Along straight conducting rod (or wire) carries a linear chargedensity of +2.0 C

Question

Along straight conducting rod (or wire) carries a linear chargedensity of +2.0C/m.This rod is totally enclosed within a thin cylindrical

shell of radius R, which carries a linear charge density of -2.0C/m. A) Describe how you would constructtwo Gaussian surfaces to calculate the electric field: one outsidethe cylindrical shell, and the other between the shell and thewire, to apply Gauss’ Law B) Write the expression for theelectric flux threading each of your Gaussian

surfaces and for the resulting electric field in each region.

Explanation / Answer

For region outside the shell, r >R:      construct a Gaussian surfacewhich is a cylinder concentric with the wire, and with a radius r> R, where R is the given radius of theshell.     r is the variable radius from thewire to the location of our gaussian surface .       The gaussian surface (cylinder) has length L. The charged wire and the charged shell BOTH lie inside ourgaussian surface. NET or total charge enclosed by our gaussian surfaceis   zero, since the two charges of +2 and -2 cancelout. gauss's law is    E (surface area of cylinder)(o ) = total chargeenclosed.     Since total charge = 0, then E = 0 at all points outside the shell, ie for r >R,    E = 0 . For region inside shell where r < R:      Gaussian suface is acylinder concentric with the wire and a radius r where r < R. The total charge enclosed by this gaussian surface is just thecharge on the wire, which is   L = +(2E-6) L . gauss's law is     o E (2 r L ) = charge enclosed = ( 2E-6) L the factor of ( 2 r L ) is the lateral surface area ofthe cylinder for our gaussian surface. The factors of L cancel out,   so   E= (2E-6) / [(2 o )r]    or using the fact that k = 1 / (4 o ), we get : E = k(4E-6) / r .     This isan inverse-square proportion between E and r inside theshell.

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