Physics II: Electricity & Magnetism

Physics II:Electricity & Magnetism

Physics II:Electricity & Magnetism

Sections 23.2 to 23.3Sections 23.2 to 23.3Sections 23.2 to 23.3Sections 23.2 to 23.3

Monday (Day 2)Monday (Day 2)

Warm-UpWarm-Up

Mon, Mar 2Mon, Mar 2 Our scenario: A positive test charge is being “dropped” in an electric field at Our scenario: A positive test charge is being “dropped” in an electric field at

location, location, aa, and will hit the other plate, , and will hit the other plate, bb. Using the following equations, . Using the following equations, derive an equation to calculate the work done by the electric field.derive an equation to calculate the work done by the electric field.

What if the charge were negative?What if the charge were negative?

Place your homework on Place your homework on mymy desk: desk: Web Assign Chapter 22 Final CopiesWeb Assign Chapter 22 Final Copies For future assignments - check online at www.plutonium-239.comFor future assignments - check online at www.plutonium-239.com

Mon, Mar 2Mon, Mar 2 Our scenario: A positive test charge is being “dropped” in an electric field at Our scenario: A positive test charge is being “dropped” in an electric field at

location, location, aa, and will hit the other plate, , and will hit the other plate, bb. Using the following equations, . Using the following equations, derive an equation to calculate the work done by the electric field.derive an equation to calculate the work done by the electric field.


Place your homework on Place your homework on mymy desk: desk: Web Assign Chapter 22 Final CopiesWeb Assign Chapter 22 Final Copies For future assignments - check online at www.plutonium-239.comFor future assignments - check online at www.plutonium-239.com

W =F ⋅d F =qE

Scenario #2: A charge in an electric field is being “dropped” from a point a, and will hit the plate at point b. Scenario #2: A charge in an electric field is being “dropped” from a point a, and will hit the plate at point b.

b

Direction of E

Direction of Motion

a+ --

b a

Warm-up ReviewWarm-up Review

Our scenario: A positive test charge is being “dropped” in an electric Our scenario: A positive test charge is being “dropped” in an electric field at location, field at location, aa, and will hit the other plate, , and will hit the other plate, bb. Using the following . Using the following equations, derive an equation to calculate the work done by the electric equations, derive an equation to calculate the work done by the electric field.field.


Our scenario: A positive test charge is being “dropped” in an electric Our scenario: A positive test charge is being “dropped” in an electric field at location, field at location, aa, and will hit the other plate, , and will hit the other plate, bb. Using the following . Using the following equations, derive an equation to calculate the work done by the electric equations, derive an equation to calculate the work done by the electric field.field.


W =F ⋅d F =qE

W = qE( )⋅d=qE⋅d=qEdcosθ

=1{ =q

e{ Ed=eEd

W = qE( )⋅d=qE⋅d=qEdcosθ

=−1{ =−q

−e{ Ed=eEd

*Recall:Electric Field and Energy

*Recall:Electric Field and Energy Using your knowledge and the law of conservation of energy, Using your knowledge and the law of conservation of energy, make a make a chartchart to to

generallygenerally Identify the direction of the electric field. (I.e. up, down, left ,right, etc)Identify the direction of the electric field. (I.e. up, down, left ,right, etc) The potential energy at (The potential energy at (aa) point ) point aa and ( and (bb) point ) point bb. (I.e. high, low, same). (I.e. high, low, same) The potential to do work at (The potential to do work at (aa) point ) point aa and ( and (bb) point ) point bb. (I.e. high, low, . (I.e. high, low,

same)same) The kinetic energy at (The kinetic energy at (aa) point ) point aa and ( and (bb) point ) point bb. (I.e. high, low, same). (I.e. high, low, same) The change in potential energy, potential to do work, and kinetic energy The change in potential energy, potential to do work, and kinetic energy

as it moves from as it moves from aa to to bb. (I.e. increase, decrease, constant). (I.e. increase, decrease, constant) The work done by the electric field (I.e. positive, negative, none)The work done by the electric field (I.e. positive, negative, none)

Using your knowledge and the law of conservation of energy, Using your knowledge and the law of conservation of energy, make a make a chartchart to to generallygenerally Identify the direction of the electric field. (I.e. up, down, left ,right, etc)Identify the direction of the electric field. (I.e. up, down, left ,right, etc) The potential energy at (The potential energy at (aa) point ) point aa and ( and (bb) point ) point bb. (I.e. high, low, same). (I.e. high, low, same) The potential to do work at (The potential to do work at (aa) point ) point aa and ( and (bb) point ) point bb. (I.e. high, low, . (I.e. high, low,

same)same) The kinetic energy at (The kinetic energy at (aa) point ) point aa and ( and (bb) point ) point bb. (I.e. high, low, same). (I.e. high, low, same) The change in potential energy, potential to do work, and kinetic energy The change in potential energy, potential to do work, and kinetic energy

as it moves from as it moves from aa to to bb. (I.e. increase, decrease, constant). (I.e. increase, decrease, constant) The work done by the electric field (I.e. positive, negative, none)The work done by the electric field (I.e. positive, negative, none)

