#$&*
course Phy 202
7/27 at 3:45**Please have back by tomorrow at 6pm**
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Problem Number 1
A charge of 20 `microC with an initial kinetic energy of 484 J and a mass of 1.44 kg is acted upon by an electric field against which it does -18 J of work. If no force other than the electrostatic forces act on the charge, what will be its final kinetic energy and speed?
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.KE= ½ k e^2/ m ?
.= 9 x 10^9 (20)^2 / 1.44 x 466
? I don’t think this formula is correct at all. I’m getting a huge number that cannot be correct.
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KE increases by 18 J when the particle does - 18 J of work against the field, so the KE ends up at 502 J.
KE = 1/2 m v^2 so
v = sqrt( 2 * KE / m) = sqrt( 2 * 502 J / (1.44 kg) ) = etc.
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Problem Number 2
What is the force on an electron (charge of magnitude 1.6 * 10^-19 C) in a uniform conducting wire 1.5 meters long, if the ends of the wire are maintained at a potential difference of 11 volts?
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.F= k Q1Q2/r^2
= 9x10^9 (1.6x10^-19 x 11)/1.5
=1.05x10^-8 N
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You don't have two point charges here, so Coulomb's Law doesn't apply.
The electric field is
E = 11 volts / (1.5 meters) = 7 volts / meter, approx.., which is also equal to about 7 Newtons / Coulomb.
The charge of an electron is 1.6 * 10^-19 Coulombs, so the force is about
F = q * E = 1.6 * 10^-19 C * 7 N / c = 1.2 * 10^-18 Newtons.
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Problem Number 3
A sunbeam with intensity 6.5 watts/m ^ 2 shines through a circular window with radius .9 meters. The direction of the sunlight is perpendicular to the plane of the window. What is the power flux through the window?
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.sin(90) x 6.5/9
.= 0.72
I’m not sure this is the correct equation either.
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Flux for a perpendicular beam is intensity * area.
Here the area is that of a circle with radius .9 meters, so you get
flux = 6.5 watts / m^2 * pi * (.9 m)^2 = etc.
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Problem Number 4
The table below depicts properties of the isotopes of the eight lightest elements. For the element of column 10, is it possible from an energy standpoint for the isotope in the fourth line to emit a neutron and change to the isotope in the third line?
Properties of Selected Particles and Light Nuclei (most common isotope listed first)
particle or atom proton neutron hydrogen helium lithium beryllium boron carbon nitrogen oxygen
atomic number 1 0 1 2 3 4 5 6 7 8
mass (amu) 1.007276 1.008665 1.007825 4.002603 6.015122 9.012182 10.01294 12 14.00307 15.99492
mass (amu) 2.014102 3.016029 7.016004 11.00931 13.00336 15.00011 16.99913
mass (amu) 3.016049 17.99916
electron mass (amu) atomic mass unit
0.000549 1.660502 * 10^-27 kgw
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.I don’t exactly understand the question or what on earth some of these numbers represent. What are there all of these different masses? What are they for?
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This problem should not have been on this test; any similar problem would be omitted.
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Problem Number 5
An electron (mass 9.11 * 10^-31 kg) with velocity directed to the North passes through a magnetic field of .006 Tesla directed vertically upward, crossed with an electric field of 42000 N/C directed either East or West. The electron passes through undeflected. Is the electric field directed East or West, and how fast is the electron moving? What force does the electron experience from each field?
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.F= 9x10^9 (0.006 x 9.11x0^-31/42000)
=1.17x10^-27 west
This is probably not the right equation, but I can’t find another one with force in it right now.
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The force on a charge moving perpendicular to a magnetic field is F = q v B. The force in an electric field is F = q E.
The particle is undeflected, so the forces are equal and opposite, and the magnitudes of the forces must therefore be equal so
q v B = q E. Solving for v we get
v = E / B.
For a negative charge moving north through an upward field the q v vector is toward the south. By the right-hand rule the magnetic force on the charge will therefore be to the west. The electric field must therefore exert a force to the east.
To exert a force to the east on a negative charge the electric must be to the west.
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Problem Number 6
A generator has a negligible internal resistance. It creates a potential difference of 2.2 volts when cranked at 1 revolution/sec, and the voltage is proportional to the cranking rate. The generator cranked at 3.2 revolutions / sec, and is connected to a series combination of two bulbs, one with a resistance of 2.2 ohms and the other with resistance 29 ohms. How much current flows in each bulb? How much power is dissipated in each?
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V=IR
I=V/R
I=3.2/2.2
=1.45 current.
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The voltage is 2.2 volts / (rev/sec) * 3.2 rev/sec = 7 volts, very approximately.
