10.1 Describing Fields Topic 10.1 is an extension of Topics 5.1 and 6.2.

Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields.

Nature of science: Paradigm shift: The move from direct, observable actions being responsible for influence on an object to acceptance of a field’s “action at a distance” required a paradigm shift in the world of science.

Understandings: • Gravitational fields • Electrostatic fields • Electric potential and gravitational potential • Field lines • Equipotential surfaces

Applications and skills: • Representing sources of mass and charge, lines of electric and gravitational force, and field patterns using an appropriate symbolism • Mapping fields using potential • Describing the connection between equipotential surfaces and field lines

Guidance: • Electrostatic fields are restricted to the radial fields around point or spherical charges, the field between two point charges and the uniform fields between charged parallel plates • Gravitational fields are restricted to the radial fields around point or spherical masses and the (assumed) uniform field close to the surface of massive celestial bodies and planetary bodies • Students should recognize that no work is done in moving charge or mass on an equipotential surface

Theory of knowledge: • Although gravitational and electrostatic forces decrease with the square of distance and will only become zero at infinite separation, from a practical standpoint they become negligible at much smaller distances. How do scientists decide when an effect is so small that it can be ignored?

Utilization: • Knowledge of vector analysis is useful for this sub-topic (see Physics sub-topic 1.3)

Aims: • Aim 9: models developed for electric and gravitational fields using lines of forces allow predictions to be made but have limitations in terms of the finite width of a line

Data booklet reference: • W = qΔVe • W = mΔVg FORCE FIELDS (The concept of the force field has been introduced in both Topic 5.1 and Topic 6.2)
The force laws F = –Gm1m2 /r 2 and F = kq1q2 /r 2 were first developed as action at a distance phenomena.
The field is the force per unit test point object (mass or positive charge) placed at a particular point.

Coulomb's Law

The force of attraction or repulsion acting between two point electric charges is directly proportional to the prouct of their chargesand inversely proportional to the square of the distance between them.

Newton's Law of Universal Gravitation

The force acting between two masses is directly proportional to each of their massesand inversely proportional to the square of the distance between them.

where q1 and q2 denote charges and the electrostatic constantk

UNIVERSAL GRAVITATIONAL CONSTANT:
G = 6.67 x 10–11 Nm2kg–2 ( The constant of proportionality )

FIELD LINE The direction and strength of a field are illustrated by a field line. The strength of a field is presented by the density of the lines.The density of field lines is considered by dividing the number of lines passing through a given area. The area at a given distance can be thought of as the area of a sphere.

EXAMPLES)
1. Sketch the electric field (E) around a charge 2. Sketch the gravitational field (g) around a mass

The direction of the field line signifies the direction of the force experienced by a particle placed in the field.
Field lines by a test object, for both g and E fields, when a test object is moved:
~ along/parallel to a field line - work will be done )
~ across/perpendicular to a field line (along the equipotential) - no work will be done.

PRACTICE 1: Sketch the gravitational field about Earth (a) as viewed close from Earth surface, (b) locally from a distance and (c) from far away.

The field lines will appear to be parallel when you get very close to a sphere (refer to the diagram 1a) and thus, we assume the field close to the Earth is uniform and thus its value is 9.8 ms-2.

PRACTICE 2: Field strength is a vector quantity so field strengths must be added vectorially.

(a) Estimate the field strength at the point shown due to A, B, C and D. Consider the planet, Earth and Moon, as shown in the diagram below.

Due to the finite speed (v = c) of the force signal required of the special theory of relativity, the action at a distance view had to be cast aside in favor of the field view. In the field view, we imagine charges and masses as capable of warping the space around them due to their presence, and other charges and masses responding only to the local curvature of space, and not the actual source of that curvature.

At a distance r there would be N field lines passing through the walls of a sphere of area 4 π r 2. The number of field lines per unit area are therefore N / 4 π r 2. The density of field lines is proportional to 1/( r2 ) which is the same as the field strength.

PRACTICE 4:
Draw electric field patterns of (a) a positive charge, (b) a negative charge (c) between two charges.
SOLUTION It simply displays the electric field permeating through space.
Remember the direction of the field arrows showed the direction of the force a test charge, or a test mass, would feel due to the local field.
(a) (b) Yes, your guess is correct !!!