*Electric Field and Energy(Positive Charge)

*Electric Field and Energy(Positive Charge)

Field: x:

cosθ = cos(0) =1

Initial Location(a) x0 = 0

Final Location(b)

Change

PotentialEnergy

High LowPE = U

= Ub - Ua –Decreases,

Potential to do Work

High LowPotential = V

= Vb - Va –Decreases,

Kinetic Energy Low HighKE = K

= Kb - Ka +Increases

*Electric Field and Energy(Negative Charge)

*Electric Field and Energy(Negative Charge)

Field: x:

cos(180) = –1

Initial Location(a) x0 = 0

Final Location(b)

Change

PotentialEnergy

High LowPE = U

= Ub - Ua –Decreases,

Potential to do Work

High LowPotential = V

= Vb - Va –Decreases,

Kinetic Energy Low HighKE = K

= Kb - Ka +Increases

Essential Question(s)Essential Question(s)

HOW DO WE DESCRIBE AND APPLY THE CONCEPT HOW DO WE DESCRIBE AND APPLY THE CONCEPT OF ELECTRIC POTENTIAL?OF ELECTRIC POTENTIAL? How do we calculate the electrical work done on a positive and How do we calculate the electrical work done on a positive and

negative charge that moves through a potential difference? negative charge that moves through a potential difference? How do we apply the Law of Conservation of Energy to determine How do we apply the Law of Conservation of Energy to determine

the speed of a charged particle that has been accelerated through the speed of a charged particle that has been accelerated through a potential difference? a potential difference?

How do we describe and apply the concept of electric fields?How do we describe and apply the concept of electric fields?





VocabularyVocabulary

Electric PotentialElectric Potential PotentialPotential Difference in PotentialDifference in Potential Potential DifferencePotential Difference VoltVolt VoltageVoltage Equipotential LinesEquipotential Lines Equipotential SurfacesEquipotential Surfaces Electric DipoleElectric Dipole


Dipole MomentDipole Moment Electron VoltElectron Volt CaCathode Ray Tubethode Ray Tube Thermionic EmissionThermionic Emission CathodeCathode AnodeAnode Cathode RaysCathode Rays OscilloscopeOscilloscope Scalar QuantityScalar Quantity


AgendaAgenda

Discuss Discuss The relation between the electric potential and the electric fieldThe relation between the electric potential and the electric field The relation between the potential energy and the electric forceThe relation between the potential energy and the electric force The relation between electric fields and gravitational fieldsThe relation between electric fields and gravitational fields Particles in uniform fieldsParticles in uniform fields How to derive the potential from the known E-field How to derive the potential from the known E-field Derive the electric potential(s) for point charges Derive the electric potential(s) for point charges

Complete Complete The Four CirclesThe Four Circles Graphic Organizers Graphic Organizers Work on Web AssignWork on Web Assign

Discuss Discuss The relation between the electric potential and the electric fieldThe relation between the electric potential and the electric field The relation between the potential energy and the electric forceThe relation between the potential energy and the electric force The relation between electric fields and gravitational fieldsThe relation between electric fields and gravitational fields Particles in uniform fieldsParticles in uniform fields How to derive the potential from the known E-field How to derive the potential from the known E-field Derive the electric potential(s) for point charges Derive the electric potential(s) for point charges

Complete Complete The Four CirclesThe Four Circles Graphic Organizers Graphic Organizers Work on Web AssignWork on Web Assign

Electrostatic Potential Energy and Potential Difference

F=qE (1) E=Fq (2)

U =qVba (3) Vba =Uq

(4)

Wba =−U =−qVba (5) U =qVba =−Wba (6)

Wba = F⋅dl (7)a

b

∫ E=1

4πε0

1r2 dq (8)

a

b

∫

Thus far, we have identified the following important equations

HOW DO WE DESCRIBE AND APPLY THE CONCEPT OF ELECTRIC POTENTIAL?

