Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 25, Problem 67PQ

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

  1. a. the plastic sphere and
  2. b. the aluminum spherical shell?
  3. c. What are the charges on the inner and outer surfaces of the aluminum spherical shell?

Chapter 25, Problem 67PQ, A solid plastic sphere of radius R1 = 8.00 cm is concentric with an aluminum spherical shell with

FIGURE P25.67

(a)

Expert Solution
Check Mark
To determine

The net charge on the plastic sphere.

Answer to Problem 67PQ

The net charge on the plastic sphere is 1.46×107C_.

Explanation of Solution

Write the expression for Gauss’s law.

    EdA=qinε0                                                                                                   (I)

Here, qin is the charge enclosed and ε0 is a constant.

Rewrite the above equation in terms of the volume charge density.

    EA=qinε0

Here, A is the area and E is the electric filed.

Substitute 4πr2 for A in the above equation.

    E(4πr2)=qinε0

Here, r is the radius of the cylinder and L is the length of the cylinder.

Conclusion:

Rearrange the above equation to find the charge in the sphere.

    qin=E(4πr2)ε0                                                                                              (II)

Substitute 0.120m for r, 9.10×104N/C for E and 8.85×1012C2/Nm2 for ε0 in the above equation.

    qin=9.10×104N/C(4π(0.120m)2)8.85×1012C2/Nm2=1.46×107C

Therefore, the net charge on the plastic sphere is 1.46×107C_.

(b)

Expert Solution
Check Mark
To determine

The net charge on the aluminum spherical shell.

Answer to Problem 67PQ

The net charge on the aluminum spherical shell is 1.68×107C_.

Explanation of Solution

Write the expression for total charge enclosed.

    qin=E(4πr2)ε0

Conclusion:

Substitute qsphere+qshell for qin, 0.340m for r, +1.70×103N/C for E and 8.85×1012C2/Nm2 for ε0 in the above equation.

    qsphere+qshell=(+1.70×103N/C)(4π(0.340m)2)(8.85×1012C2/Nm2)=2.19×108C

Substitute 1.46×107C for qsphere in the above equation and rearrange it to find the charge on the aluminum shell.

    1.46×107C+qshell=2.19×108Cqshell=2.19×108C(1.46×107C)=1.68×107C

Therefore, the net charge on the aluminum spherical shell is 1.68×107C_.

(c)

Expert Solution
Check Mark
To determine

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 1.46×107C_ and 2.2×108C_ respectively.

Explanation of Solution

Write the expression for charge on the inner surface of the aluminum spherical shell.

    Qinner shell surface=qsphere                                                                                      (III)

Write the expression for charge on the inner surface of the aluminum spherical shell.

    Qouter shell surface=qshellQinner shell surface                                                                      (IV)

Conclusion:

Substitute 1.46×107C for qsphere in equation (III).

    Qinner shell surface=(1.46×107C)=1.46×107C

Substitute 1.46×107C for Qinner shell surface and 1.68×107C for qshell in equation (IV).

    Qouter shell surface=1.68×107C1.46×107C=2.2×108C

Therefore, the charge on the inner and outer surface of the aluminum spherical shell are 1.46×107C_ and 2.2×108C_ respectively.

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Chapter 25 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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