28. The problem is stated as:(adsbygoogle = window.adsbygoogle || []).push({});

A long, thin wire carrying 5.6nC/m runs down the center of a long, thin-walled, hollow pipe with radius 1.0cm carrying -4.2nC/m spread uniformly over its surface. (a)Find the electric field 0.50 cm from the wire. (b)Find the electric field 1.5 cm from the wire.

*Ok, I'm not exactly sure how to go at this problem. I'm really confused with gaussian surfaces, and the equations in general.

The following is my attempt at the 0.5cm away from the wire.

Enclose the wire in a guassian cylinder with length [itex]l[/itex].

[tex]

\lambda (q_{wire})=5.6 \times 10^{-9} \frac{C}{m}

[/tex]

Cylinder properties:

[tex]

r=0.0005m

[/tex]

[tex]

dA=2\pi=2\pi(0.0005m)

[/tex]

Now to the electric field:

[tex]

E = \frac{\lambda l}{dA* \epsilon_{0}) l}=\frac{\lambda}{dA \epsilon_{0}}

[/tex]

[tex]

= \frac{5.6 \times 10^{-9} \frac{C}{m}}

{2 \pi (0.0005m)\frac{1}{4 \pi k}} = \frac{(5.6 \times 10^{-9})(C)(4 \pi k)}{2 \pi (0.0005m)m} = \frac{2.24 \times 10^{-8} \pi C k}{0.001 \pi m^{2}} = \frac{2.2 \times 10^{-5} C (9.0 \times 10^{9} N m^{2}))}{m^{2} C^{2}}

[/tex]

[tex]

= 1.98 \times 10^{5} \frac{N}{C}

[/tex]

Is this right? Can I just assume that since I am "enclosing" the surface and not worrying about the ends that the electric field is equal to what I got? Or, do I need to worry about the surface charge of the "long hollow thin walled pipe" ?

What about the electric field 1.5cm away from the wire?

I'm thinking this is wrong, and that I need to worry about the pipe, due to the superposition principal... but ... ahh hell, I'm just confused.

thanks in advance.

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# Homework Help: Gaussian surface, and uniform charge density

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