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Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter21: Electric Fields
Section: Chapter Questions
Problem 103A
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*I only need answer in the guide question

An isolated charged conducting sphere of radius 14.0 cm creates an electric field of 7.31 ? 104 ?/? at a distance 10.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?



16

Solution:
? =
??? ?2
; ? =
(7.31 ? 104 ?/?)(10.0 ? 10−2 ?)2 (8.99 ? 109 ?⋅ ?2/?2)
= 81.3126 ??
a. ? = ? ?
= 81.3126 ?? 4?(14.0 ? 10−2 ?)2 ? = 0.3301 ??/?2 (answer)

b. ? = 4???? = 4?(8.85 ? 10−12 ?/?)(14.0 ? 10−2 ?) ? = 15.5697 ?? (answer)

Guide Question: What will happen to the electric field between plates of capacitor on the center and near the edges?

Learners-Materials3rd-c X
FA MIL-QUARTER-I-LEARNING-
A MIL-QUARTER-I-HANDOUT
O file:///C:/Users/MOJAL/Downloads/Learners-Materials3rd-quarter.pdf
16
of 23
+
E Fit to page
O Page view
A) Read aloud
2 Add notes
C-4πε,R
Sample Problem:
An isolated charged conducting sphere of radius 14.0 cm creates an electric field of 7.31 x 104 N/c
at a distance 10.0 cm from its center. (a) What is its surface charge density? (b) What is its
capacitance?
15
Solution:
keq
E =
(7.31 x 10* N/C)(10.0 x 10-2 m)²
(8.99 x 109 N · m² /C²)
= 81.3126 nC
81.3126 nc
а.
O =
47(14.0 x 10-2 m):
o = 0.3301 µC/m? (answer)
b. C = 4ne,r = 4n(8.85 x 10-12 F/m)(14.0 x 10-2 m)
C = 15.5697 pF (answer)
Guide Question: What will happen to the electric field between plates of capacitor on the center
and near the edges?
SUPPLEMENTARY WORKSHEET 3
GENERAL PHYSICS 1
Competency with code: (STEM_GP12EMIIlb-13), (STEM_GP12EMIII6-14), (STEM_GP12EM-Ilb -15)
and (STEM_GP12EM-Illc -17).
Lessons: Gauss's Law, Electric Field, Electric Potential and Electric Potential Energy
Objectives:
Transcribed Image Text:Learners-Materials3rd-c X FA MIL-QUARTER-I-LEARNING- A MIL-QUARTER-I-HANDOUT O file:///C:/Users/MOJAL/Downloads/Learners-Materials3rd-quarter.pdf 16 of 23 + E Fit to page O Page view A) Read aloud 2 Add notes C-4πε,R Sample Problem: An isolated charged conducting sphere of radius 14.0 cm creates an electric field of 7.31 x 104 N/c at a distance 10.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance? 15 Solution: keq E = (7.31 x 10* N/C)(10.0 x 10-2 m)² (8.99 x 109 N · m² /C²) = 81.3126 nC 81.3126 nc а. O = 47(14.0 x 10-2 m): o = 0.3301 µC/m? (answer) b. C = 4ne,r = 4n(8.85 x 10-12 F/m)(14.0 x 10-2 m) C = 15.5697 pF (answer) Guide Question: What will happen to the electric field between plates of capacitor on the center and near the edges? SUPPLEMENTARY WORKSHEET 3 GENERAL PHYSICS 1 Competency with code: (STEM_GP12EMIIlb-13), (STEM_GP12EMIII6-14), (STEM_GP12EM-Ilb -15) and (STEM_GP12EM-Illc -17). Lessons: Gauss's Law, Electric Field, Electric Potential and Electric Potential Energy Objectives:
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