Relation between Electric Potential and Electric Field

By substituting equation (7) into (6):

U =−Wba =− F⋅dla

b

∫U =− F⋅dl

a

b

∫ (9)

By dividing both sides by q gives:

Uq

=−Fq⋅dl

a

b

∫

Vba =− E⋅dla

b

∫ (10)



From our warm-ups, it is interesting to note that the field will do the same amount of work as confirmed by our warm-up, regardless whether it is a positive charge moving with field or and a negative charge moving in the †opposite direction (†reestablished as a b ):

Work can then be related to the potential by:

Wba =−U =− − F⋅dla

b

∫⎡⎣⎢

⎤⎦⎥= qE⋅dl

a

b

∫ = qEdla

b

∫ cosθ;

+e:Wba = qEdla

b

∫ cos 0°( )1

1 24 34= qEdl

a

b

∫ =qE dla

b

∫d

{=q

+e{ Ed=eEd

−e:Wba† =− qEdl

a

b

∫ cos 180°( )−1

1 24 34=− qEdl

a

b

∫ =−qE dla

b

∫d

{=−q

−e{ Ed=eEd



F=qE (1) E=1

4πε0

1r2 dq

a

b

∫ (8)

U =− F⋅dla

b

∫ (9) Vba =− E⋅dla

b

∫ (10)

We now have the following equations


Calculating the Potential Difference, Vba, in a Uniform Field

(1) Because the field and the direction of motion are the same, cos θ = 1,

(2) Because E is uniform, it can be pulled out of the integral.


Vba =− E⋅dla

b

∫ = E⋅dla

b

∫ = Edla

b

∫ cosθ;

Vba =− Edla

b

∫ cos 0°( )1

1 24 34=−E dl

a

b

∫ =−E dla=0

b=d

∫

Vba = −Ed

(3) The negative sign implies that the potential is decreasing as it moves a distance, d, in the electric field.

Relation to Gravity using the Potential Energy, Uba, in a Uniform Field

(1) Because the field (the positive direction) and the direction of motion are the same, cos θ = 1,

(2) Because the Gravitational Field is uniform at the surface of the Earth, it can be pulled out of the integral.


Uba =− FG ⋅dla

b

∫ = FG ⋅dlhi

hf

∫ = FG dlhi

hf

∫ cosθ;

Uba =− FG dlhi

hf

∫ cos 0°( )1

1 24 34=−FG dl

hi

hf

∫ =−FG l[ ]hi

hf

Uba = −FG hf −hi( ) where hf > hi

(3) By using the field direction as the positive direction, hi is higher above the surface of the Earth and hf is closer to the ground.

Application: Potential Difference in a Uniform EApplication: Potential Difference in a Uniform E Two parallel plates are charged to a voltage (Two parallel plates are charged to a voltage (akaaka. .

potential difference) of 50 V. If the separation potential difference) of 50 V. If the separation between the plates is 5.0 cm, calculate the electric between the plates is 5.0 cm, calculate the electric field between them.field between them.

Two parallel plates are charged to a voltage (Two parallel plates are charged to a voltage (akaaka. . potential difference) of 50 V. If the separation potential difference) of 50 V. If the separation between the plates is 5.0 cm, calculate the electric between the plates is 5.0 cm, calculate the electric field between them.field between them.

Vba =−Ed⇒ E=−Vba

d=−

Vb −Va

d=−

0 V −50 V( )0.05 m

E= 1000Vm

Note that Va = 50 V and Vb = 0 V, therefore the E field is in the direction of the motion. We could have also established Va = 0 V and Vb = –50 V or Va = 25 V and Vb = –25 V, etc.


Point ChargePoint Charge


Determination of E:Positive Point ChargeDetermination of E:Positive Point Charge

+

E

At radius r,

A1

r

E⋅dA—∫

dAsphere

=EA1 cosθ =EA1 =E 4π r2( ) =

Qencl

ε 0

Therefore, Epositive point charge =1

4πε0

Qr2

What if the charge were negative?


Determination of E:Negative Point ChargeDetermination of E:Negative Point Charge

–

E

At radius r,

A1

r

E⋅dA—∫

dAsphere

=EA1 cosθ

−1{

=−EA1

=−E 4π r2( )

=Qencl

ε 0


=−Q

ε 0

Therefore, Enegative point charge =1

4πε0

Qr2

Why is E positive for the positive and negative charge?

Determination of V from E:Point ChargeDetermination of V from E:Point Charge


E =1

4πε0

Qr2 ; Vba =− E⋅dl

a

b

∫

+

Era

dr

rb

=− E dr

ra

rb

∫ cosθ=1

{ =− E drra

rb

∫

Vba =−1

4πε0

Qr2 dr

ra

rb

∫ =−Q

4πε 0

1

r2 drra

rb

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ ra

rb

=Q

4πε 0

1

rb

−1

ra

⎛

⎝⎜⎞

⎠⎟

It is common to choose

the potential to be 0 at ∞.Thεrεforε, lεt Vb =0 at rb =∞

Vba =Vb −Va

−Va

V{ = −

1

4πε 0

Q

ra

r{⇒ V =

1

4πε 0

Q

rat rb =∞

Note: If rb is a finite distance, then

at rb = ∞

⇒ U = qV =1

4πε 0

qQ

r

Electric Potential Due to Point Charges

These plots show the potential due to (a) positive and (b) negative charge.


