## 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.

1. anonymous

patience my friend

2. TuringTest

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. TuringTest

forgot the p too...

5. anonymous

I actually already did that and it didn't work. Online homework stuff.*

6. TuringTest

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

no, i think you got it. I must be making a mistake somewhere in my calculations. ill doublecheck it.

8. anonymous

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

9. anonymous

i got -6.8073uC/m

10. TuringTest

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

*or in the body of the insulator rather

12. anonymous

13. TuringTest

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

That dot in the middle is the infinite line.

15. TuringTest

yeah I get that, but does it carry charge of its own?

16. anonymous

Yeah that line in the middle is charged with a lambda1= 7uC/m

17. TuringTest

but you want the total linear charge? should be $\lambda=\lambda_1+\pi\rho(b^2-a^2)$I would think...

18. anonymous

I thought only conductors surfaces were affected because their electrons can move while insulators can't

19. anonymous

well lambda for just the shell.

20. TuringTest

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

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

I don't think I even have a grasp of what lambda, rho really mean.

23. anonymous

yeah the given lambda is only for the middle line and doesn't work for lambda 2

24. TuringTest

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

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

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

Thanks aquaheper!