Find the electric field everywhere of an infinite uniform line charge with total charge Q. Sol. Using Gauss's Law: DĒ - dA = EA = Since, its the line charge we use the area of a cylinder surrounding the line charge E*2TT L= But all the charged get enclosed by the cylinder area, so qenc = Q %3D Deriving, we get: E = (1/2 TE0 )*( IrL)

Find the electric field everywhere of an infinite uniform line charge with total charge Q. Sol. Using Gauss's Law: DĒ - dA = EA = Since, its the line charge we use the area of a cylinder surrounding the line charge E2TT L= But all the charged get enclosed by the cylinder area, so qenc = Q %3D Deriving, we get: E = (1/2 TE0 )( IrL)

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 49P: Determine if approximate cylindrical symmetry holds for the following situations. State why or why...

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Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we seek electric field at a point 5 cm from the center of the rod. (b) A 10-cm long copper of radius 1 cm is charged with +500 nC of charge and we seek electric field at a point 5 cm from the center of the rod. (c) A 150-cm wooden rod is glued to a 150-cm plastic rod to make a 300 cm long rod, which is then painted with a charged paint so that one obtains a uniform charge density. The radius of each rod is 1 cm, and we seek an electric field at a point that is 4 cm from the center of the rod. (d) Same rod as (c), but we seek electric field at a point that is 500 cm from the center of the rod.
Find the electric field everywhere of an infinite uniform line charge with total charge Q. Sol. Using Gauss's Law: (contour integral E with rightwards harpoon with barb upwards on top) * (stack d A with rightwards harpoon with barb upwards on top) = ___________ /epsilon0 EA = ___________ / epsilon0 Since, its the line charge we use the area of a cylinder surrounding the line charge E*2pi ___________ L= ___________ / epsilon0 But all the charged get enclosed by the cylinder area, so qenc = Q Deriving, we get: E = (1/2(pi*epsilon0))*(___________ /rL) **see picture for the symbols used
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