Physics 1P22/1P92 Review

Chapter 20 Electric Forces and Fields

Key Concepts

electric charge•principle of conservation of charge•charge polarization, both permanent and induced•good electrical conductors vs. good electrical insulators•Coulomb's law for the force exerted by one charged particle on another

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the electric field concept; representation of an electric field using field lines or field vectors

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electric field of a point charge (formula and field pattern, both for a positive point charge and a negative point charge)

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field pattern for a constant (i.e., uniform) electric field•field pattern for an electric dipole•properties of electric field lines in the vicinity of conductors (20.6)•

Review Problems

CP 13 A small glass bead has been charged to +20 nC. A tiny ball bearing 1.0 cm above the bead feels a 0.018 N downward electric force. Determine the charge on the ball bearing.

Solution: Use Coulomb's law.

Sunday, September 01, 2013

1:43 PM

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The magnitude of the charge on the ball bearing is 10 nC. Because the charge is attractive, and the charge on the glass bead is positive, the charge on the ball bearing is negative. Thus, the charge on the

ball bearing is 10 nC.

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CP 59 A +10 nC charge is located at (x, y) = (0 cm, 10 cm) and

a 5.0 nC charge is located at (x, y) = (5.0 cm, 0 cm). Where would

a 10 nC charge need to be located in order that the electric field at the origin be zero?

Solution: Let (x, y) represent the coordinates of the 10 nC charge. The strategy is to write an expression for the total electric field at the origin created by the three charges, then equate this expression to zero, and solve for the unknown coordinates. Note the use of the superposition principle.

The net electric field created by the three charges at the origin is

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CP 64 Two 3.0 g spheres on 1.0-m-long threads repel each other after being equally charged, as shown in the figure. Determine the charge q.

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Solution: Draw a free-body diagram for one of the spheres. I'll choose to draw a free-body diagram for the sphere on the right:

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Chapter 21 Electric Potential

Key concepts:

electric potential•electric potential energy•the electron-volt (eV), a convenient unit of energy when dealing with atomic or subatomic phenomena

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a point charge○

a constant electric field○

electric potential formula for: •

connection between electric potential and electric field•properties of equipotential surfaces•properties of a conductor in electrostatic equilibrium•capacitance•capacitance of a parallel-plate capacitor•capacitance of a parallel-plate capacitor with a dielectric medium between the plates

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energy stored in an electric field•

Review Problems

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Page 708, CP 66 An alpha particle and an antiproton are

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released from rest a great distance apart. They are oppositely charged, so they accelerate towards each other. What are their speeds when they are 2.5 nm apart?

Solution:

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Chapter 22 Current and Resistance

Key Concepts

what is an electric current; the reality, and our simplified model•definition of current; "conventional current"•the principle of conservation of current; Kirchhoff's junction law•the "charge escalator" model of a battery•resistivity and resistance of a wire or other circuit element•Ohm's law•the ideal wire model for analyzing simple circuits•energy and power in simple circuits•

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Chapter 23 Electric Circuits

Review Problems

CP 41 The total charge a household battery can supply is

given in units of mA hr. For example, a 9.0 V alkaline battery is

rated at 450 mA hr, meaning that such a battery could supply a 1 mA current for 450 hr, a 2 mA current for 225 hr, etc. How much energy, in joules, is this battery capable of supplying?

Key Concepts

circuit diagrams; the symbols used to represent circuit elements on circuit diagrams

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Kirchhoff's junction law; Kirchhoff's loop law•series and parallel circuits•equivalent resistance for a series circuit•equivalent resistance for a parallel circuit•analyzing simple circuits•analyzing more complex circuits•capacitors in series and parallel --- OMIT•

Review Problems

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Chapter 24 Magnetic Fields and Forces

Key Concepts

magnetic fields, magnetic field lines•magnetic field patterns for a long straight wire, a circular loop, and a solenoid

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using the formulas for the magnetic field for a long straight wire, a circular loop, and a solenoid

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force exerted by a magnetic field on a moving charged particle•possible paths of charged particles in magnetic fields•applications: aurorae, flow meters, mass spectrometer•force exerted by a magnetic field on a current-carrying wire•magnetic force between parallel current-carrying wires•torque exerted by a magnetic field on a current loop; electric motors•OMIT Section 24.8•

Review Problems

CP 50 An antiproton is moving in the combined electric and magnetic fields shown in the figure. (a) Determine the magnitude and direction of the antiproton's acceleration at this instant.(b) Repeat Part (a) if the direction of the velocity is reversed.

