Diameter Electric Field Side View Front View Diameter = 3.50 m Flectric Field 58 0 N/C

Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter25: Electromagnetic Induction
Section: Chapter Questions
Problem 28A
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Below is a spherical shell (it's hollow but completely enclosed, like a balloon).  There is an electric field that enters the shell on the right hand side as shown and continues through the shell.  There is a charge enclosed by the shell with a charge of (purple dot shown in second image).  The green line outlines the same shell - but it is removed in the second image so that you can see the charge enclosed.

The electric field is perpendicular to the widest point of the balloon horizontally.  This is enough angle information to answer the question.

Diameter = 1.10 m

Electric Field = 51.0 N/C

Charge Enclosed = 1.40 nC

Calculate the electric flux through the surface of the entire closed shell due to the external electric field 51.0 N/C and the enclosed charge 1.40 nC.  Your units should be Nm2/C.

Diameter
Electric Field
Side View
Front View
Diameter = 3.50 m
Electric Field = 58.0 N/C
Transcribed Image Text:Diameter Electric Field Side View Front View Diameter = 3.50 m Electric Field = 58.0 N/C
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