Concept explainers
A solid plastic sphere of radius R1 = 8.00 cm is concentric with an aluminum spherical shell with inner radius R2 = 14.0 cm and outer radius R3 = 17.0 cm (Fig. P25.67). Electric field measurements are made at two points: At a radial distance of 34.0 cm from the center, the electric field has magnitude 1.70 × 103 N/C and is directed radially outward, and at a radial distance of 12.0 cm from the center, the electric field has magnitude 9.10 × 104 N/C and is directed radially inward. What are the net charges on
- a. the plastic sphere and
- b. the aluminum spherical shell?
- c. What are the charges on the inner and outer surfaces of the aluminum spherical shell?
FIGURE P25.67
(a)
The net charge on the plastic sphere.
Answer to Problem 67PQ
The net charge on the plastic sphere is
Explanation of Solution
Write the expression for Gauss’s law.
Here,
Rewrite the above equation in terms of the volume charge density.
Here,
Substitute
Here,
Conclusion:
Rearrange the above equation to find the charge in the sphere.
Substitute
Therefore, the net charge on the plastic sphere is
(b)
The net charge on the aluminum spherical shell.
Answer to Problem 67PQ
The net charge on the aluminum spherical shell is
Explanation of Solution
Write the expression for total charge enclosed.
Conclusion:
Substitute
Substitute
Therefore, the net charge on the aluminum spherical shell is
(c)
The charge on the inner and outer surface of the aluminum spherical shell.
Answer to Problem 67PQ
The charge on the inner and outer surface of the aluminum spherical shell are
Explanation of Solution
Write the expression for charge on the inner surface of the aluminum spherical shell.
Write the expression for charge on the inner surface of the aluminum spherical shell.
Conclusion:
Substitute
Substitute
Therefore, the charge on the inner and outer surface of the aluminum spherical shell are
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Chapter 25 Solutions
Physics for Scientists and Engineers: Foundations and Connections
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