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anonymous

  • 4 years ago

If you have the p (charge per volume) value for a thick charged insulating cylindrical shell how do you find the linear density? The problem doesn't give the length of the shell but it does have an infinite charged line running through the axis of the shell.

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  1. anonymous
    • 4 years ago
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    patience my friend

  2. TuringTest
    • 4 years ago
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    the volume of a cylindrical shell is \[V=\pi\ell(r_o^2-r_i^2)\]where ro is the outer radius and r is the inner radius. Charge density p would be given by\[p=\frac qV=\frac q{\pi\ell(r_o^2-r_i^2)}\]charge per unit length would then be\[\lambda=\frac q\ell=\pi(r_o^2-r_i^2)\]at least that's what I figure :D

  3. TuringTest
    • 4 years ago
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    *ri is the inner radius

  4. TuringTest
    • 4 years ago
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    forgot the p too...

  5. anonymous
    • 4 years ago
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    I actually already did that and it didn't work. Online homework stuff.*

  6. TuringTest
    • 4 years ago
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    What I meant was: the volume of a cylindrical shell is \[V=\pi\ell(r_o^2-r_i^2)\]where ro is the outer radius and r is the inner radius. Charge density p would be given by\[p=\frac qV=\frac q{\pi\ell(r_o^2-r_i^2)}\]charge per unit length would then be\[\lambda=\frac q\ell=\pi p(r_o^2-r_i^2)\]but if that's wrong, perhaps I'm not understanding the question...

  7. anonymous
    • 4 years ago
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    no, i think you got it. I must be making a mistake somewhere in my calculations. ill doublecheck it.

  8. anonymous
    • 4 years ago
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    no, still not working. My units are right. Here p= -588uC/m^3 r0=0.043m ri=0.026m

  9. anonymous
    • 4 years ago
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    i got -6.8073uC/m

  10. TuringTest
    • 4 years ago
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    what do you mean by a line of charge running through the axis? is there more charge than just that on the surface?

  11. TuringTest
    • 4 years ago
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    *or in the body of the insulator rather

  12. anonymous
    • 4 years ago
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  13. TuringTest
    • 4 years ago
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    so is lambda1 a separate source of charge from that in the body of the shell? if so we need to add it or something

  14. anonymous
    • 4 years ago
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    That dot in the middle is the infinite line.

  15. TuringTest
    • 4 years ago
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    yeah I get that, but does it carry charge of its own?

  16. anonymous
    • 4 years ago
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    Yeah that line in the middle is charged with a lambda1= 7uC/m

  17. TuringTest
    • 4 years ago
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    but you want the total linear charge? should be \[\lambda=\lambda_1+\pi\rho(b^2-a^2)\]I would think...

  18. anonymous
    • 4 years ago
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    I thought only conductors surfaces were affected because their electrons can move while insulators can't

  19. anonymous
    • 4 years ago
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    well lambda for just the shell.

  20. TuringTest
    • 4 years ago
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    I'm confused the problem says the cylinder is an insulator, which means it can have charge inside (hence we have a charge/volume ration) Also lambda1 looks like a separate source of charge from rho. If lambda1 and the line in the middle are imaginary, then I would think it is the first answer I gave If lambda1 is its own source of charge, then I would think the second answer If you are given a number for lambda though, are you sure that doesn't work for linear charge density?

  21. anonymous
    • 4 years ago
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    Now i am confused heres the problem word for word: An infinite line of charge with linear density λ = 7 μC/m is positioned along the axis of a thick insulating cylindrical shell of inner radius a = 2.6 cm and outer radius b = 4.3 cm. The insulating shell is uniformly charged with a volume density of ρ = -588 μC/m3.1)What is λ2, the linear charge density of the insulating shell?

  22. anonymous
    • 4 years ago
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    I don't think I even have a grasp of what lambda, rho really mean.

  23. anonymous
    • 4 years ago
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    yeah the given lambda is only for the middle line and doesn't work for lambda 2

  24. TuringTest
    • 4 years ago
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    rho is charge per unit volume lambda is charge per unit length so I would think that is my earlier answer\[\frac q\ell=\lambda_2=\pi\rho(b^2-a^2)=\pi(-588\times10^{-6})(0.043^2-0.026^2)\]if not that, then maybe include the contribution of the center line of charge\[\lambda_2=\lambda_1+\pi\rho(b^2-a^2)=7\times10^{-6}+\pi(-588\times10^{-6})(0.043^2-0.026^2)\]if not either of those then I'm really not sure :/

  25. anonymous
    • 4 years ago
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    Yeah I have tried them both several times (not sure why because it won't change). Maybe SMARTPHYSICS messed up the problem but damn!

  26. anonymous
    • 4 years ago
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    it's probably to late but It's cross sectional area times the charge / volume i.e. (volume of outside cylinder - volume of inside cylinder)(charge per unit volume) = line charge of the cylindrical shell.

  27. anonymous
    • 4 years ago
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    Thanks aquaheper!

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