Steps to Determine the Potential from the known E-Field Created by a Uniform Charge Distributions

Steps to Determine the Potential from the known E-Field Created by a Uniform Charge Distributions

Graphical:Graphical: In relation to the known E-field, determine and draw the initial and final displacement locations.In relation to the known E-field, determine and draw the initial and final displacement locations. Determine the angle between the direction of motion and the electric field (Determine the angle between the direction of motion and the electric field (EE••ddll).). Mathematical: Mathematical: Write the formula for Write the formula for EE.. Write the formula for potential (V= -∫Write the formula for potential (V= -∫EE••ddll ). ). Calculate the cos Calculate the cos θθ between the direction of motion and the electric field . between the direction of motion and the electric field . Set up the integral by determining location of the initial and final displacements.Set up the integral by determining location of the initial and final displacements. ††Solve the integral by using the predetermined locations of the initial and final displacementsSolve the integral by using the predetermined locations of the initial and final displacements Write the answer in a concise manner. Verify that the answer makes Write the answer in a concise manner. Verify that the answer makes physical physical sense.sense. If possible, set the one of the potential to zero at an infinite location away. (ex. If possible, set the one of the potential to zero at an infinite location away. (ex. VVbb= 0 at = 0 at rrbb = ∞). = ∞).

Also, note that the negative signs may cancel leaving a positive quantity.Also, note that the negative signs may cancel leaving a positive quantity. Write the answer for the new Write the answer for the new absoluteabsolute potential. potential.

††See the instructor, AP Calculus BC students, or Schaum’s Mathematical Handbook.See the instructor, AP Calculus BC students, or Schaum’s Mathematical Handbook.

Graphical:Graphical: In relation to the known E-field, determine and draw the initial and final displacement locations.In relation to the known E-field, determine and draw the initial and final displacement locations. Determine the angle between the direction of motion and the electric field (Determine the angle between the direction of motion and the electric field (EE••ddll).). Mathematical: Mathematical: Write the formula for Write the formula for EE.. Write the formula for potential (V= -∫Write the formula for potential (V= -∫EE••ddll ). ). Calculate the cos Calculate the cos θθ between the direction of motion and the electric field . between the direction of motion and the electric field . Set up the integral by determining location of the initial and final displacements.Set up the integral by determining location of the initial and final displacements. ††Solve the integral by using the predetermined locations of the initial and final displacementsSolve the integral by using the predetermined locations of the initial and final displacements Write the answer in a concise manner. Verify that the answer makes Write the answer in a concise manner. Verify that the answer makes physical physical sense.sense. If possible, set the one of the potential to zero at an infinite location away. (ex. If possible, set the one of the potential to zero at an infinite location away. (ex. VVbb= 0 at = 0 at rrbb = ∞). = ∞).

Also, note that the negative signs may cancel leaving a positive quantity.Also, note that the negative signs may cancel leaving a positive quantity. Write the answer for the new Write the answer for the new absoluteabsolute potential. potential.

††See the instructor, AP Calculus BC students, or Schaum’s Mathematical Handbook.See the instructor, AP Calculus BC students, or Schaum’s Mathematical Handbook.


GO: The Four Circles(aka. You’re givin’ me fevu?)GO: The Four Circles(aka. You’re givin’ me fevu?)

Relationships between Relationships between FF,, E E,, V V, and, and U U Relationships between Relationships between FF,, E E,, V V, and, and U U

F=1

4πε0

q1q2

r2 →E=

Fq

−q2E=

14πε0

q1r2

−r↓U=− F

a

b

∫ ⋅dl −r↓V=− E

a

b

∫ ⋅dl

U =1

4πε0

q1q2

r→

V=Uq

−q2Vba =

14πε0

q1r


SummarySummary

Using the following electrostatic equations, develop their gravitational counterparts for force, Using the following electrostatic equations, develop their gravitational counterparts for force, gravitational field, potential energy and gravitational potential.gravitational field, potential energy and gravitational potential.

HW (Place in your agenda): HW (Place in your agenda): Web Assign 23.3 - 23.4Web Assign 23.3 - 23.4

Future assignments:Future assignments: Electrostatics Lab #4 Report (Due in 2 Classes)Electrostatics Lab #4 Report (Due in 2 Classes)

Using the following electrostatic equations, develop their gravitational counterparts for force, Using the following electrostatic equations, develop their gravitational counterparts for force, gravitational field, potential energy and gravitational potential.gravitational field, potential energy and gravitational potential.

