10- chapter 23 A nonconducting disk of radius R= 2 m has a uniform positive surface charge density = 12 nC/m². Which of the following integral define the x component of the electric field at x=2 m on the axis of the disk (in SI units, = 3.14, k=9x10° N-m²/C²)? Option1 339 To 2rdr (y² +r³j/z Option2 2rdr (4+²y/² 9x 10° 10¹ L7 Option3 678 2rdr (4+R)/2 18 [²3 339 Option4 2rdr 678 m Options 2rdr

10- chapter 23 A nonconducting disk of radius R= 2 m has a uniform positive surface charge density = 12 nC/m². Which of the following integral define the x component of the electric field at x=2 m on the axis of the disk (in SI units, = 3.14, k=9x10° N-m²/C²)? Option1 339 To 2rdr (y² +r³j/z Option2 2rdr (4+²y/² 9x 10° 10¹ L7 Option3 678 2rdr (4+R)/2 18 [²3 339 Option4 2rdr 678 m Options 2rdr

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter23: Electric Fields

Section: Chapter Questions

Problem 23.38P: A uniformly charged disk of radius 35.0 cm carries charge with a density of 7.90 10-3 C/m2....

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A solid insulating sphere of radius R has a nonuniform charge density that varies with r according to the expression = Ar2, where A is a constant and r R is measured from the center of the sphere. (a) Show that the magnitude of the electric field outside (r R) the sphere is E = AR5/50r2. (b) Show that the magnitude of the electric field inside (r R) the sphere is E = Ar3/50. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4r2dr.
Suppose the conducting spherical shell of Figure 15.29 carries a charge of 3.00 nC and that a charge of -2.00 nC is at the center of the sphere. If a = 2.00 m and b = 2.40 m. find the electric field at (a) r = 1.50 m, (b) r = 2.20 m, and (c) r = 2.50 m. (d) What is the charge distribution on the sphere?
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Example 23.3 derives the exact expression for the electric field at a point on the axis of a uniformly charged disk. Consider a disk of radius R = 3.00 cm having a uniformly distributed charge of +5.20 C. (a) Using the result of Example 23.3, compute the electric field at a point on the axis and 3.00 mm from the center. (b) What If? Explain how the answer to part (a) compares with the field computed from the near-field approximation E = /20. (We derived this expression in Example 23.3.) (c) Using the result of Example 23.3, compute the electric field at a point on the axis and 30.0 cm from the center of the disk. (d) What If? Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.20-C charged particle at a distance of 30.0 cm.
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