chapter 23 electric potential - wordpress.com · 2017-09-26 · 12 copyright © 2009 pearson...

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Copyright © 2009 Pearson Education, Inc.

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Lecture PowerPoints

Physics for Scientists and Engineers, with Modern

Physics, 4th edition

Giancoli

Chapter 23


Chapter 23Electric Potential

2


• Electric Potential Energy and Potential Difference

• Relation between Electric Potential and Electric Field

• Electric Potential Due to Point Charges

• Potential Due to Any Charge Distribution

• Equipotential Surfaces

Units of Chapter 23


• E Determined from V

• Electrostatic Potential Energy; the Electron Volt

• Cathode Ray Tube: TV and Computer Monitors, Oscilloscope

E

Units of Chapter 23

3


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

The work done by the electric field is given by

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

23-1 Electrostatic Potential Energy and Potential Difference

qEdWUU ab

qEdFdW


Electric potential is defined as potential energy per unit charge:

Unit of electric potential: the volt (V):

1 V = 1 J/C.


q

UV a

a

4


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


+ve charge moves from high to low potential

-ve charge moves from low to high potential

q

W

q

UUVVVV baab

abba



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?

5


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?

The electron will move towards the positive plate if released, thereby increasing its kinetic energy. Its potential energy must therefore decrease. However, it is moving to a region of higher potential V; the potential is determined only by the existing charge distribution and not by the point charge. U and V have different signs due to the negative charge.

0baab

ba

VVV

UU


Analogy between gravitational and electrical potential energy:


(a) Two rocks are at the same height. The larger rock has more potential energy.

(b) Two charges have the same electric potential. The 2Q charge has more potential energy.

(a) (b)

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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?

7



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?

e- accelerates towards the +ve platewith a decrease in potential energy:

U = qVba

Loss in potential energy = gain in kinetic energy.(a) q = -e = -1.610-19 CU = qVba = -1.610-195000 = -8.010-16 J



Suppose an electron in a cathode ray tube is accelerated from rest through a potential difference

Vb – Va = Vba = +5000 V.

(b) What is the speed of the electron (m = 9.1 × 10-31 kg) as a result of this acceleration?

m/s 104.2101.9

5000106.122

00

-

0 :Energy ofon Conservati

731

19ba

baab2

21

m

qVv

KqVVVqmv

UK

UK

initial

8


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:

EdV

EddEdVVV

ba

b

a

b

aabba ll

E

9



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.

V/m 100005.0

50

- Since ba

d

VE

EdV

E



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.

10


Example 23-4: Charged conducting sphere.

Determine the potential at a distance r from the center of a uniformly charged conducting sphereof radius r0 for

(a) r > r0The total charge on the sphere is Q

r

QV

V

rVdrd

rr

Q

r

drQdVV

r

Q

a

0

b0

r

r 2

r

r0

ab

20

4

1

charge)any from distance large aat

0conventionarbitary an is (this

at 0let and Set

11

44

4

1E that Law sGauss' using found We

b

a

b

a

l

l

E



Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for

(b) r = r0

The total charge on the sphere is Q

000 4

1 : approaches As

r

QVrr

11



Determine the potential at a distance r from the center of a uniformly charged conducting sphere of radius r0 for

(c) r < r0The total charge on the sphere is Q

004

1 :surfaceat as same theis V So

Theorem) s(Gauss'

field electric is theresphere theinside pointsFor

r

QV

NO



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:

12



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.


Example 23-5: Breakdown voltage.

The breakdown of air occurs for electric fields of about3 × 106 V/m. (Note that this is for dry air – in humid conditions this drops to about 2 × 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 cm (r = 0.5 cm = 5x10-3 m).

kV 15V 15000103105 (b)

(a)

63

0

200

200

V

ErVQ

Er

Q

Vrk

r

kQEand

r

kQV

13


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:

14


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?

J 1.08500.0

10201099.8103

C 1020Q C; 103qLet

m; 500.0

4

1&

696

6-6-

abab

0

ab

W

rrr

kQ

r

kQqW

r

kQ

r

QV

VVqUW



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.