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CP 60 A mass spectrometer is designed to separate atoms of carbon to determine the fraction of different isotopes. There are three main isotopes of carbon, with the following atomic masses:

Solution: The direction of the electric force on the antiproton is upwards, and the direction of the magnetic force on the antiproton is downwards. The magnitudes of the forces are calculated as follows:

The atoms of carbon are singly ionized and enter a mass spectrometer with magnetic field strength B = 0.200 T at a speed of 150 km/s. The ions move along a semicircular path and exit through an exit slit. How far from the entrance will the beams of the different isotope ions end up?

(b) The magnetic force has the same magnitude as in Part (a) but is now directed upwards. Thus, the net force on the antiproton is

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Chapter 25 Electromagnetic Induction and Electromagnetic Waves

Key Concepts

electromagnetic induction; motional emf•magnetic flux•Faraday's law of induction•Lenz's law•electromagnetic waves•intensity and energy density of an electromagnetic waves•polarization of an electromagnetic wave•the electromagnetic spectrum•

up?

Solution:

Thus, the distances for each of the three isotopes is

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Review Problems

CP 5 A 50 g horizontal metal bar, 12 cm long, is free to slide up and down between two tall, vertical metal rods that are 12 cm apart. A 0.060 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a 1.0 Ohm resistor is connected across the two rods at the top. Then the bar is dropped. Determine the terminal speed at which the bar falls. Assume the bar remains horizontal and in contact with the rods at all times.

Solution: The bar reaches terminal speed when the net force on it is zero. This occurs when the downward gravitational force on the bar is balanced by the upward magnetic force on the bar.

CP 17 The circuit in the figure is a square 5.0 cm on a side. The magnetic field increases steadily from 0 T to 0.50 T in 10 ms. Determine the current in the resistor.

Solution: First determine the induced emf.

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CP 28 A radio antenna broadcasts a 1.0 MHz radio wave with 25 kW of power. Assume that the radiation is emitted uniformly in all directions.(a) Determine the wave's intensity 30 km from the antenna.(b) Determine the electric field amplitude at this distance.

Solution:

CP 52 A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60 degrees away from vertical increases from 0.50 T to 1.50 T in 0.60 s. Determine the induced emf in the coil.

Solution:

Is this induced emf in the same direction as the battery, or in the opposite direction? Use Lenz's law to determine this.

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Chapter 26 Alternating Current

Key Concepts

alternating current•peak quantities and rms quantities•electrical transformers•use of transformers to facilitate cross-country power transmission•household electricity•calculating the cost of electricity•biological effects and electrical safety•OMIT 26.5 to 26.7 inclusive•

Review Problems

CP 10 A soldering iron uses an electric current in a wire to heat the tip. A transformer with 100 turns on the secondary coil provides 50 W at an rms voltage of 24 V.(a) Determine the resistance of the wire in the soldering iron.(b) Determine the number of turns in the primary coil.(c.) Determine the current in the primary coil.

Only the component of the magnetic field PARALLEL to the axis of the loop matters.

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Solution:

CP 12 A neon sign transformer has a 450 W AC output with an rms voltage of 15 kV when connected to a normal household outlet. There are 500 turns of wire in the primary coil.(a) Determine the number of turns of wire in the secondary coil.(b) Determine the current in both the primary and secondary coil when the transformer is running at full power.

Solution:

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Chapter 28 Quantum Physics

Key Concepts

CP 20 The manufacturer of an electric table saw claims that it has a 3.0 hp motor. It is designed to be used on a normal 120 V outlet with a 15 A circuit breaker. Is this claim reasonable? Explain.

Solution: Determine the maximum power possible from the normal outlet:

CP 24 A fisherman has netted a torpedo ray. As he picks it up, this electric fish creates a short-duration 50 V potential difference between his hands. His hands are wet with salt water, and so his skin resistance is a very low 100 Ohms. Determine the current that passes through his body and whether the fisherman feels it.

Solution: From figures in the textbook, the total internal resistance of the fisherman's two arms is 620 Ohms. Adding in the skin resistance to get the total resistance, we obtain 720 Ohms. Thus, the current through the fisherman is

The claim is not reasonable, because the circuit breaker will trip at a maximum power of 2.4 hp. One would need a beefier circuit to carry a larger current so that the saw could deliver 3 hp of power.

