question 31 asks to find the electric field due to a uniformlycharged straight f
ID: 1762716 • Letter: Q
Question
question 31 asks to find the electric field due to a uniformlycharged straight filament in two ways. A.) through a cylindericalsurface that surrounds the filament and B.) At the the surface ofthe cylinderical surface. My question is what is the differencebetween A. & B conceptually and mathamticaly. For part A one must create a gussian surface that is thesame shape as the cylinder. For part B one uses the cylinder as the gussian surfacemultiplyed by the E field produced by the filiament?? I beleive my confusion because of dimensions of the gussian suface andthe cylinder? question 31 asks to find the electric field due to a uniformlycharged straight filament in two ways. A.) through a cylindericalsurface that surrounds the filament and B.) At the the surface ofthe cylinderical surface. My question is what is the differencebetween A. & B conceptually and mathamticaly. For part A one must create a gussian surface that is thesame shape as the cylinder. For part B one uses the cylinder as the gussian surfacemultiplyed by the E field produced by the filiament?? I beleive my confusion because of dimensions of the gussian suface andthe cylinder?Explanation / Answer
electric field through gaussian surface which is sameshape as cylinder then E = 2ke /r where = Q/L plug values we get = E =5.14*104 N/C radially out ward in the seconed case we caliculate flux through this cylinder E = EA cos where = 0 becaues from the flux defination when =90 there is no flux finally we caliculate flux through entirecylindrecal surface E = EA cos = 5.14*104 N/C2 (0.100m ) (0.020m ) cos00 solving we get E =646Nm2/C the imaginary surface enclosing the position ofcharges is defiined as gausian surface it is selected in accordance our conveince this surface passes through that point at which E is tobe determined in Part A we ca liculate E is at gausian surface. I hope this will help you thank you. solving we get E =646Nm2/C the imaginary surface enclosing the position ofcharges is defiined as gausian surface it is selected in accordance our conveince this surface passes through that point at which E is tobe determined in Part A we ca liculate E is at gausian surface. I hope this will help you thank you.Related Questions
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