Explanation Given: E is the electric field due to a long, uniformly charged rod of radius R with charge density δ per unit length. Formula used: Gauss’s law: ∬ S E . d S = Total charge enclosed by S ∈ 0 Calculation: Let us consider the cylinder for a length L . Thus, we can write the surface area of the cylinder as: S u r f a c e a r e a = 2 π d L Also, since we know that charge per unit length. Thus, total charge will be δ L . Thus, we can write using the Gauss’s Law: ⇒ ∬ S E ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 18.3, Problem 28E

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$E (d) = \frac{δ}{2 π \in_{o} d}; d > R$

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Chapter 18.3, Problem 27E

Chapter 18.3, Problem 29E

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Chapter 18 Solutions

Calculus - Standalone book

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To show E ( d ) = δ 2 π ∈ o d ; d > R

Solution Summary: The author explains that E is the electric field due to a long, uniformly charged rod of radius R with charge density per unit length.

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To show

$E (d) = \frac{δ}{2 π \in_{o} d}; d > R$

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Chapter 18 Solutions

To show E ( d ) = δ 2 π ∈ o d ; d &gt; R

Solution Summary: The author explains that E is the electric field due to a long, uniformly charged rod of radius R with charge density per unit length.

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To show E(d)=δ2π∈od; d>R

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Chapter 18 Solutions

To show E ( d ) = δ 2 π ∈ o d ; d > R

To show

$E (d) = \frac{δ}{2 π \in_{o} d}; d > R$