PPC 2021 Part 3 - Lecture Notes PDF

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C (L10-L12) Physics Preparatory Course Lesson 10 / 11: Gravitational Fields Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University 26 Jul 2021

All things are difficult before they are easy. Thomas Fuller

Key things to committed to long term memory

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Newton and Apple

𝑊 = 𝑚𝑔

https://www.flickr.com/photos/hiki ngartist/6217869031/

Earth

Giancoli Fig 6.2 The gravitational force one object exerts on a second object is directed toward the first object, and is equal and opposite to the force exert by the second on the second object on the first.

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Newton’s Law of Gravitation Newton’s Law of gravitation:

Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line joining the two particles. The magnitude of the gravitational force can be written as 𝐹𝐺 =

𝐺𝑚1 𝑚2 𝑟2

where 𝑚1 and 𝑚2 are the masses of the two particles and 𝑟 is the distance between them, and 𝐺 is a universal constant which must be measured experimentally and has the same numerical value for all objects. The value of 𝑮 = 𝟔. 𝟔𝟕 × 𝟏𝟎−𝟏𝟏 𝑵 𝒎𝟐 /𝒌𝒈. Note here that Big “G” is universal and do not change with location. Small 𝒈 changes with location and is 9.8 m/s2 near surface of Earth.

Giancoli Fig 6.3 Cavendish Experiment

Some Examples Giancoli pg 141, Example 6-1: Can you attract another person gravitationally? A 50-kg person and a 70-kg person are sitting on a bench close to each other. Estimate the magnitude of the gravitational force each exerts on the other.

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Newton’s Law of Gravitation – a little problem

𝐹=

𝐺𝑚1 𝑚2 𝑟2

What is the value of 𝑟 for this case?

https://www.flickr.com/photos/hiki ngartist/6217869031/

The total force exerted on the apple due to Earth is equivalent to the force due to all mass of Earth concentrated in its own center.

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EXERCISE 1 You are given that the radius of the Earth is 6400 km and that the mass of Earth is 5.98 × 1024 kg. a) Determine the gravitational force acting on an apple with mass 0.120 kg near the surface of Earth. b) How does this gravitational force compare to the weight of the apple?

c) Determine the gravitational force on the apple when it is 6400 km above the surface of the Earth,

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Why does the moon not crash into Earth?

he cannon ball is kept in orbit by its peed—it is continually falling, but the arth curves from underneath it.

Knight Fig 8-12

Force along radial direction: 𝐹𝐺 =

𝑚𝑣 2 ; 𝑟

𝐹𝐺 = 𝑚𝑟𝜔2

Giancoli pg 146 Fig 6.12: Gravitational force is causes the satellite to deviate from a straight line. At the right distance from Earth’s centre and velocity, the satellite moves in a circular path.

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Newton’s Law of Gravitation Giancoli pg 142, An extension of Example 6-2: Spacecraft at 2rE.

What is the force of gravity acting on a 2000-kg spacecraft when it orbits two Earth radii from the Earth’s center (that is, a distance rE = 6380 km above the Earth’s surface)? The mass of the Earth is mE = 5.98 x 1024 kg.

If the spacecraft is orbiting in a circle at constant speed, what is the centripetal force it experiences?

To stay in orbit, what speed must the spacecraft be moving? How long does it take to complete one round?

If we wish to have a satellite orbiting with a period of 24 hr, how far from the Earth’s center should the satellite be? At what speed should it be moving in the orbit? Is energy needed to keep it in motion?

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Newton’s Law of Gravitation If we wish to have a satellite orbiting with a period of 24 hr, how far from the Earth’s center should the satellite be? At what speed should it be moving in the orbit? Is energy needed to keep it in motion?

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EXERCISE 1

Actual radius of Earth is about 6400 km.

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3. [G6.28] Two satellites orbit Earth at altitudes of 5000 km and 15,000 km. Which satellite is faster, and by what factor?

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Concept of field

Probably your first encounter… magnetic fields.

http://en.wikipedia.org/wiki/Magnetic_field#/media/File:Magnet0873.png http://en.wikipedia.org/wiki/Magnetic_field#/media/File: VFPt_cylindrical_magnet_thumb.svg

Closer field lines → Greater field strength → Greater force

In gravitation,

Larger field lines density → Greater field strength → Greater force

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Gravitational Field Strength Gravitational force per unit mass. 𝐹𝑔 = 𝑚𝑔 =

𝑔=

𝑔=

𝐺𝑀 𝑟2

𝐺𝑀𝑚 𝑟2

𝐺𝑀 𝑟2

More Examples Giancoli Example 6-4: Gravity on Everest. Estimate the effective value of g on the top of Mt. Everest, 8850 m above sea level. That is, what is the acceleration due to gravity at this altitude? (𝑟𝐸 = 6380 km, 𝑀𝐸 = 5.98 × 1024 kg)

Google home: Oil exploration, gravimeter 11

Gauss’s Law for Gravitational Field

𝑔=

𝐺𝑀 ⇒ 𝑟2

At 𝑟𝐸 ,

At 2𝑟𝐸 ,

At any 𝑟 = 𝑟𝐸 ,

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EXERCISE 1. [G6.6] Calculate the effective value of g, the acceleration of gravity, at (a) 6400 m, and (b) 6400 km, above the Earth’s surface.

