A long straight conducting rod (or wire) carries a linearcharge density of +2.0C

Question

A long straight conducting rod (or wire) carries a linearcharge density of +2.0C/m. This rod is totallyenclosed within a thin cylindrical shell of radius R, which carriesa linear charge density of -2.0C/m. (A) Describe how you would construct two Gaussian surfaces tocalculate the electric field: one outside the cylindrical shell,and the other between the shell and the wire, to apply Gauss'Law. (B) Write the expression for the electric flux threading eachof your Gaussian surfaces and for the resulting electric field ineach region. Any help is greatly appreciated! A long straight conducting rod (or wire) carries a linearcharge density of +2.0C/m. This rod is totallyenclosed within a thin cylindrical shell of radius R, which carriesa linear charge density of -2.0C/m. (A) Describe how you would construct two Gaussian surfaces tocalculate the electric field: one outside the cylindrical shell,and the other between the shell and the wire, to apply Gauss'Law. (B) Write the expression for the electric flux threading eachof your Gaussian surfaces and for the resulting electric field ineach region. Any help is greatly appreciated!

Explanation / Answer

You would construct imaginary cylindrical surfaces with theappropriate radii Inside the outer cylinder there is a field due only to thewire because the inner would enclose none of the charge on the outer cylinder = Q / o = L Q / 0 = E A = E 2 L r So E = / (2 0r)    inside the outer cylinder The outer cylinder would enclose no net charge so the fieldwould be zero (Incidentally the field outside of a uniformly chargedcylindrical shell is also E = / (2 0r) = E A = E 2 L r So E = / (2 0r)    inside the outer cylinder The outer cylinder would enclose no net charge so the fieldwould be zero (Incidentally the field outside of a uniformly chargedcylindrical shell is also E = / (2 0r)

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