4 points?

Which of these configurations gives V = 0 at all points on the x axis?

D) all of the above E) none of the above

A)

x

+2C

-2C

+1C

-1CB)

x

+2C

-1C

+1C

-2CC)

x

+2C

-1C

-2C

+1C

ConcepTest 23.5bConcepTest 23.5b Equipotential Surfaces IIEquipotential Surfaces II

Only in case (A), where opposite charges lie

directly across the x axis from each other, do

the potentials from the two charges above the

x axis cancel the ones below the x axis.

Which of these configurations gives V = 0 at all points on the x axis?


A)

x

+2C

-2C

+1C

-1CB)

x

+2C

-1C

+1C

-2CC)

x

+2C

-1C

-2C

+1C

ConcepTest 23.5bConcepTest 23.5b Equipotential Surfaces IIEquipotential Surfaces II

Which of these configurations gives V = 0 at all points on the y axis?


ConcepTest 23.5cConcepTest 23.5c Equipotential Surfaces IIIEquipotential Surfaces III

A)

x

+2C

-2C

+1C

-1CB)

x

+2C

-1C

+1C

-2CC)

x

+2C

-1C

-2C

+1C

Which of these configurations gives V = 0 at all points on the y axis?


ConcepTest 23.5cConcepTest 23.5c Equipotential Surfaces IIIEquipotential Surfaces III

A)

x

+2C

-2C

+1C

-1CB)

x

+2C

-1C

+1C

-2CC)

x

+2C

-1C

-2C

+1C

Only in case (C), where opposite charges lie

directly across the y axis from each other, do

the potentials from the two charges above the

y axis cancel the ones below the y axis.

Follow-up:Follow-up: Where is Where is VV = 0 for configuration #2? = 0 for configuration #2?

Which two points have Which two points have the the samesame potential? potential?

A) A and C

B) B and E

C) B and D

D) C and E

E) no pair

A

C

B DE Q

ConcepTest 23.6ConcepTest 23.6 Equipotential of Point ChargeEquipotential of Point Charge

Since the potential of a point charge is:

only points that are at the same distancesame distance from charge Q are at the same potentialsame potential. This is true for points C and E.

They lie on an equipotential surfaceequipotential surface.

Which two points have Which two points have the the samesame potential? potential?

A) A and C

B) B and E

C) B and D

D) C and E

E) no pair

A

C

B DE Q

rQ

kV

ConcepTest 23.6ConcepTest 23.6 Equipotential of Point ChargeEquipotential of Point Charge

Follow-up:Follow-up: Which point has the smallest potential? Which point has the smallest potential?

Which requires the most work,

to move a positive charge from

P to points 1, 2, 3 or 4 ? All

points are the same distance

from P.

A) P 1

B) P 2

C) P 3

D) P 4

E) all require the same amount of work

P1

2

3

E

4

ConcepTest 23.7aConcepTest 23.7a Work and Electric Potential IWork and Electric Potential I

For path #1path #1, you have to push the

positive charge againstagainst the E field,

which is hard to dohard to do. By contrast,

path #4 is the easiest, since the

field does all the work.

Which requires the most work,

to move a positive charge from



from P.

A) P 1

B) P 2

C) P 3

D) P 4


P1

2

3

E

4

ConcepTest 23.7aConcepTest 23.7a Work and Electric Potential IWork and Electric Potential I

Which requires zero work, to

move a positive charge from



from P.

A) P 1

B) P 2

C) P 3

D) P 4


P1

2

3

E

4

ConcepTest 23.7bConcepTest 23.7b Work and Electric Potential IIWork and Electric Potential II

For path #3path #3, you are moving in a

direction perpendicular to the field

lines. This means you are moving

along an equipotential, which

requires no work (by definition).

Which requires zero work, to

move a positive charge from



from P.

A) P 1

B) P 2

C) P 3

D) P 4


P1

2

3

E

4

ConcepTest 23.7bConcepTest 23.7b Work and Electric Potential IIWork and Electric Potential II

Follow-up:Follow-up: Which path requires the least work? Which path requires the least work?


Ch. 23 Assignments

• Group problems #’s 4, 10, 12, 30, 34, 36, 42

• Homework problems #’s 1, 9, 11, 13, 15, 27, 29, 33, 49, 51, 57

• Finish reading Ch. 23 and begin looking over Ch. 24.

copyright © 2009 pearson education, inc. chapter 23 electric potential

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