2. [G6.2] Calculate the acceleration due to gravity on the Moon. The Moon’s radius is 1.74 × 106 𝑚 and its mass is 7.35 × 1022 𝑘𝑔.

3. [G6.14] Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, and that Mars’ radius is 3400 km, determine the mass of Mars.

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Physics Preparatory Course Lesson 11 / 12: Electric Fields Dr Ho Shen Yong Lecturer, School of Physical and Mathematical Sciences Nanyang Technological University 28 Jul 2021

A people that values its privileges above its principles soon loses both. Dwight D. Eisenhower (1890-1969), Inaugural Address, January 20, 1953

Key things to committed to long term memory

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Introduction to Electrostatics

In mechanics (dynamics), calculations often involve the physical property mass of objects. In electricity and magnetism, the key physical property of objects that is of interest is electric charge. The subject of electricity and magnetism can be divided into three realms: 1) the charges that we study are at rest with respect to each other and to observers like us; 2) the charges are moving with a relatively constant average speed in the circuitry – “electricity”. Another phenomena that involves charges moving at constant velocity is the generation of magnetic fields. 3) the charges are accelerating – fields can be generated and energy radiated into space. Gravitational force is important on scale of planets and our everyday life. However, electromagnetic interaction is the dominant force at the microscopic level down to the atomic scale. It is the force responsible for holding the electrons and the nuclei together in atoms and all forms of molecular bonding – thus responsible for all of chemical and biological phenomena. Looking at our present reliance on electrical appliances, we can also safely say that the understanding and application of electricity and magnetism is one of the cornerstone in the modernization of human civilization. [email protected] The unit of charge: Coulomb (C) The SI unit of charge is Coulomb (C) which is defined as the amount of charge that flows in 1 second when there is a steady current of 1 ampere. The electron is an elementary charge carrier and has a charge of −𝑒 where 𝑒 = 1.60217733 × 10−19 𝐶. The charge of a proton is +𝑒. Thus, we see that one C is made up of many elementary charges.

Giancoli pg 561 Fig 21.3: Simple model of atom

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Examples of Daily Occurrences and Applications 1. 2. 3. 4. 5.

Lightning Electrical igniters at gas stove Charging of cellphones, camera, flash etc Computer Keyboards Capacitive Touchscreens Aspects of Electrostatics

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Visualizing Charge (Knight) Pause to Ponder: A rod attracts a positively charged hanging ball. The rod is A. Positive. B. Negative. C. Neutral. D. Either A or C. E. Either B or C.

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Charge Polarization (Knight) Charge polarization is the slight separation of the positive and negative charges in a neutral object when a charged object is brought near.

The polarization force arises because the charges in a neutral object are slightly separated, not because the objects are oppositely charged. The polarization force between a charged object and a neutral one is always attractive.

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Pause to Ponder: Metal spheres 1 and 2 are touching. Both are initially neutral. a. The charged rod is brought near. b. The charged rod is then removed. c. The spheres are separated. Afterward, the charges on the sphere are: A.

Q1 is + and Q2 is +

B.

Q1 is + and Q2 is –

C.

Q1 is – and Q2 is +

D.

Q1 is – and Q2 is –

E.

Q1 is 0 and Q2 is 0

Pause to Ponder: Metal spheres 1 and 2 are touching. Both are initially neutral. a. The charged rod is brought near. b. The spheres are separated. c. The charged rod is then removed. Afterward, the charges on the sphere are: A.

Q1 is + and Q2 is +

B.

Q1 is + and Q2 is –

C.

Q1 is – and Q2 is +

D.

Q1 is – and Q2 is –

E.

Q1 is 0 and Q2 is 0

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Atomic view of charging (Knight)

Molecular ions can be created when one of the bonds in a large molecule is broken.

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Newton’s Law of Gravitation and Coulomb’s Law Newton’s Law of Gravitation

Coulomb’s Law on Electrostatic forces Coulomb’s Law:

Gauss’s Law which was discussed for gravitational fields is also similarly applicable to electric fields.

