73 A nonconducting solid sphere has a uni- form volume charge density p. Let 7 be the vector from the center of the sphere to a gen- eral point P within the sphere. (a) Show that the electric field at P is given by E = pr13eg. (Note that the result is independent of the ra- dius of the sphere.) (b) A spherical cavity is hollowed out of the sphere, as shown in Fig. 23- 60. Using superposition concepts, show that the electric field at all points within the cavity is uniform and equal to E = päl3eg, where a is the position vector from the center of the sphere to the center of the cavity. Figure 23-60 Problem 73. to

College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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73 A nonconducting solid sphere has a uni-
form volume charge density p. Let 7 be the
vector from the center of the sphere to a gen-
eral point P within the sphere. (a) Show that
the electric field at P is given by E = pr13eg.
(Note that the result is independent of the ra-
dius of the sphere.) (b) A spherical cavity is
hollowed out of the sphere, as shown in Fig. 23-
60. Using superposition concepts, show that
the electric field at all points within the cavity
is uniform and equal to E = päl3eg, where a is the position vector
from the center of the sphere to the center of the cavity.
Figure 23-60
Problem 73.
to
Transcribed Image Text:73 A nonconducting solid sphere has a uni- form volume charge density p. Let 7 be the vector from the center of the sphere to a gen- eral point P within the sphere. (a) Show that the electric field at P is given by E = pr13eg. (Note that the result is independent of the ra- dius of the sphere.) (b) A spherical cavity is hollowed out of the sphere, as shown in Fig. 23- 60. Using superposition concepts, show that the electric field at all points within the cavity is uniform and equal to E = päl3eg, where a is the position vector from the center of the sphere to the center of the cavity. Figure 23-60 Problem 73. to
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