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Finding magnitude of electric field of a cylinder

  • Thread starter ecsx00
  • Start date
  • #1
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Homework Statement


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Homework Equations


E = 1/4([itex]\pi[/itex] [itex]\epsilon_{0}[/itex]) * [itex]\frac{p}{r^2}[/itex]



The Attempt at a Solution


E = 2[itex]\pi[/itex][itex]r_{0}[/itex] [itex]\epsilon_{0}[/itex] = pl/[itex]\epsilon_{0}[/itex]
= [itex]\frac{pr}{2\pi r_{0} \epsilon_{0}}[/itex]

I am going by what I know about Gauss Law and using a similar format for the Electric field equation for a infinite charge line in a cylinder.

I fixed it little by little and I left off at:
E = [itex]\frac{r}{2 \epsilon_{0}}[/itex]
The hint it gives me is that i am missing p but putting p in the numerator or denominator will say it is not dependent on p.
I probably did something wrong in the process or used a wrong equation.
 

Answers and Replies

  • #2
834
2
What you have written is gibberish. Is this a uniformly charged cylinder?

You have to start at Gauss's law:

[tex]\oint E \cdot dA = Q_\text{enc}/\epsilon_0[/tex]

And then invoke cylindrical symmetry to say that

[tex]|E|A = Q_\text{enc}/\epsilon_0[/tex]

What is the area [itex]A[/itex] of the Gaussian surface? What is the charge enclosed by this surface?
 

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