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Coulomb's law (discovered in 1785) The magnitude of the force (F) between two electrically charged bodies, which are small compared with their separation (r), is inversely proportional to r2 and proportional to the product of their charges (Q1 and Q2): 𝑄𝑄

𝐹 = 𝑘 122 𝑟 where the constant of proportionality k, in vacuum, expressed in SI units is given by 1 𝑘= 4𝜋𝜀 where 𝜖 is the permittivity. It is dependent on the medium in which the charges are in. For example, the permittivity of water at room temperature is about 80 times the permittivity of free space (vacuum) and thus the attractive forces between Na+ and Cl- is much weaker in water. The permittivity of free space (vacuum) 𝐶2 𝜀𝑜 = 8.85418781762 × 10−12 . 𝑁 ∙ 𝑚2 For the purpose of simple computation in vacuum, we just use 1 𝑘 = 4𝜋𝜀 = 8.99 × 109 𝑁 ∙ 𝑚2/𝐶2 𝑜

Thus, Coulomb’s law can be written as 𝑄𝑄

𝐹 = 4𝜋𝜀1 𝑟22. 𝑜

Taken from Knight, pg 801, Fig 26.17

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Coulomb's law (discovered in 1785) EXERCISE: Which of the three right-hand charges experiences the largest force? A. q B. 2q C. 4q D. q and 2q are tied E. q and 4q are tied

EXERCISE: In each of the following cases, an identical small, positive charge is placed at the black dot. In which case is the force on the small charge the largest?

EXERCISE: In each of the following cases, an identical small, positive charge is placed at the black dot. In which case is the force on the small charge the largest? (All charges shown are of equal magnitude.)

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Example Giancoli page 565, Conceptual Example 21-1: Which charge exerts the greater force? Two positive point charges, Q1 = 50 μC and Q2 = 1 μC, are separated by a distance 𝑙 . Which is larger in magnitude, the force that Q1 exerts on Q2 or the force that Q2 exerts on Q1?

Giancoli pg 566 Example 21-2: Three charges in a line Three charged particles are arranged in a line, as shown. Calculate the net electrostatic force on particle 3 (the -4.0 μC on the right) due to the other two charges.

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Exercise 1. [G21.12] Particles of charge +75, +48 and −85 𝜇𝐶 are placed in a line. The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two.

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Exercise 1

2

3

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Giancoli pg 566 Example 21-3: Electric force using vector components. Calculate the net electrostatic force on charge 𝑄3 shown in the figure due to the charges 𝑄1 and 𝑄2 .

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Exercise 2. [G21.13] Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m as shown. Calculate the magnitude and direction of the net force on each due to the other two.

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Electric field of a point charge The electric field exist in the region of space around a charged object, the source charge Q. To study the properties of the electric field, a very small positive test charge qo is placed in this region to probe the electric field without distorting it. The test charge qo will experience a force 𝐹റ 𝑒 due to the source charge. The electric field 𝐸 of the charge Q at a point is defined as 𝐹റ 𝐸 ≡ lim 𝑒 𝑞𝑜→0 𝑞𝑜 Remarks:

1. 2. 3. 4.

5.

𝐸 has SI units N/C. The direction of 𝐸 is the direction of force the positive test charge experiences. The electric field is a property of the charge Q, independent of the test charge. We can write the force 𝐹റ 𝑒 acting on any charge q in an electric field 𝐸 as 𝐹റ𝑒 = 𝑞𝐸. This is true for any electric field. For a point charge Q, 𝐹റ𝑒 𝑄 = 𝑟Ƹ 𝐸= [email protected] 𝑞𝑜 4𝜋𝜀𝑜𝑟2

The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.

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Electric fields of various systems of charges

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Exercise 1. [G21.21] What are the magnitude and direction of the electric force on an electron in a uniform electric field of strength 1920 N/C that points due east?

2. [G21.22] A proton is released in a uniform electric field, and it experiences an electric force of 2.18 × 10−14 𝑁 toward the south. What are the magnitude and direction of the electric field?

3. [G21.23] Determine the magnitude and direction of the electric field 16.4 cm directly above an isolated 33.0 × 10−6 𝐶 charge.

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Exercise 4. [G21.35] Determine the direction and magnitude of the electric field at the point 𝑃 in the figure. The charges are separated by a distance 2𝑎, and point P is a distance 𝑥 from the midpoint between the two charges. Express your answer in terms of 𝑄, 𝑥, 𝑎, and 𝑘 .

5. [G21.41] You are given two unknown point charges, 𝑄1 and 𝑄2 . At a point on the line joining them, one-third of the way from 𝑄1 to 𝑄2 the electric field is zero as shown. What is the ratio 𝑄1 /𝑄2 ?

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Exercise 6. [G21.88] A point charge (𝑚 = 1.0 𝑔) at the end of an insulating cord of length 55 cm is observed to be in equilibrium in a uniform horizontal electric field of 15,000 N/C, when the pendulum’s position is as shown, with the charge 12 cm above the lowest (vertical) position. If the field points to the right as shown in the figure, determine the magnitude and sign of the point charge.

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Exercise 7. [G21.85] Suppose electrons enter a uniform electric field midway between two plates at an angle 𝜃𝑜 to the horizontal, as shown in the figure. The path is symmetrical, so they leave at the same angle 𝜃𝑜 and just barely miss the top plate. What is 𝜃𝑜 ? Ignore fringing of the field.

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