Solution Found!
Answer: It was shown in Example 21.11 (Section 21.5) that
Chapter 22, Problem 4E(choose chapter or problem)
Problem 4E
It was shown in Example 21.11 (Section 21.5) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E=\lambda / 2 \pi \in_{0} r\).Consider an imaginary cylinder with radius \(r=0.250\) m and length \(l=0.400\) m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda=3.00 \mu \mathrm{C} / \mathrm{m}\) m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r=0.500\) m? (c) What is the flux through the cylinder if its length is increased to \(l=0.800\) m?
Equation Transcription:
Text Transcription:
E=lambda/2pie_0r
r=0.250
l=0.400
lambda=3.00uc/m
r=0.500
l=0.800
Questions & Answers
QUESTION:
Problem 4E
It was shown in Example 21.11 (Section 21.5) that the electric field due to an infinite line of charge is perpendicular to the line and has magnitude \(E=\lambda / 2 \pi \in_{0} r\).Consider an imaginary cylinder with radius \(r=0.250\) m and length \(l=0.400\) m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is \(\lambda=3.00 \mu \mathrm{C} / \mathrm{m}\) m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to \(r=0.500\) m? (c) What is the flux through the cylinder if its length is increased to \(l=0.800\) m?
Equation Transcription:
Text Transcription:
E=lambda/2pie_0r
r=0.250
l=0.400
lambda=3.00uc/m
r=0.500
l=0.800
ANSWER:
Solution 4E
Step 1 :
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