Mastering Physics Solutions: Mass Spectrometer

Mastering Physics Solutions: Mass Spectrometer

On February 23, 2013, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = u = sqrt((2 * V * q) / m)
Part B = m/q = (R * B0)^2 / 2V

J. J. Thomson is best known for his discoveries about the nature of cathode rays. Another important contribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio of mass m to (positive) charge q of an ion may be accurately determined in a mass spectrometer. In essence, the spectrometer consists of two regions: one that accelerates the ion through a potential V and a second that measures its radius of curvature in a perpendicular magnetic field. The ion begins at potential V and is accelerated toward zero potential. When the particle exits the region with the electric field it will have obtained a speed u.

With what speed u does the ion exit the acceleration region?
After being accelerated, the particle enters a uniform magnetic field of strength B0 and travels in a circle of radius R (determined by observing where it hits on a screen–as shown in the figure). The results of this experiment allow one to find m/q in terms of the experimentally measured quantities such as the particle radius, the magnetic field, and the applied voltage. What is m/q?

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Mastering Physics Solutions: Speed of an Electron in an Electric Field

Mastering Physics Solutions: Speed of an Electron in an Electric Field

On February 20, 2013, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = 8,577,000 m/s Click to use the calculator/solver for this part of the problem

Two stationary positive point charges, charge 1 of magnitude 4.00 nC and charge 2 of magnitude 1.95 nC, are separated by a distance of 57.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

What is the speed v(final) of the electron when it is 10.0 cm from charge 1?

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Mastering Physics Solutions: Visualizing Electric Fields

Mastering Physics Solutions: Visualizing Electric Fields

On February 9, 2013, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = Option D
Part B = Field lines cannot cross each other, The field lines should be parallel because of the sheet’s symmetry
Part C = Option B
Part D = The field lines should be smooth curve, The field lines should always end on negative charges or at infinity
Part E = QA = +7q, QB = -3q

Electric field lines are a tool used to visualize electric fields. A field line is drawn beginning at a positive charge and ending at a negative charge …

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Mastering Physics Solutions: Forces in a Three-Charge System

Mastering Physics Solutions: Forces in a Three-Charge System

On February 8, 2013, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = -2.87 * 10^-5 N Click to use the calculator/solver for this part of the problem

Consider two point charges located on the x axis: one charge, q1 = -11.5 nC, is located at x1 = -1.685 m; the second charge, q2 = 40.0 nC, is at the origin (x = 0.0000).

What is the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1 and q2 at x = -1.120 m?

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Mastering Physics Solutions: Question 15.20

Mastering Physics Solutions: Question 15.20

On January 29, 2012, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = the net number is zero

What can you say about the net number of electric field lines passing through a Gaussian surface located completely within the region between a set of oppositely charged parallel plates?

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Mastering Physics Solutions: Question 15.16

Mastering Physics Solutions: Question 15.16

On January 29, 2012, in Chapter 15: Electric Charge, Forces, and Fields, by Mastering Physics Solutions

Part A = zero

In electrostatic equilibrium, is the electric field just below the surface of a charged conductor the same value as the field just above the surface, zero, dependent on the amount of charge on the conductor, or given by kq/R2?

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