The resistance of the series is 2.2 ohms + 29 ohms = 31.2 ohms, so the current is 7 volts / (31.2 ohms) = .23 amps, approx..
The circuit is series so the same current must go through both bulbs.
power = current * voltage, voltage = current * resistance, so power = current * current * resistance = current^2 * resistance:
P = I^2 * R = (.23 amps)^2 * 31.2 ohms = 1.5 watts, very approximately.
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Problem Number 7
Find an expression for the electric field strength at distance r from the axis of a capacitor consisting of two co-axial cylinders 8 meters long, with diameters 1.19 cm and 1.84 cm, with the inner cylinder carrying charge - 11.1 `microCoulombs and the outer cylinder carrying charge + 11.1 `microCoulombs. You will find three expressions, depending on whether r is less than 1.19 cm, greater than 1.84 cm or between the two.
• Estimate the average electric field between the cylinders, then determine the approximate work done per unit charge to move a charge from the inner cylinder to the outer. (University Physics students find the exact amount of work, using an integral).
• What is the voltage between the cylinders?
• What is the capacitance of this capacitor?
• What would be the approximate magnitude of the effect on the work, voltage and capacitance if the inner cylinder was not changed, but the outer cylinder shrunk in such a way as to be 10 times closer to the inner?
.E= kQ/r^2
This applies to a point charge and won't work here.
E= 9x10^9(11.1)/1.19^2
1.2x10^9
What do you mean by depending on what r is?
V=IR
Is there a different equation that can be used to find voltage possibly? I don’t have any of these variables.
What is the capacitance? I don’t recall going over that. What chapter is it in?
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The total flux of -11.1 microCoulombs of charge is
flux = 4 pi k Q = 4 pi * 9 * 10^9 N m^2 / C^2 * (-11.1 * 10^-6 C) = -1.2 * 10^6 N m^2 / C.
A cylinder of length 8 m and radius r has surface area 2 pi r * 8 m = 16 pi * r meters.
If 1.19 cm < r < 1.84 cm a cylindrical Gaussian surface coaxial with the actual cylinders will enclose the charge of the inner cylinder, so the flux through the Gaussian surface is -1.2 * 10^6 N m^2 / C. Dividing this by the area 16 pi * r meters we find the field
E = (-1.2 * 10^6 N m^2 / C ) / (16 pi * r meters) = -1/r * 2.4 * 10^4 N m / C.
Estimating the average electric field to be the field halfway between the cylinders, we use r = (1.19 cm + 1.84 cm) / 2 = 1.52 cm = .015 m, approx., and get
E_ave = -1 / (.015 m) * 2.4 * 10^4 N m / C = -1.6 * 10^6 N / C, again very approximately.
Voltage is the work per unit charge to move charge from one point to the other. Moving against an average field of -1.6 * 10^6 N / C a test charge Q would experience average force F = Q E = Q * -1.6 * 10^6 N / C. Moving from the inner to the outer cylinder, a distance of .65 cm or .0065 m, the charge would do work `dW = F * `ds = Q * 1.6 * 10^6 N / C * .0065 meters against the field.
So the work per unit charge would be (Q * 1.6 * 10^6 N / C * .0065 meters) / Q = 1 * 10^4 N m / C = 1 * 10^4 J / C, or 1 * 10^4 volts.
The voltage would therefore be about 10^4 volts, or 10 000 volts.
The capacitance would therefore be
C = Q / V = 11.1 * 10^-6 C / (10 000 volts) = 1.1 * 10^-9 C / volt = 1.1 * 10^-9 Farad.
If the outer cylinder was moved 10 times closer the average electric field would increase by about 25%, since r will decrease from about 1.5 m to about 1.2 m. However the distance the test charge has to move through the field becomes 1/10 as great, so the work done will become about 1/10 as great and the voltage about 1/10 as great.
Ignoring the effect of the increased field, the voltage becomes about 1/10 as great, so the capacitance would increase by a factor of about 10.
The 25% increase in average electric field would bring the factor up to about 12 or 13.
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Problem Number 8
Find the magnetic flux through a square wire loop with side .13 meters, in the presence of a magnetic field with strength .0081 Tesla, when the field makes an angle of 49 degrees with a perpendicular to the plane of the loop.
What is a magnetic flux. We haven’t gone over magnetism yet for this test, have we?
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No but we have gone over flux, so this should be an easy problem.
Flux = field * area when field is perpendicular to the loop. If not, multiply by the cosine of the angle between the field and the perpendicular. In this case we get
flux = .0081 Tesla * (.13 meters)^2 = etc.. The result will be in units of Tesla * m^2.
#$&*
course Phy 202
7/27 at 3:45**Please have back by tomorrow at 6pm**
.