PRACTICE 5: Justify the statement “the electric field strength is uniform between two parallel plates.”
SOLUTION
Hint: Sketch the electric field lines between two parallel plates and demonstrate that the electric field lines have equal density everywhere between the plates.
The field lines are appeared to be parallel in between two parallel plates.

Gravitational Potential, Vg
Energy per unit test point mass that the mass has as a result of the gravitational field.
Vg = energy/mass = W/m = - GM / r [ Unit: J / kg ] The amount of work required to bring a test mass of one kilogram from infinity to a given point.

Worked examples ( K.A. Tsokos p 399~401 )
10.2 Calculate the gravitational potential at point P. (Masses and distances are shown on the diagram on text page 399.

V = V1 + V2 = -8.6 X 10 ^-4 J / kg
10.3 The mass of the Moon is about 81 times smaller than that of the Earth. The distance between the Earth and the Moon is about d=3.8 x 10^8 m. The mass of the Earth is 5.97 x 10^24kg.
(a) Determine the distance from the centre of the Earth of the point on the line joining the Earth to the Moon where the combined gravitational field strength of the Earth and the Moon is zero. The distance at point where the field strength between them is zero is 3.4 x 10^8 m.
(b) Calculate the combined gravitational potential at that point. Vg = - 1.28 x 10^6 J/kg
(c) Calculate the potential energy when a 2500 kg probe is placed at that point. Ep = mVg = - 3.2 x 10^9 J

Potential difference – The gravitational force As masses experience the gravitational force, when one mass is moved in the vicinity of another, work W is done (recall that work is a force times a displacement). Gravitational Potential Difference: DVg = W /m

We define the potential difference Vg between two points A and B as the amount of work ( W ) done per unit mass m in moving a point mass from A to B.
( Note that the units of Vg are J/kg )

Gravitational potential and gravitational potential energy The Fizzics Organization Published on 21 Oct 2014

PRACTICE 6: A mass of m = 500 kg is moved from point A, having a gravitational potential of 75.0 J kg-1 to point B, having a gravitational potential of 25.0 J kg-1. (a) What is the potential difference undergone by the mass? (b) What is the work done in moving the mass from A to B? SOLUTION: (a) ΔVg = VB – VA = 25.0 – 75.0 = - 50.0 J kg-1 (b) W = mΔVg = 500 x -50.0 = - 25000 J

Electric Potential, Ve
Energy per unit test point charge that the charge has as a result of the electric field.
Ve = energy/charge = W/q = kQ / r [ Unit: J / C, (V) ]

Worked examples ( K.A. Tsokos p 405, 407 )
10.5 The hydrogen atom has a single proton and a single electron.
(a) Find the electric potential a distance of 0.50 x 10^-10 m from the proton of the hydrogen atom. The proton has a charge 1.6 x 10^-19, equal and opposite to that of the electron. Ve = kQ / r = 28.7 = 29 V
(b) Use your answer to a to calculate the electric potential energy between the proton in a hydrogen atom and an electron orbiting the proton at a radius 0.50 x 10^10 m Ep = q Ve = 4.6 x 10^-18 J
10.6 Two unequal positive charges +Q and +q are placed at a distance x (Fig.10.16). Which one of the graphs shows the variation with distance x from the larger charge of the electric potential Ve along the line joining the centres of the charges?

SOLUTION: B Potential difference – The electric force As electric charges experience the electric force, when one charge is moved in the vicinity of another, work W is done (recall that work is a force times a displacement). Electrostatic Potential Difference: DVe = W /q

We define the potential difference Ve between two points A and B as the amount of work ( W ) done per unit charge q in moving a point charge from A to B.
( Note: The units of Ve are J/C which are volts V )

PRACTICE 7: A battery has a potential difference of 3V. How much energy is required to move 2mC of charge from the negative to the positive end? U = qV = ( 2x 10^-6) (3V) = 10^-6 J
PRACTICE 8: A charge of q = +15.0 mC is moved from point A, having a voltage (potential) of 25.0 V to point B, having a voltage (potential) of 18.0 V.
(a) What is the potential difference undergone by the charge?
(b) What is the work done in moving the charge from A to B?
SOLUTION:
(a) ΔVe = VB – VA = 18.0 – 25.0 = -7.0 V
(b) W = qΔVe = 15.0 x10-6 x (-7.0) = -1.1 x 10-4 J

Electric fields

Make your own and double check if all equations are correct from
text or compare with www.physbot.co.uk

Equipotential surfaces – The gravitational field
The contour map of Mt. Everest ?