HW (Place in your agenda): HW (Place in your agenda): Web Assign 23.3 - 23.4Web Assign 23.3 - 23.4

Future assignments:Future assignments: Electrostatics Lab #4 Report (Due in 2 Classes)Electrostatics Lab #4 Report (Due in 2 Classes)


F=1

4πε0

q1q2

r2 →E=

Fq

−q2E=

14πε0

q1r2

−r↓U=− F

a

b

∫ ⋅dl −r↓V=− E

a

b

∫ ⋅dl

U =1

4πε0

q1q2

r→

V=Uq

−q2Vba =

14πε0

q1r

Tuesday (Day 3)Tuesday (Day 3)

Warm-UpWarm-Up

Tues, Mar 3Tues, Mar 3 Suppose an electron in the picture tube of a television set is accelerated from Suppose an electron in the picture tube of a television set is accelerated from

rest through a potential difference rest through a potential difference VVbaba = +5000 V. (= +5000 V. (aa) What is the change in ) What is the change in potential energy of the electron (potential energy of the electron (mmee = 9.1 x 10= 9.1 x 10-31-31 kg)? ( kg)? (bb) What is the final ) What is the final speed of the electron? (speed of the electron? (cc) Repeat for a proton () Repeat for a proton (mmpp = 1.67 x 10= 1.67 x 10-27-27 kg) that kg) that accelerates through a potential difference of accelerates through a potential difference of VVbaba = -5000 V.= -5000 V.

Place your homework on Place your homework on mymy desk (if applicable): desk (if applicable): Web Assign: Problems 23.3 & 23.4Web Assign: Problems 23.3 & 23.4 For future assignments - check online at www.plutonium-239.comFor future assignments - check online at www.plutonium-239.com

Tues, Mar 3Tues, Mar 3 Suppose an electron in the picture tube of a television set is accelerated from Suppose an electron in the picture tube of a television set is accelerated from

rest through a potential difference rest through a potential difference VVbaba = +5000 V. (= +5000 V. (aa) What is the change in ) What is the change in potential energy of the electron (potential energy of the electron (mmee = 9.1 x 10= 9.1 x 10-31-31 kg)? ( kg)? (bb) What is the final ) What is the final speed of the electron? (speed of the electron? (cc) Repeat for a proton () Repeat for a proton (mmpp = 1.67 x 10= 1.67 x 10-27-27 kg) that kg) that accelerates through a potential difference of accelerates through a potential difference of VVbaba = -5000 V.= -5000 V.

Place your homework on Place your homework on mymy desk (if applicable): desk (if applicable): Web Assign: Problems 23.3 & 23.4Web Assign: Problems 23.3 & 23.4 For future assignments - check online at www.plutonium-239.comFor future assignments - check online at www.plutonium-239.com

Application: TV ParticlesApplication: TV Particles

Suppose an electron in the Suppose an electron in the picture tube of a television set picture tube of a television set is accelerated from rest is accelerated from rest through a potential difference through a potential difference VVbaba = +5000 V. (= +5000 V. (aa) What is the ) What is the change in potential energy of change in potential energy of the electron (the electron (mmee = 9.1 x 10= 9.1 x 10-31-31 kg)? (kg)? (bb) What is the final speed ) What is the final speed of the electron? (of the electron? (cc) Repeat for ) Repeat for a proton (a proton (mmpp = 1.67 x 10= 1.67 x 10-27-27 kg) kg) that accelerates through a that accelerates through a potential difference of potential difference of VVbaba = -= -5000 V.5000 V.

Suppose an electron in the Suppose an electron in the picture tube of a television set picture tube of a television set is accelerated from rest is accelerated from rest through a potential difference through a potential difference VVbaba = +5000 V. (= +5000 V. (aa) What is the ) What is the change in potential energy of change in potential energy of the electron (the electron (mmee = 9.1 x 10= 9.1 x 10-31-31 kg)? (kg)? (bb) What is the final speed ) What is the final speed of the electron? (of the electron? (cc) Repeat for ) Repeat for a proton (a proton (mmpp = 1.67 x 10= 1.67 x 10-27-27 kg) kg) that accelerates through a that accelerates through a potential difference of potential difference of VVbaba = -= -5000 V.5000 V.

Application: TV Particles(the electron)Application: TV Particles(the electron)

U =qVba = −1.6 x 10−19 C( ) 5000 V( )

U = −8.0 x 10−16 J


Suppose an electron in the picture tube of a television set is accelerated from rest through a potential difference Vba = +5000 V. (a) What is the change in potential energy of the electron (me = 9.1 x 10-31 kg)? (b) What is the final speed of the electron?

K + U =0⇒ K =−U ⇒ 12 mvf

2 −12 mv i

2=−qVba

⇒ vf =−2qVba

m=

−2 −1.6 x 10−19 C( ) 5000 V( )

9.1 x 10−31 kg

vf = 4.2 x 107 m s → 0.14c

Application: TV Particles(the proton)Application: TV Particles(the proton)

U =qVba = 1.6 x 10−19 C( ) −5000 V( )

U = −8.0 x 10−16 J


Suppose a proton in the picture tube of a television set is accelerated from rest through a potential difference Vba = -5000 V. (a) What is the change in potential energy of the proton (mp = 1.67 x 10-27 kg)? (b) What is the final speed of the proton?

