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JasonHathaway

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[itex][/itex]

In fact, there are two problems, and I want to know whether my solutions are right or not.

[itex]E=\frac{\rho_{l}}{2\pi\epsilon b}[/itex] (Electric field of a uniformly charged line)

[itex]\rho_{s}=\frac{Q}{A}[/itex]

[itex]\psi (Flux)=Q (Charge)[/itex]

And because the two line are in the same distance from the point, and they have same density:

[itex]E=2\frac{150×10^{-9}}{2\pi\epsilon 4}=134.8(N/m)[/itex]

## Homework Statement

In fact, there are two problems, and I want to know whether my solutions are right or not.

**1-**Two charged line with density of 15 n c/m along the x and y axes ([itex]x\pm\infty[/itex], [itex]y\pm\infty[/itex]), Find the Electric field at:**(a)**(0,0,4)**(b)**(0,5,4).**2-**A cylinder with radius [itex]\rho=8cm[/itex] and surface charge distribution of [itex]\rho_{s}=5e^{-20|z|} nc/m^{2}[/itex], determine the electrical flux which crosses a part of this cylinder limited by [itex]1<z<5[/itex], [itex]\frac{pi}{6}<\phi<\frac{pi}{2}[/itex]## Homework Equations

[itex]E=\frac{\rho_{l}}{2\pi\epsilon b}[/itex] (Electric field of a uniformly charged line)

[itex]\rho_{s}=\frac{Q}{A}[/itex]

[itex]\psi (Flux)=Q (Charge)[/itex]

## The Attempt at a Solution

**1-****(a)**For the uniformly charged line, I am only interested in the distance between the line and the point, which is 4.And because the two line are in the same distance from the point, and they have same density:

[itex]E=2\frac{150×10^{-9}}{2\pi\epsilon 4}=134.8(N/m)[/itex]

**(b)**The distance is [itex]\sqrt{(4)^{2}+5^{2}}[/itex], and then apply the previous formula?**2-**[itex]\int^\frac{\pi}{2}_\frac{\pi}{6} \int^{0.05}_{0.01}=5e^{-20|z|} z dz d\phi[/itex] ?
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