The closer the lines, the steeper the terrain(the greater difference in energy level). The farther apart the lines, the flatter the terrain.

Previously, we see that ΔVg = – gΔh which tells us that if h is constant, so is Vg. Thus each elevation line clearly represents a plane of constant potential, an equipotential surface.
• No work is done in moving mass on an equipotential surface

Image K.A. Tsokos page 409

Equipotential surfaces – The electric field

EXAMPLE:
Sketch in the equipotential surfaces Ve (the E-field lines) of (a) a uniform electric field, (b) a point charge and (c) dipole.
SOLUTION: Just make sure your surfaces are perpendicular to the field lines.

PRACTICE: Identify this equipotential surface.
SOLUTION:
·This is an electric dipole.
·The positive charge is the peak and the negative charge is the pit.
·The test charge will “fall” from the peak to the pit. Why?
·What would a negative charge do if released in the pit?

Equipotentials images Tim Kirk IB Study Guides p114

(a) a charge and a mass

(b) two same point charges

(c) two equal and opposite charges

(d) Equipotential lines between charged parallel plates

PRACTICE 10: Draw a 3D contour map of the equipotential surface surrounding a negative charge.
SOLUTION:
A positive test charge placed on the hill will roll downhill (Refer to the right side of image above). Just as with the gravitational equipotential surface, no work is done by the electric field if a charge is moved about on an electrostatic equipotential surface.

27th Feb. 2017 Your task is to investigate and produce an educational video or other forms of presentation on Fields at Work. You can make power point slides but they should include a video clip that describes Fields or one of the bullet points from below. You must include your understanding of:

• Aim 2: Newton’s law of gravitation and Coulomb’s law form part of the structure known as “classical physics”. This body of knowledge has provided the methods and tools of analysis up to the advent of the theory of relativity and the quantum theory.
• Aim 9: models developed for electric and gravitational fields using lines of forces allow predictions to be made but have limitations in terms of the finite width of a line.

• Field lines
• Equipotential surfaces
• No work is done in moving charge or mass on an equipotential surface
• Difference between Potential and potential energy
• Electric potential and gravitational potential
• Potential difference
• Total Energy in the system
• Relevant diagrams

You will upload your completed work on ManageBac or share your creation with Ms. Lee before the end of the lesson and will present your work on Thursday.

10.2 Fields at Work Topic 10.2 is an extension of Topics 5.1, 6.1 and 6.2.This subtopic has a lot of stuff in it. Sometimes the IBO organizes their stuff that way. Live with it!

Essential idea: Similar approaches can be taken in analyzing electrical and gravitational potential problems.

Nature of science: Communication of scientific explanations: The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate, analyze and report findings to a general public used to scientific discoveries based on tangible and discernible evidence.

Understandings: • Potential and potential energy • Potential gradient • Potential difference • Escape speed • Orbital motion, orbital speed and orbital energy • Forces and inverse-square law behavior

Guidance: • Orbital motion of a satellite around a planet is restricted to a consideration of circular orbits (links to 6.1 and 6.2) • Both uniform and radial fields need to be considered • Students should recognize that lines of force can be two-dimensional representations of three-dimensional fields • Students should assume that the electric field everywhere between parallel plates is uniform with edge effects occurring beyond the limits of the plates.

Utilization: • The global positioning system depends on complete understanding of satellite motion • Geostationary / polar satellites • The acceleration of charged particles in particle accelerators and in many medical imaging devices depends on the presence of electric fields (see Physics option sub-topic C.4)

Aims: • Aim 2: Newton’s law of gravitation and Coulomb’s law form part of the structure known as “classical physics”. This body of knowledge has provided the methods and tools of analysis up to the advent of the theory of relativity and the quantum theory. • Aim 4: the theories of gravitation and electrostatic interactions allows for a great synthesis in the description of a large number of phenomena

Data booklet reference:

Potential gradient Potential gradient is the local rate of change of the potential with respect to displacement.