K + U =0⇒ K =−U ⇒ 12 mvf

2 −12 mv i

2=−qVba

⇒ vf =−2qVba

m=

−2 1.6 x 10−19 C( ) −5000 V( )

1.67 x 10−27 kg

vf = 9.8 x 105 m s → v=0.003c

Essential Question(s)Essential Question(s)









VocabularyVocabulary





AgendaAgenda

Derive the electric potential(s) for a(n):Derive the electric potential(s) for a(n): Conducting sphereConducting sphere Concentric conducting spheresConcentric conducting spheres

Discuss the determination of potential for multiple point chargesDiscuss the determination of potential for multiple point charges Work on Web AssignWork on Web Assign

Derive the electric potential(s) for a(n):Derive the electric potential(s) for a(n): Conducting sphereConducting sphere Concentric conducting spheresConcentric conducting spheres

Discuss the determination of potential for multiple point chargesDiscuss the determination of potential for multiple point charges Work on Web AssignWork on Web Assign

Important Note #1Important Note #1

We will now have the ability to sum up all of the point We will now have the ability to sum up all of the point charges to determine the total potential for any charges to determine the total potential for any charge distribution.charge distribution.

We will now have the ability to sum up all of the point We will now have the ability to sum up all of the point charges to determine the total potential for any charges to determine the total potential for any charge distribution.charge distribution.

Potential due to a point charge: V =1

4πε0

Qr

For multiple point charges: Vnet = V ii=1

n

∑ =1

4πε0

Qi

rii=1

n

∑

For continuous chargεs: Vnet = dV0

Vtot

∫ =1

4πε0

dqr0

Qtot

∫

But we will save that for another day . . .But we will save that for another day . . .


Important Note #2Important Note #2

We will now have the ability to sum up all of the point We will now have the ability to sum up all of the point charges to determine the total potential energy for charges to determine the total potential energy for any charge distribution.any charge distribution.

We will now have the ability to sum up all of the point We will now have the ability to sum up all of the point charges to determine the total potential energy for charges to determine the total potential energy for any charge distribution.any charge distribution.

Potential Energy due to a point charges: U =1

4πε0

qQr

For multiple point charges: Unet = U ii=1

n

∑ =1

4πε0

qiqj

riji< j

n

∑ =12

14πε0

qiqj

riji, ji≠j

n

∑

For continuous chargεs: Unet = dU0

Utot

∫ =1

8πε0

1rij

dqi dqj0

Qtot

∫0

Qtot

∫

But again we will save that for another day . . .But again we will save that for another day . . .


Spherical ConductorSpherical Conductor


Spherical ConductorSpherical Conductor

+

+

+

++

+

+

+

E

A1

r1

At radius r1 r > r0( ),

E⋅dA—∫

dAsphere

=EA1 cosθ =EA1=E 4π r2( ) =

Qencl

ε 0

Therefore, Eoutside =1

4πε0

Qr12

To calculate the electric field inside of a

conductor, we look at radius r2 r < r0( ),

r0

A2

r2

E⋅dA—∫ =EA2 cosθ =EA2=E 4π r2

( ) =Qencl

ε 0Because the enclosed charge inside

of a conductor is zero, Einside =0


+

+

+

++

+

+

+

E

r0

dr

rb

ra

E =1

4πε0


a

b

∫ =− E dr

ra

rb

∫ cosθ=1

{ =− E drra

rb

∫

Vba =−1

4πε0

Qr2 dr

ra

rb

∫ =−Q

4πε 0

1

r2 drra

rb

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ ra

rb

=Q

4πε 0

1

rb

−1

ra

⎛

⎝⎜⎞

⎠⎟



Vba =Vb −Va

−Va

V{ = −

1

4πε 0

Q

ra

r{

⇒ Voutside =1

4πε 0

Q

rat r > r0


Determination of V from E: Outside a Spherical Conductor (r > r0)

Determination of V from E: Outside a Spherical Conductor (r > r0)

+

+

+

++

+

+

+

E

dAsphere

r0

dr

rb

ra

E =1

4πε0


a

b

∫ =− E dr

ra

rb

∫ cosθ=1

{ =− E drra

rb

∫

Vba =−1

4πε0

Qr2 dr

ra

rb

∫ =−Q

4πε 0

1

r2 drra

rb

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ ra

rb

=Q

4πε 0

1

rb

−1

ra

⎛

⎝⎜⎞

⎠⎟



Vba =Vb −Va

⇒ Vsurface =1

4πε 0

Q

r0

at r =r0


Determination of V from E: Spherical Conductor Surface (r r0)

Determination of V from E: Spherical Conductor Surface (r r0)

Determination of V from E: Inside a Spherical Conductor (r < r0) Determination of V from E: Inside a Spherical Conductor (r < r0)

+

+

+

++

+

+

+

E

Einside=0r0

drrb

ra

Einside =0; Vba =− E⋅dla

b

∫ =− E ⋅drra

rb

∫ = 0[ ]ra =r

rb =r0

Vba =0

Because there is no change in Vba , (Vba =0),

between rb = r0 and and any point within

the conductor, the potential inside must be

equal to the potential at the surface.

