Gauss's Law and a long thin wire

[yang09]

Homework Statement

A long thin wire, hundreds of meters long, carries a uniformly distributed charge of -7.2 microCoulombs per meter of length. Estimate the magnitude and direction of the electric field at points (a) 5.0 m and (b) 1.5m perpendicular from the center of the wire.

Homework Equations

Electric flux = Integral(E*dA)
Electric flux = Charge enclosed(Q) divided by Eo
Therefore, equating the two equations, you get:
Q/Eo = Integral(E*dA)

The Attempt at a Solution

The answer to part a(5.0m) is -2.6 X 10^4 N/C, but I do not know how they got that answer. I equated the two equations and solved for E, but I got 2.6 X 10^7 as an answer, which is wrong.

 

[pgardn]

Homework Statement

A long thin wire, hundreds of meters long, carries a uniformly distributed charge of -7.2 microCoulombs per meter of length. Estimate the magnitude and direction of the electric field at points (a) 5.0 m and (b) 1.5m perpendicular from the center of the wire.

Homework Equations

Electric flux = Integral(E*dA)
Electric flux = Charge enclosed(Q) divided by Eo
Therefore, equating the two equations, you get:
Q/Eo = Integral(E*dA)

The Attempt at a Solution

The answer to part a(5.0m) is -2.6 X 10^4 N/C, but I do not know how they got that answer. I equated the two equations and solved for E, but I got 2.6 X 10^7 as an answer, which is wrong.

So using Gauss what equation did you come up with that describes the E-field around a very long thin wire?

E = ? For example we know the e-field around a point charge is E = KQ/r^2 and of course the equation is using K instead of 1/4pi(espilon o). So E = what around a very long wire? You have the linear charge density and you can make a gaussian surface that will allow you to describe the e field.