Gravitational field intensity = - Potential gradient, g = - ∆ Vg / ∆ r

Its units are J/Kg m, (N/Kg, m/s^2)

Electric field intensity = - Potential gradient, E= - ∆ Ve / ∆ r

Its units are N/C, (V/m)

How the force changes over distance can be shown from the gradient of the potential vs position graph.
10.4 The graph shows the variation of the gravitational potential V with distance r away from the centre of a dense compact planet of radius 2 x 10^9 m. Use the graph to calculate the work required to move a probe of mass 3400 kg from the surface to a distance of 7.5 x 10^6 m from the centre of the planet. (page 401)

SOLUTION: W = m ΔVg = 6.5 x 10^13 J Conservative forces
We call any force which does work independent of path a conservative force. Only conservative forces have associated potential energy functions. Gravitational force is a conservative force.
EXAMPLE: Show that for a conservative force ΔEp = - W = – F d cos Θ where F is a conservative force.
SOLUTION:

Conservation of energy ΔEP +ΔEK = 0 and the work-kinetic energy theorem W = ΔEK. Then - ΔEK= ΔEP = - W
From the definition of work W = F·d·cos Θ and we can thus write ΔEP = – F d cos Θ

The electrostatic force and the gravitational force are both examples of conservative forces. Thus both forces have potential energy functions. Fg= –G m1m2 /r 2 where G = 6.67× 10-11N m 2 kg −2 ( Universal law of gravitation ) Fe= k q1 q2 /r 2 where k = 8.99 x 109 N m 2 C −2 ( Coulomb's law )

PRACTICE 9: A mass m moves upward a distance Δh without accelerating. (a) What is the change in potential energy of the mass-Earth system? (b) What is the potential difference undergone by the mass? SOLUTION: (a) F = mg , d = Δh and Θ = 180o so that ΔEP= - W = – F ·d ·cos Θ = -(mg) Δh cos(180 o) = mgΔh (b) W = – ΔEP so that W = - mg Δh, thus m ΔVg = W = -mg Δh, ΔVg = -gΔh

PRACTICE 8: A mass m moves upward a distance Δh without accelerating. (c) What is the gravitational field strength g in terms of Vg and Δh? (d) What is the potential difference experienced by the mass in moving from h = 1.25 m to h = 3.75 m ? Use g = 9.81 m/s^2 SOLUTION: ΔVg = -gΔh (c) From ΔVg = -gΔh, g= - ΔVg/Δh, thus field strength = - potential difference / position change (d) From ΔVg = -gΔh, ΔVg = - (9.81)(3.75 - 1.25) = - 24.5 J kg^-1

Topic 10.1 is an extension of Topics 5.1 and 6.2.Electric charges and masses each influence the space around them and that influence can be represented through the concept of fields.

Nature of science:Paradigm shift: The move from direct, observable actions being responsible for influence on an object to acceptance of a field’s “action at a distance” required a paradigm shift in the world of science.

Understandings:• Gravitational fields

• Electrostatic fields

• Electric potential and gravitational potential

• Field lines

• Equipotential surfaces

Applications and skills:• Representing sources of mass and charge, lines of electric and gravitational force, and field patterns using an appropriate symbolism

• Mapping fields using potential

• Describing the connection between equipotential surfaces and field lines

Guidance:• Electrostatic fields are restricted to the radial fields around point or spherical charges, the field between two point charges and the uniform fields between charged parallel plates

• Gravitational fields are restricted to the radial fields around point or spherical masses and the (assumed) uniform field close to the surface of massive celestial bodies and planetary bodies

• Students should recognize that no work is done in moving charge or mass on an equipotential surface

Theory of knowledge:• Although gravitational and electrostatic forces decrease with the square of distance and will only become zero at infinite separation, from a practical standpoint they become negligible at much smaller distances. How do scientists decide when an effect is so small that it can be ignored?

Utilization:• Knowledge of vector analysis is useful for this sub-topic (see Physics sub-topic 1.3)

Aims:• Aim 9: models developed for electric and gravitational fields using lines of forces allow predictions to be made but have limitations in terms of the finite width of a line

Data booklet reference:• W = qΔVe

• W = mΔVg

FORCE FIELDS (The concept of the force field has been introduced in both Topic 5.1 and Topic 6.2)

The force laws F = –Gm1m2 /r 2 and F = kq1q2 /r 2 were first developed as

phenomena.action at a distanceThe field is the force per unit test point

object (mass or positive charge)placed at a particular point.The force of attraction or repulsion acting between two point electric charges isdirectly proportional to the prouct of their chargesand inversely proportional to the square of the distance between them.The force acting between two masses isdirectly proportional to each of their massesand inversely proportional to the square of the distance between them.eo = 8.85 x10 -12 C2m-2N-1 (PERMITTIVITY of free space)Images from x-engineer.org and Hyperphysics

G = 6.67 x 10–11 Nm2kg–2 ( The constant of proportionality )

The direction and strength of a field are illustrated by a field line. The strength of a field is presented by the density of the lines.The density of field lines is considered by dividing the number of lines passing through a given area. The area at a given distance can be thought of as the area of a sphere.