.
.
Problem Number 1
A charge of 20 `microC with an initial kinetic energy of 484 J and a mass of 1.44 kg is acted upon by an electric field against which it does -18 J of work. If no force other than the electrostatic forces act on the charge, what will be its final kinetic energy and speed?
.
.KE= ½ k e^2/ m ?
.= 9 x 10^9 (20)^2 / 1.44 x 466
? I don’t think this formula is correct at all. I’m getting a huge number that cannot be correct.
.
KE increases by 18 J when the particle does - 18 J of work against the field, so the KE ends up at 502 J.
KE = 1/2 m v^2 so
v = sqrt( 2 * KE / m) = sqrt( 2 * 502 J / (1.44 kg) ) = etc.
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.
.
.
.
Problem Number 2
What is the force on an electron (charge of magnitude 1.6 * 10^-19 C) in a uniform conducting wire 1.5 meters long, if the ends of the wire are maintained at a potential difference of 11 volts?
.
.F= k Q1Q2/r^2
= 9x10^9 (1.6x10^-19 x 11)/1.5
=1.05x10^-8 N
.
You don't have two point charges here, so Coulomb's Law doesn't apply.
The electric field is
E = 11 volts / (1.5 meters) = 7 volts / meter, approx.., which is also equal to about 7 Newtons / Coulomb.
The charge of an electron is 1.6 * 10^-19 Coulombs, so the force is about
F = q * E = 1.6 * 10^-19 C * 7 N / c = 1.2 * 10^-18 Newtons.
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Problem Number 3
A sunbeam with intensity 6.5 watts/m ^ 2 shines through a circular window with radius .9 meters. The direction of the sunlight is perpendicular to the plane of the window. What is the power flux through the window?
.
.sin(90) x 6.5/9
.= 0.72
I’m not sure this is the correct equation either.
.
Flux for a perpendicular beam is intensity * area.
Here the area is that of a circle with radius .9 meters, so you get
flux = 6.5 watts / m^2 * pi * (.9 m)^2 = etc.
.
.
.
.
.
Problem Number 4
The table below depicts properties of the isotopes of the eight lightest elements. For the element of column 10, is it possible from an energy standpoint for the isotope in the fourth line to emit a neutron and change to the isotope in the third line?
Properties of Selected Particles and Light Nuclei (most common isotope listed first)
particle or atom proton neutron hydrogen helium lithium beryllium boron carbon nitrogen oxygen
atomic number 1 0 1 2 3 4 5 6 7 8
mass (amu) 1.007276 1.008665 1.007825 4.002603 6.015122 9.012182 10.01294 12 14.00307 15.99492
mass (amu) 2.014102 3.016029 7.016004 11.00931 13.00336 15.00011 16.99913
mass (amu) 3.016049 17.99916
electron mass (amu) atomic mass unit
0.000549 1.660502 * 10^-27 kgw
.
.I don’t exactly understand the question or what on earth some of these numbers represent. What are there all of these different masses? What are they for?
.
This problem should not have been on this test; any similar problem would be omitted.
.
.
.
.
.
.
Problem Number 5
An electron (mass 9.11 * 10^-31 kg) with velocity directed to the North passes through a magnetic field of .006 Tesla directed vertically upward, crossed with an electric field of 42000 N/C directed either East or West. The electron passes through undeflected. Is the electric field directed East or West, and how fast is the electron moving? What force does the electron experience from each field?
.
.F= 9x10^9 (0.006 x 9.11x0^-31/42000)
=1.17x10^-27 west
This is probably not the right equation, but I can’t find another one with force in it right now.
.
.
.
The force on a charge moving perpendicular to a magnetic field is F = q v B. The force in an electric field is F = q E.
The particle is undeflected, so the forces are equal and opposite, and the magnitudes of the forces must therefore be equal so
q v B = q E. Solving for v we get
v = E / B.
For a negative charge moving north through an upward field the q v vector is toward the south. By the right-hand rule the magnetic force on the charge will therefore be to the west. The electric field must therefore exert a force to the east.
To exert a force to the east on a negative charge the electric must be to the west.
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Problem Number 6
A generator has a negligible internal resistance. It creates a potential difference of 2.2 volts when cranked at 1 revolution/sec, and the voltage is proportional to the cranking rate. The generator cranked at 3.2 revolutions / sec, and is connected to a series combination of two bulbs, one with a resistance of 2.2 ohms and the other with resistance 29 ohms. How much current flows in each bulb? How much power is dissipated in each?
.
.
.
V=IR
I=V/R
I=3.2/2.2
=1.45 current.
.
.
The voltage is 2.2 volts / (rev/sec) * 3.2 rev/sec = 7 volts, very approximately.