Currents this large will definitely be felt. A brief pulse this large may be harmless, but if it were prolonged it would be potentially lethal.

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Chapter 28, CP 7 Electrons are emitted when a metal is illuminated by light with a wavelength less than 388 nm but for no greater wavelength. Determine the metal's work function.

Solution:

key puzzles leading to the introduction of quantum ideas: blackbody radiation, photoelectric effect, atomic structure, atomic spectra

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Planck's introduction of the hypothesis that energy is collected in bundles before being radiated by a glowing object

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Einstein's explanation of the photoelectric effect; the photon hypothesis

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solving problems involving the photoelectric effect•de Broglie's matter wave hypothesis; wave-particle duality•quantization of energy for a bound particle; the "particle in a box," a toy example

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energy levels for a bound particle; energy transitions ("quantum jumps")

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Heisenberg's uncertainty principle•

Review Problems

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CP 8 Electrons in a photoelectric effect experiment emerge from a copper surface with a maximum kinetic energy of 1.10 eV. Determine the light's wavelength.

Solution: Look up the work function for copper; it is 4.65 eV.

CP 9 You need to design a photodetector that can respond to the entire range of visible light. Determine the maximum possible work function of the cathode.

Solution: The smallest desired frequency (which corresponds with the largest desired wavelength) should be able to just eject electrons at the maximum possible work function. Thus,

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Chapter 28, CP 11 Zinc has a work function of 4.3 eV. (a) Determine the longest wavelength of light that will release an electron from a zinc surface. (b) A 4.7 eV photon strikes the surface and an electron is emitted. Determine the maximum possible speed of the electron.

Solution:

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CP 15 The spacing between atoms in graphite is approximately 0.25 nm. Determine the energy of an x-ray photon with this wavelength.

Solution:

CP 29 Determine the kinetic energy of an electron with a de Broglie wavelength of 1.0 nm.

Solution:

CP 35 Determine the length of a box in which the minimum energy of an electron in the n = 1 level has the same energy as a photon with a wavelength of 600 nm.

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Chapter 29 Atoms and Molecules

Key Concepts

discrete atomic spectra; emission spectra, absorption spectra•the Geiger-Marsden experiment•Rutherford's atomic model•Bohr's atomic model•Bohr's model of a hydrogen atom•electron spin•quantum numbers for atomic electrons•Pauli exclusion principle and the periodic table•absorption and emission spectra of atoms•OMIT Sections 29.8 and 29.9•

Review Problems

CP 8 The figure shows an energy level diagram for a simple atom. Determine the wavelengths that appear in the atom's emission and absorption spectra.

wavelength of 600 nm.

Solution: The energy of a photon with wavelength 600 nm is

The energy levels for a particle in a box are

For an electron in the n = 1 level, we obtain

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Solution: For the emission spectrum, determine the energy differences:

Now calculate the corresponding photon wavelengths:

Only the first and third wavelengths appear in the absorption spectrum, because atoms are ordinarily in their ground states. All three wavelengths appear in the emission spectrum.

CP 9 An electron with kinetic energy 2.0 eV collides with an atom whose energy-level diagram is the same as in the previous problem. (a) Is the incident electron able to stimulate the atom to an excited state? Explain. (b) If the answer to part (a) is yes, then determine the electron's kinetic energy after the collision.

Solution: (a) Yes. Atoms are normally in their ground states, and it is possible for the electron in the atom's ground state to absorb 1.5 eV of energy from the incident electron to jump into the atom's n = 2 state.

(b) 2.0 eV 1.5 eV = 0.5 eV

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CP 19 Determine all possible wavelengths of photons that can be emitted by transitions of electrons from the n = 4 level of a hydrogen atom.

Solution:

CP 33 An electron with a speed of 5.00 × 106 m/s collides with an atom. The collision excites the atom from its ground state (0 eV) to a state with an energy of 3.80 eV. Determine the speed of the electron after the collision.

Solution: Determine the initial kinetic energy of the electron. Then note that the electron loses 3.80 eV of kinetic energy by transfer to the atom's electron. Finally, use the remaining kinetic energy of the electron to determine its final speed.

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The final kinetic energy of the electron is therefore 71.2 3.8 = 67.4 eV. The corresponding final speed of the electron is

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