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Chapter 24

BIO120H1 Chapter 24: chap 24 solutions Serway.pdf

Department
Biology
Course Code
BIO120H1
Professor
Jennifer Harris
Chapter
24

Page:
of 18
1
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
24
Gauss’s Law
CHAPTER OUTLINE
24.1 Electric Flux
24.2 Gauss’s Law
24.3 Application of Gauss’s Law to Various Charge Distributions
24.4 Conductors in Electrostatic Equilibrium
* An asterisk indicates a question or problem new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
OQ24.1 (i) Answer (a). The field is cylindrically radial to the filament, and is nowhere
zero at any face of the gaussian surface.
(ii) Answer (b). The flux is zero through the two faces pierced by the filament
because the field is parallel to those surfaces.
OQ24.2 Answer (c). The outer wall of the conducting shell will become polarized to
cancel out the external field. The interior field is the same as before.
OQ24.3 Answer (e). The symmetry of a charge distribution and of its field is the same.
Gauss’s law applies to these charge distributions because (a) has cylindrical
symmetry, (b) has translational symmetry, (c) has spherical symmetry, and (d)
has spherical symmetry.
OQ24.4 (i) Answer (c). Equal amounts of flux pass through each of the six faces of the
cube.
(ii) Answer (b). Move the charge to very close below the center of one face, so
that half the flux passes through that face and half the flux passes through the
other five faces.
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2 Gauss’s Law
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
OQ24.5 Answer (b). The electric flux through a closed surface equals where q is
the total charge contained within the surface:
OQ24.6 (i) Answer (e). The shell becomes polarized.
(ii) Answer a. The net charge on the shell’s inner and outer surfaces is zero.
(iii) Answer (c). The charge has been transferred to the outer surface of the
conductor.
(iv) Answer (c). The charge has been transferred to the outer surface of the
conductor.
(v) Answer (a). The charge has been transferred to the outer surface of the
conductor.
OQ24.7 (i) Answer (c). Because the charge distributions are spherically symmetric,
both spheres create equal fields at exterior points, like particles at the
centers of the spheres.
(ii) Answer (e). The field within the conductor is zero. The field a distance r
from the center of the insulator is proportional to r, so it is 4/5 of its value
at the surface.
OQ24.8 Answer (c). The electric field inside a conductor is zero.
OQ24.9 (a) The ranking is A > B > D > C. Let q represent the charge of the insulating
sphere. The field at A is The field at B is
The field at C is zero. The field at D is
(b) The ranking is B = D > A > C. The flux through the 4-cm sphere is (4/5)3q/
. The flux through the 8-cm sphere and through the 16-cm sphere is q/
because they enclose the same amount of charge. The flux through the
12-cm sphere is 0 because the field is zero inside the conductor.
OQ24.10 (i) Answer (a). The field is perpendicular to the sheet, and is nowhere zero
at any face of the gaussian surface.
(ii) Answer (c). The flux is nonzero through the top and bottom faces because
the field is perpendicular to them, and zero through the other four faces
because the field is parallel to them.
OQ24.11 The ranking is C > A = B > D. The total flux is proportional to the enclosed
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Chapter 24 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
charge: 3Q > Q = Q > 0.
ANSWERS TO CONCEPTUAL QUESTIONS
CQ24.1 (a) If the volume charge density is nonzero, the field cannot be uniform in
magnitude. Consider a gaussian surface in the shape of a rectangular box
with two faces perpendicular to the direction of the field. It encloses some
charge, so the net flux out of the box is nonzero. The field must be
stronger on one side than on the other. The field cannot be uniform in
magnitude.
(b) Now the volume contains no charge. The net flux out of the box is zero.
The flux entering is equal to the flux exiting. The field must be uniform in
magnitude along any line in the direction of the field. It can vary between
points in a plane perpendicular to the field lines.
ANS. FIG. CQ24.1
CQ24.2 The electric flux through a closed surface is proportional to the total charge
contained within the surface: (a) the flux is doubled because the charge is
doubled, (b) the flux remains the same because the charge is the same, (c) the
flux remains the same because the charge is the same, (d) the flux remains the
same because the charge is the same, (e) the flux becomes zero because the
charge inside the surface is zero.
CQ24.3 The net flux through any gaussian surface is zero. We can argue it two ways.
Any surface contains zero charge, so Gauss’s law says the total flux is zero.
The field is uniform, so the field lines entering one side of the closed surface
come out the other side and the net flux is zero.
CQ24.4 Gauss’s law cannot be used to find the electric field at different points on a
surface if the field is not constant over that surface. If the symmetry of an
electric field allows us to say that where E is an
unknown constant on the surface, then we can use Gauss’s law. When electric
field is a general unknown function E(x, y, z), there can be no such
simplification.
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