The resistance of the series is 2.2 ohms + 29 ohms = 31.2 ohms, so the current is 7 volts / (31.2 ohms) = .23 amps, approx..
The circuit is series so the same current must go through both bulbs.
power = current * voltage, voltage = current * resistance, so power = current * current * resistance = current^2 * resistance:
P = I^2 * R = (.23 amps)^2 * 31.2 ohms = 1.5 watts, very approximately.
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Problem Number 7
Find an expression for the electric field strength at distance r from the axis of a capacitor consisting of two co-axial cylinders 8 meters long, with diameters 1.19 cm and 1.84 cm, with the inner cylinder carrying charge - 11.1 `microCoulombs and the outer cylinder carrying charge + 11.1 `microCoulombs. You will find three expressions, depending on whether r is less than 1.19 cm, greater than 1.84 cm or between the two.
• Estimate the average electric field between the cylinders, then determine the approximate work done per unit charge to move a charge from the inner cylinder to the outer. (University Physics students find the exact amount of work, using an integral).
• What is the voltage between the cylinders?
• What is the capacitance of this capacitor?
• What would be the approximate magnitude of the effect on the work, voltage and capacitance if the inner cylinder was not changed, but the outer cylinder shrunk in such a way as to be 10 times closer to the inner?
.E= kQ/r^2
This applies to a point charge and won't work here.
E= 9x10^9(11.1)/1.19^2
1.2x10^9
What do you mean by depending on what r is?
V=IR
Is there a different equation that can be used to find voltage possibly? I don’t have any of these variables.
What is the capacitance? I don’t recall going over that. What chapter is it in?
.
.
The total flux of -11.1 microCoulombs of charge is
flux = 4 pi k Q = 4 pi * 9 * 10^9 N m^2 / C^2 * (-11.1 * 10^-6 C) = -1.2 * 10^6 N m^2 / C.
A cylinder of length 8 m and radius r has surface area 2 pi r * 8 m = 16 pi * r meters.
If 1.19 cm < r < 1.84 cm a cylindrical Gaussian surface coaxial with the actual cylinders will enclose the charge of the inner cylinder, so the flux through the Gaussian surface is -1.2 * 10^6 N m^2 / C. Dividing this by the area 16 pi * r meters we find the field
E = (-1.2 * 10^6 N m^2 / C ) / (16 pi * r meters) = -1/r * 2.4 * 10^4 N m / C.
Estimating the average electric field to be the field halfway between the cylinders, we use r = (1.19 cm + 1.84 cm) / 2 = 1.52 cm = .015 m, approx., and get
E_ave = -1 / (.015 m) * 2.4 * 10^4 N m / C = -1.6 * 10^6 N / C, again very approximately.
Voltage is the work per unit charge to move charge from one point to the other. Moving against an average field of -1.6 * 10^6 N / C a test charge Q would experience average force F = Q E = Q * -1.6 * 10^6 N / C. Moving from the inner to the outer cylinder, a distance of .65 cm or .0065 m, the charge would do work `dW = F * `ds = Q * 1.6 * 10^6 N / C * .0065 meters against the field.
So the work per unit charge would be (Q * 1.6 * 10^6 N / C * .0065 meters) / Q = 1 * 10^4 N m / C = 1 * 10^4 J / C, or 1 * 10^4 volts.
The voltage would therefore be about 10^4 volts, or 10 000 volts.
The capacitance would therefore be
C = Q / V = 11.1 * 10^-6 C / (10 000 volts) = 1.1 * 10^-9 C / volt = 1.1 * 10^-9 Farad.
If the outer cylinder was moved 10 times closer the average electric field would increase by about 25%, since r will decrease from about 1.5 m to about 1.2 m. However the distance the test charge has to move through the field becomes 1/10 as great, so the work done will become about 1/10 as great and the voltage about 1/10 as great.
Ignoring the effect of the increased field, the voltage becomes about 1/10 as great, so the capacitance would increase by a factor of about 10.
The 25% increase in average electric field would bring the factor up to about 12 or 13.
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Problem Number 8
Find the magnetic flux through a square wire loop with side .13 meters, in the presence of a magnetic field with strength .0081 Tesla, when the field makes an angle of 49 degrees with a perpendicular to the plane of the loop.
What is a magnetic flux. We haven’t gone over magnetism yet for this test, have we?
"
No but we have gone over flux, so this should be an easy problem.
Flux = field * area when field is perpendicular to the loop. If not, multiply by the cosine of the angle between the field and the perpendicular. In this case we get
flux = .0081 Tesla * (.13 meters)^2 = etc.. The result will be in units of Tesla * m^2.