Consider a wire with linear charge density . Assuming that for symmetry reasons the field is directed at every point orthogonally to the wire axis and depends only on the distance rfrom the wire, we can conclude that the flux of the electric field through the coaxial cylinde r (Gauss surface) is: O O(E) = Ah O O(E) = 27 r h E(r) + 27 r? E(r) O (E) = 27 r h E(r) O (E) = 20 Ar |3D

icon
Related questions
Question
Hello, if possible i want a step by step solution for this question
Consider a wire with linear charge density Assuming that for symmetry reasons the field is directed at every point orthogonally to the
wire axis and depends only on the distance rfrom the wire, we can conclude that the flux of the electric field through the coaxial cylinde
r (Gauss surface) is:
O (E) = Ah
O (E) = 27 rh E(r) + 27 r E(r)
O (E) = 27 r h E(r)
O (E) =
|3|
Ar
Transcribed Image Text:Consider a wire with linear charge density Assuming that for symmetry reasons the field is directed at every point orthogonally to the wire axis and depends only on the distance rfrom the wire, we can conclude that the flux of the electric field through the coaxial cylinde r (Gauss surface) is: O (E) = Ah O (E) = 27 rh E(r) + 27 r E(r) O (E) = 27 r h E(r) O (E) = |3| Ar
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

Blurred answer