(Note: Vinside ≠0 unlεss Qencl =0)

⇒1

r0

=1

r


⇒ r = r0

Va =1

4πε0

Qrar0{

⇒ V =1

4πε 0

Q

r0

at r < r0

Plots of E vs. r and of V vs. r: for a single Spherical ConductorPlots of E vs. r and of V vs. r: for a single Spherical Conductor

r0

r0

Eoutside =1

4πε0

Qr2 at r > r0

Voutside =1

4πε0

Qr

at r > r0

at r =r0

Vsurface =1

4πε0

Qr0

at r =r0

Einside = 0 at r < r0

Vinside =1

4πε0

Qr0

at r < r0

E ∝ 1r2

V ∝ 1r

Concentric Conducting Spherical ShellsConcentric Conducting Spherical Shells


Concentric Conducting Spherical ShellsConcentric Conducting Spherical Shells


+

+

+

++

+

+

+ dAsphere

–

–

–

––

–

–

–

ra

rb

At radius r3 r > rb( ),

A3

E⋅dA—∫

r3

=EA3 cosθ =EA3 =E 4π r2( ) =

Qencl

ε 0Because the enclosed charge outside

of a conductor is zero, Eoutside =0

At radius r2 ra < r < rb( ),

A2

r2

E⋅dA—∫ =EA2 cosθ =EA2 =E 4π r2

( ) =Qencl

ε 0

Therefore, Ebetween =1

4πε0

Qr12

At radius r1 r < ra( ),

E

r1

E⋅dA—∫

A1

=EA1 cosθ =EA2 =E 4π r2( ) =

Qencl

ε 0Because the enclosed charge inside

of a conductor is zero, Einside =0

Determination of V from E: Outside Concentric Spheres (r ≥ rb) Determination of V from E: Outside Concentric Spheres (r ≥ rb)


dr

+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

E

Eoutside=0

Eoutside =0; Vba =− E⋅dla

b

∫

rcrd

=− E ⋅drrc

rd

∫ = 0[ ]rc→r

rd

Vba = 0 at r > rb (Note: Voutside =0 unlεss Qencl ≠0)

At radius r > rb,

Determination of V from E: Between Concentric Spheres (r rb) Determination of V from E: Between Concentric Spheres (r rb)


+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

E

At radius r → rb,

Einside=0

E =1

4πε0

Qr2 ;

Vba =− E⋅dla

b

∫

dr

rc

rd

=− E dr

rc

rd

∫ cosθ=1

{ =− E drrc

rd

∫

Vba =−1

4πε0

Qr2 dr

rc

rd

∫ =−Q

4πε 0

1

r2 drrc

rd

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ rc

rd

=Q

4πε 0

1

rd

−1

rc

⎛

⎝⎜⎞

⎠⎟

It is common to choose the potential to be 0 at ∞.Thεrεforε, lεt Vd =0 at rd =∞

Vba =Vb −Va

⇒ V =1

4πε 0

Q

rb

at rd =∞

Determination of V from E: Between Concentric Spheres (r ra) Determination of V from E: Between Concentric Spheres (r ra)


+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

E

At radius r → ra,

Einside=0

E =1

4πε0

Qr2 ;

Vba =− E⋅dla

b

∫

drrc

rd

=− E dr

rc

rd

∫ cosθ=1

{ =− E drrc

rd

∫

Vba =−1

4πε0

Qr2 dr

rc

rd

∫ =−Q

4πε 0

1

r2 drrc

rd

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ rc

rd

=Q

4πε 0

1

rd

−1

rc

⎛

⎝⎜⎞

⎠⎟

It is common to choose the potential to be 0 at ∞.Thεrεforε, lεt Vd =0 at rd =∞

Vba =Vb −Va

⇒ V =1

4πε 0

Q

ra

at rd =∞


Determination of V from E: Between Concentric Spheres (ra< r < rb)

Determination of V from E: Between Concentric Spheres (ra< r < rb)