EXAMPLES)

1. Sketch the electric field (E) around a charge

2. Sketch the gravitational field (g) around a mass

Image from www.wikipremed.com

EXAMPLE 3: Describe the difference between electric field and gravitational field.

Electric Field Hockey from Phet

The direction of the field line signifies the direction of the force experienced by a particle placed in the field.

Field lines by a test object, for both g and E fields, when a test object is moved:

~ along/parallel to a field line - work will be done )

~ across/perpendicular to a field line (along the equipotential) - no work will be done.

PRACTICE 1: Sketch the gravitational field about Earth (a) as viewed close from Earth surface, (b) locally from a distance and (c) from far away.

SOLUTION:

## PRACTICE 2: Field strength is a vector quantity so field strengths must be added vectorially.

(a) Estimate the field strength at the point shown due to A, B, C and D. Consider the planet, Earth and Moon, as shown in the diagram below.Image from gradegorilla.com

(b) Calculate the resultant field strength at positions A, B, C and D.

## PRACTICE 3

## Graphical representation

A gravitational field can be shown with a field strength vs distance from centre of Earth graph.SOLUTION:

Image from www.patana.ac.th

Due to the finite speed (v = c) of the force signal required of the special theory of relativity, the

view had to be cast aside in favor of the field view. In the field view, we imagine charges and masses as capable of warping the space around them due to their presence, and other charges and masses responding only to the local curvature of space, and not the actual source of that curvature.action at a distanceAt a distance r there would be N field lines passing through the walls of a sphere of area 4 π r 2. The number of field lines per unit area are therefore N / 4 π r 2. The density of field lines

is proportional to 1/( r2)which is the same as the field strength.Image from www.physicsclassroom.com

PRACTICE 4:

Draw electric field patterns of (a) a positive charge, (b) a negative charge (c) between two charges.

SOLUTION

It simply displays the electric field permeating through space.

Remember the direction of the field arrows showed the direction of the force a test charge, or a test mass, would feel due to the local field.

(a) (b) Yes, your guess is correct !!!

(c)

Image source from www.patana.ac.th

PRACTICE 5: Justify the statement “the electric field strength is uniform between two parallel plates.”

SOLUTION

Hint: Sketch the electric field lines between two parallel plates and demonstrate that the electric field lines have equal density everywhere between the plates.

The field lines are appeared to be parallel in between two parallel plates.

Gravitational Potential, Vg

Energy per unit test point mass that the mass has as a result of the gravitational field.

Vg = energy/mass = W/m = - GM / r [ Unit: J / kg ]

The amount of work required to bring a test mass of one kilogram from infinity to a given point.

( K.A. Tsokos p 399~401 )Worked examples10.2 Calculate the gravitational potential at point P. (Masses and distances are shown on the diagram on text page 399.

V = V1 + V2 = -8.6 X 10 ^-4 J / kg

10.3 The mass of the Moon is about 81 times smaller than that of the Earth. The distance between the Earth and the Moon is about d=3.8 x 10^8 m. The mass of the Earth is 5.97 x 10^24kg.

(a) Determine the distance from the centre of the Earth of the point on the line joining the Earth to the Moon where the combined gravitational field strength of the Earth and the Moon is zero.

The distance at point where the field strength between them is zero is 3.4 x 10^8 m.

(b) Calculate the combined gravitational potential at that point.

Vg = - 1.28 x 10^6 J/kg

(c) Calculate the potential energy when a 2500 kg probe is placed at that point.

Ep = mVg = - 3.2 x 10^9 J

Potential difference – The gravitational force

As masses experience the gravitational force, when one mass is moved in the vicinity of another, work W is done (recall that work is a force times a displacement).

Gravitational Potential Difference:

DVg = W /m

( Note that the units of Vg are J/kg )

Gravitational potential and gravitational potential energy The Fizzics Organization Published on 21 Oct 2014

PRACTICE 6: A mass of m = 500 kg is moved from point A, having a gravitational potential of 75.0 J kg-1 to point B, having a gravitational potential of 25.0 J kg-1.

