College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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The figure is a section of a conducting rod of radius R₁ = 1.30 mm and length L = 12.80 m inside a thin-walled coaxial conducting
cylindrical shell of radius R₂ = 11.2R₁ and the (same) length L. The net charge on the rod is Q₁ = +3.40 × 10-12 C; that on the shell is Q2 =
-2.03Q₁. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r = 2.09R₂?
What are (c) E and (d) the direction at r = 5.24R₁? What is the charge on the (e) interior and (f) exterior surface of the shell?
(a) Number
(b)
(c) Number
(d)
(e) Number
(f) Number
Units
Units
Units
Units
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Transcribed Image Text:The figure is a section of a conducting rod of radius R₁ = 1.30 mm and length L = 12.80 m inside a thin-walled coaxial conducting cylindrical shell of radius R₂ = 11.2R₁ and the (same) length L. The net charge on the rod is Q₁ = +3.40 × 10-12 C; that on the shell is Q2 = -2.03Q₁. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r = 2.09R₂? What are (c) E and (d) the direction at r = 5.24R₁? What is the charge on the (e) interior and (f) exterior surface of the shell? (a) Number (b) (c) Number (d) (e) Number (f) Number Units Units Units Units
Expert Solution
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Step 1: Gauss 's law

“Since you have posted a question with multiple sub parts, we will provide the solution only to the first three sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately.”


Gauss's law states that the electric flux through a closed surface equals the total charge enclosed by the surface divided by the permittivity of free space (straight epsilon ₀).

Using Gauss's law:

The electric field is radially outward and constant on the Gaussian surface. We take the Gaussian surface as a cylinder of length L, coaxial with the given cylinders, and of larger radius r than either.

straight E left parenthesis 2 straight pi rL right parenthesis equals straight Q subscript enc over straight epsilon subscript 0
straight E equals fraction numerator straight Q subscript enc over denominator left parenthesis 2 straight pi straight epsilon subscript 0 rL right parenthesis end fraction

Given data:

  • straight R subscript 1 equals 1.30 space mm equals 1.3 cross times 10 to the power of negative 3 end exponent space straight m
  • straight L equals 12.80 space straight m
  • straight R subscript 2 equals 11.2 straight R subscript 1 equals 11.2 cross times 1.3 cross times 10 to the power of negative 3 end exponent straight m equals 1.456 cross times 10 to the power of negative 2 end exponent straight m
  • straight Q subscript 1 equals plus 3.40 cross times 10 to the power of negative 12 end exponent straight C
  • straight Q subscript 2 equals negative 2.03 straight Q subscript 1 equals negative 2.03 cross times left parenthesis plus 3.40 cross times 10 to the power of negative 12 end exponent straight C right parenthesis equals negative 6.902 cross times 10 to the power of negative 12 end exponent straight C
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