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Question: on the second picture why do we use length instead...

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(On the second picture) why do we use length instead of area while calculating Q and (on the third picture) why surface b lies within the confuctor of the outer cylinder, In the question it says that outer cylinder is insulator?
0p390wwtiegg.com/Romework-help/University-Physics-14th-edition-chapter-22-problem-39P-solution-978013396929 E Zookal Study TEXTBOOK SOLUTIONS EXPERT ORA PRACTICE Search home / study / science/physics / calculus based physics/calculus based physics solutions manuals/university physics/ 14th edition University Physics |(14th Edition) Chapter 22, Problem 39P 0 (e Bookmarks)Show all steps: ON An hor Problem The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length A. Calculate the electric field (a) at any point between the cylinders a distance r from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r 0 tor 2c. (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder My T Step-by-step solutiorn Universit Physics Step 1 of 4 A
IEX I BOOR SOLUTIONS EXPERT Q&A PRACTICE NEW! Chapter 22. Problem 39P(Bookmarks)Show all steps: ON a) Let / be the length of the charged cylinder and Q be the charge on the cylinder. The charge enclosed by the cylinder is equal to the linear charge density multiplied by the length. Let us draw a Gaussian surface at a < r < b as cylinder of radius r with coaxial to charged cylinder. The total surface area of this Gaussian surface is A-2rl Here, l is the length of the Gaussian surface and a < r < b . Let E be the magnitude of electric field at a distance r from the axis of cylinder. This electric field causes by the charge which is enclosed by Gaussian surface. Apply Gausss law for Gaussian surface of radius r E(2m)-왜 The charge enclosed by the Gaussian surface is equal to the charge on the charged cylinder which is inside the Gaussian surface. =왜 Substitute al for Q in equation E(2xrl)
Step 4 of 4 d) The surface b lies within the conductor of the outer cylinder. Here, electric field is zero; therefore charge enclosed is also zero. Therefore, On 0 in that region The surface charge encloses the charge λί on the inner conductor. Hence, the charge enclosed must be -2l on the inner surface of the outer conductor. The charge per unit length of the on inner surface of outer cylinder is The outer surface carries no net charge. Therefore, the charge per unit length on the outer surface must be equal to +2 Comment Was this solution helpful? 33 1
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