15



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

V 107.530.0

1050

60.0

1050100.9

.

soscalar a is potential electric that Note

566

9

2A

2

1A

1A2A1A

r

Q

r

QkVVV

account into taken be to havenot does direction



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

Could use same method as part (a) but note that point B is equidistant to 2 equal magnitude charges (one + & one -) so that the potentials due the both charges will cancel VB = 0This would be true for all points along the perpendicular bisector.

16


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.

charge.point single afor result theiswhich

4

QV toreduces thisR x if that Note

4

11

4

1

4

1

0

220

2200

x

Rx

Qdq

Rxr

dqV

17


23-4 Potential Due to Any Charge Distribution

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 xfrom its center.


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

charge.point for as

4 toreducemust this if that Note

22

4

2

4

1

4

1

22 : thicknessof ringon Charge

Qdensity Charge

ring.circular for result using

diskover integrate and thicknessand

radius of rings thin intodisk theDivide

00

20

2200

0 22

200

0 22200

2200

20

20

20

0

0

x

Q Rx

xRxR

QRx

R

QRx

RdR

R

Q

Rx

dq

r

dqV

R

QRdR

πR

πRdRQσdAdqdrdq

R

dR

R

RR

R

R

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

Electric field (RED) lines are perpendicular to equipotentials (GREEN).

The surface of a conductor is an equipotential.

23-5 Equipotential Surfaces



900cos

havemust ialequipotentan along

path the0 reregion whe aIn

0 surface ialequipotentan On

cos

l

ll

dE

V

dEdV

E

Equipotential lines are perpendicular to the electric field lines.

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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.



Example 23-10: Point charge equipotential surfaces.

Q = 4.0 × 10-9 C

V1 = 10 V, V2 = 20 V, and V3 = 30 V.

m 2.130

100.4100.9:V 30

m 8.120

100.4100.9:V 20

m 3.610

100.4100.9:V 10

99

33

99

22

99

11

rV

rV

rV

V

kQr

r

kQV

20


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).

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23-7 E Determined from VIf 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:

E

.directions and for Similarly

constant. held and with derivative ie.

a is thatNote

zy

zyx

Vderivative partial


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.

E

E

22


23-7 E Determined from V

Example 23-11: E for ring.

E

E

2204

1

Rx

QV

We have shown that for a ring V at point P is:

The only component of is in the x directionE

23

2204

1

Rx

Qx

x

VEx

This is the same result that we obtained in chapter 21


Example 23-11: E for disk.E

21.chapter in obtained weasresult same theis This

density. charge surface where

22

disk, the toclose very pointsFor

12

2

disk charged a todue potential for theresult our From

20

0200

0

22200

20

2200

R

Q

R

QE

Rx

Rx

x

R

Q

x

VE

xRxR

QV

x

x

23


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.


24


Question: Calculate the velocity of an electron when accelerated through a potential difference of 1000 V.

m/s 101.9101.9

106.1100022

J) convert to toneed(

energykineticofeV1000 gainsElectron

731

19

e

2e2

1

m

Kv

vmK

1 eV = 1.610-19 Jme = 9.110-31 kg



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.

The total energy of the electron in atom is potential + kinetic. After separation the total energy is zero.

25


Example 23-12: Disassembling a hydrogen atom.

hydrogen.for energy ionization measured theisWhich

eV -13.6J 102.1810529.01085.88

106.1

884

8

44

Force Electric Force lCentripeta

orbitcircular in is e Assume

44

181012

219

0

2

0

2

0

2

total

0

22

e21

e0

22

20

22e

-

0

2

0

21

total

r

e

r

e

r

eKUE

r

evmK

rm

ev

r

e

r

vm

r

e

r

QQU

KUE

r = 0.529 × 10-10 m

e = 1.610-19 C

0 = 8.8510-12 C2/N.m2


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

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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.


27


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



Summary of Chapter 23

• Electric potential is potential energy per unit charge:

• Potential difference between two points:

• Potential of a point charge:

28


• Equipotential: line or surface along which potential is the same.

• Electric dipole potential is proportional to 1/r2.

• To find the field from the potential:

Summary of Chapter 23

chapter 23 electric potential - wordpress.com · 2017-09-26 · 12 copyright © 2009 pearson...

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