(a) What is the potential difference undergone by the mass?

(b) What is the work done in moving the mass from A to B?

SOLUTION:

(a) ΔVg = VB – VA = 25.0 – 75.0 = - 50.0 J kg-1

(b) W = mΔVg = 500 x -50.0 = - 25000 J

Electric Potential, Ve

Energy per unit test point charge that the charge has as a result of the electric field.

Ve = energy/charge = W/q = kQ / r [ Unit: J / C, (V) ]

( K.A. Tsokos p 405, 407 )Worked examples10.5 The hydrogen atom has a single proton and a single electron.

(a) Find the electric potential a distance of 0.50 x 10^-10 m from the proton of the hydrogen atom. The proton has a charge 1.6 x 10^-19, equal and opposite to that of the electron.

Ve = kQ / r = 28.7 = 29 V

(b) Use your answer to a to calculate the electric potential energy between the proton in a hydrogen atom and an electron orbiting the proton at a radius 0.50 x 10^10 m

Ep = q Ve = 4.6 x 10^-18 J

10.6 Two unequal positive charges +Q and +q are placed at a distance x (Fig.10.16). Which one of the graphs shows the variation with distance x from the larger charge of the electric potential Ve along the line joining the centres of the charges?

SOLUTION: B

Potential difference – The electric force

As electric charges experience the electric force, when one charge is moved in the vicinity of another, work W is done (recall that work is a force times a displacement).

Electrostatic Potential Difference:

DVe = W /q

( Note: The units of Ve are J/C which are volts V )

PRACTICE 7: A battery has a potential difference of 3V. How much energy is required to move 2mC of charge from the negative to the positive end?

U = qV = ( 2x 10^-6) (3V) = 10^-6 J

PRACTICE 8: A charge of q = +15.0 mC is moved from point A, having a voltage (potential) of 25.0 V to point B, having a voltage (potential) of 18.0 V.

(a) What is the potential difference undergone by the charge?

(b) What is the work done in moving the charge from A to B?

SOLUTION:

(a) ΔVe = VB – VA = 18.0 – 25.0 = -7.0 V

(b) W = qΔVe = 15.0 x10-6 x (-7.0) = -1.1 x 10-4 J

## Electric fields

Make your own and double check if all equations are correct from

text or compare with

www.physbot.co.uk

Equipotential surfaces – The gravitational field

The contour map of Mt. Everest ?

The closer the lines, the steeper the terrain(the greater difference in energy level). The farther apart the lines, the flatter the terrain.Previously, we see that ΔVg = – gΔh which tells us that if h is constant, so is Vg. Thus each elevation line clearly represents

, ana plane of constant potentialequipotential surface.• No work is done in moving mass on an equipotential surface

Image K.A. Tsokos page 409

Equipotential surfaces – The electric field

Sketch in the equipotential surfaces Ve (the E-field lines) of (a) a uniform electric field, (b) a point charge and (c) dipole.

SOLUTION: Just make sure your surfaces are perpendicular to the field lines.

SOLUTION:

·This is an electric dipole.

·The positive charge is the peak and the negative charge is the pit.

·The test charge will “fall” from the peak to the pit. Why?

·What would a negative charge do if released in the pit?

PRACTICE 10: Draw a 3D contour map of the equipotential surface surrounding a negative charge.

SOLUTION:

A positive test charge placed on the hill will roll downhill (Refer to the right side of image above). Just as with the gravitational equipotential surface,

is done by the electric field if a charge is moved about on an electrostaticno worksurface.equipotential27th Feb. 2017

Your task is to investigate and produce an educational video or other forms of presentation on Fields at Work. You can make power point slides but they should include a video clip that describes Fields or one of the bullet points from below. You must include your understanding of:

• Aim 2: Newton’s law of gravitation and Coulomb’s law form part of the structure known as “classical physics”. This body of knowledge has provided the methods and tools of analysis up to the advent of the theory of relativity and the quantum theory.

• Aim 9: models developed for electric and gravitational fields using lines of forces allow predictions to be made but have limitations in terms of the finite width of a line.

• Field lines

• Equipotential surfaces

• No work is done in moving charge or mass on an equipotential surface

• Difference between Potential and potential energy

• Electric potential and gravitational potential

• Potential difference

• Total Energy in the system

• Relevant diagrams

You will upload your completed work on ManageBac or share your creation with Ms. Lee before the end of the lesson and will present your work on Thursday.