+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

Eoutside=0

E

E =1

4πε0

Qr2 ; Vcd =− E⋅dl

c

d

∫

drrc

rd

=− E dr

rc

rd

∫ cosθ=1

{ =− E drrc

rd

∫

Vcd =−1

4πε0

Qr2 dr

rc

rd

∫ =−Q

4πε 0

1

r2 drrc

rd

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ rc

rd

=Q

4πε 0

1

rd

−1

rc

⎛

⎝⎜⎞

⎠⎟

Vcd =Q

4πε0

rc −rdrcrd

⎛

⎝⎜⎞

⎠⎟at ra < rc < rd < rb

dr

Note: If rc → ra

ra

and rd → rb,rb

then

Vba =Q

4πε0

ra −rbra rb

⎛

⎝⎜⎞

⎠⎟

Determination of V from E: Center of Concentric Spheres (r < ra) Determination of V from E: Center of Concentric Spheres (r < ra)


+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

E

Einside=0

At radius r < ra, Einside =0;

Vba =− E⋅dla

b

∫

dr

rd

rc

=− E ⋅drrc

rd

∫ = 0[ ]rc =r

rd =ra

Vba =0

Because there is no change in Vba , (Vba =0),

between rd = ra and any point within

the conductor, the potential inside must be

equal to the potential at the surface.

(Note: Vinside ≠0 unlεss Qenclosing conductor =0)

⇒1

ra

=1

r⇒ r = ra at r < ra

⇒ V =1

4πε 0

Q

ra

at r < ra

Determination of V from E: Between Concentric Spheres (r rm) Determination of V from E: Between Concentric Spheres (r rm)


+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

E

At radius r → rm,

Einside=0

E =1

4πε0

Qr2 ;

Vdm =− E⋅dld

m

∫

rm

dr

rm

rd

=− E dr

rm

rd

∫ cosθ=1

{ =− E drrm

rd

∫

Vdm =−1

4πε0

Qr2 dr

rc

rd

∫ =−Q

4πε 0

1

r2 drrc

rd

∫

=Q

4πε 0

1

r⎡⎣⎢

⎤⎦⎥ rm

rd

=Q

4πε 0

1

rd

−1

rm

⎛

⎝⎜⎞

⎠⎟

It is common to choose the potential to be 0 at ∞.Lεt Vd =0 at rd =∞, thεrεforε, Vdm =0

Vdm =Vd

=0{

−Vm =0 ⇒ Vm = 0 at rd =∞ How can V = 0 if you have

to move past a negative shell? Negative work is done as it approaches,

and positive work is done as it continues past → ∞; thε + πlatε ⇒ +work (always)


Determination of rm where V = 0: Between Concentric Spheres (ra< r < rb)

Determination of rm where V = 0: Between Concentric Spheres (ra< r < rb)

Vba

+

+

+

++

+

+

+

–

–

–

––

–

–

–

ra

rb

E

Eoutside=0

Vm =0

rm

There exists a point, rm ,

between ra and rb where Vm∞ =0

Q

4πε0

1rb−1ra

⎛

⎝⎜⎞

⎠⎟

=Vbm +Vma

=Q

4πε 0

1

rb

−1

rm

⎛

⎝⎜⎞

⎠⎟+

Q

4πε 0

1

rm

−1

ra

⎛

⎝⎜⎞

⎠⎟

where Vbm =Vma, ∴ Vba = 2Vma

⇒ Vma = 12 Vba

1

2

Q

4πε0

1rb−1ra

⎛

⎝⎜⎞

⎠⎟=

Q4πε0

1rm

−1ra

⎛

⎝⎜⎞

⎠⎟

⇒1

rm

−1

ra

=1

2rb

−1

2ra

⇒1

rm

=1

2rb

−1

2ra

+1

ra

⇒1

rm

=1

2rb

+1

2ra

⇒1

rm

=2ra + 2rb

4rarb

rm =2rarbra + rb

for Concεntric Sπhεrical

Shεlls, ra < rb

E(r) for Concentric Spherical ShellsE(r) for Concentric Spherical Shells

E

rrb

rmra

Eb

Ea

rm =2rarbra + rb

Eoutside = 0

at r > rb

Eb =1

4πε0

Qrb2

at r =rb

Ebetween =1

4πε0

Qr2

at ra < r < rb

Ea =1

4πε0

Qra2

at r =ra

Ecenter = 0

at r < r0E ∝ 1

r2

Areara→ rm=Arearm→rb

V(r) for Concentric Spherical ShellsV(r) for Concentric Spherical Shells

V

Va

Vb

ra rm rb

r

Voutside = 0

at r > rb

Vb =1

4πε0

Qrb

at r =rb

at ra < r < rb

Va =1

4πε0

Qra

at r =ra

Vcenter

=1

4πε0

Qra

at r < ra

Vm = 0

at r =rm

V ∝ 1r

Physics II: Electricity & Magnetism

Documents

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