10.2 Fields at Work

Topic 10.2 is an extension of Topics 5.1, 6.1 and 6.2.This subtopic has a lot of stuff in it. Sometimes the IBO organizes their stuff that way. Live with it!Essential idea: Similar approaches can be taken in analyzing electrical and gravitational potential problems.

Nature of science:Communication of scientific explanations: The ability to apply field theory to the unobservable (charges) and the massively scaled (motion of satellites) required scientists to develop new ways to investigate, analyze and report findings to a general public used to scientific discoveries based on tangible and discernible evidence.

Understandings:• Potential and potential energy

• Potential gradient

• Potential difference

• Escape speed

• Orbital motion, orbital speed and orbital energy

• Forces and inverse-square law behavior

Guidance:• Orbital motion of a satellite around a planet is restricted to a consideration of circular orbits (links to 6.1 and 6.2)

• Both uniform and radial fields need to be considered

• Students should recognize that lines of force can be two-dimensional representations of three-dimensional fields

• Students should assume that the electric field everywhere between parallel plates is uniform with edge effects occurring beyond the limits of the plates.

Utilization:• The global positioning system depends on complete understanding of satellite motion

• Geostationary / polar satellites

• The acceleration of charged particles in particle accelerators and in many medical imaging devices depends on the presence of electric fields (see Physics option sub-topic C.4)

Aims:• Aim 2: Newton’s law of gravitation and Coulomb’s law form part of the structure known as “classical physics”. This body of knowledge has provided the methods and tools of analysis up to the advent of the theory of relativity and the quantum theory.

• Aim 4: the theories of gravitation and electrostatic interactions allows for a great synthesis in the description of a large number of phenomena

Data booklet reference:Potential gradient

Potential gradient is the local rate of change of the potential with respect to displacement.

10.4 The graph shows the variation of the gravitational potential V with distance r away from the centre of a dense compact planet of radius 2 x 10^9 m. Use the graph to calculate the work required to move a probe of mass 3400 kg from the surface to a distance of 7.5 x 10^6 m from the centre of the planet. (page 401)

SOLUTION: W = m ΔVg = 6.5 x 10^13 J

Conservative forces

We call any force which does work independent of path a conservative force. Only conservative forces have associated potential energy functions. Gravitational force is a conservative force.

EXAMPLE: Show that for a conservative force ΔEp = - W = – F d cos Θ where F is a conservative force.

SOLUTION:

K= ΔEP = - WFrom the definition of work W = F·d·cos Θ and we can thus write ΔEP = – F d cos Θ

The electrostatic force and the gravitational force are both examples of conservative forces. Thus both forces have potential energy functions.

Fg= –G m1m2 /r 2 where G = 6.67× 10-11N m 2 kg −2 ( Universal law of gravitation )

Fe= k q1 q2 /r 2 where k = 8.99 x 109 N m 2 C −2 ( Coulomb's law )

PRACTICE 9: A mass m moves upward a distance Δh without accelerating. (a) What is the change in potential energy of the mass-Earth system? (b) What is the potential difference undergone by the mass?

SOLUTION:

(a) F = mg , d = Δh and Θ = 180o so that ΔEP= - W = – F ·d ·cos Θ = -(mg) Δh cos(180 o) = mgΔh

(b) W = – ΔEP so that W = - mg Δh, thus m ΔVg = W = -mg Δh, ΔVg = -gΔh

PRACTICE 8: A mass m moves upward a distance Δh without accelerating. (c) What is the gravitational field strength g in terms of Vg and Δh? (d) What is the potential difference experienced by the mass in moving from h = 1.25 m to h = 3.75 m ? Use g = 9.81 m/s^2

SOLUTION:

ΔVg = -gΔh

(c) From ΔVg = -gΔh, g= - ΔVg/Δh, thus field strength = - potential difference / position change

(d) From ΔVg = -gΔh, ΔVg = - (9.81)(3.75 - 1.25) = - 24.5 J kg^-1

My Solar System Phet

Kepler's law applet

Escape speed

speed to escape gravitational force of a mass such as planet’s gravity/speed to reach zero gravitational field

V = s.root/ ( 2GM/r ) 1/2

Universal Gravitational Potential Energy Ian Page Published on 25 Jan 2016