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The charge within a small volume dv is dq=ρdv. the integral of ρdv over a cylinder of length l is the total charge q=λl within the cylinder. use this fact to to determine the constant ρ0 in terms of λ and r. hint: let dv be a cylindrical shell of length l, radius r, and thickness dr. what is the volume of such a shell?

Respuesta :

Demiurgos
The volume of the shell that you described would be:
[tex]dV=2L\pi r dr[/tex]
Now we can rewrite the given integral:
[tex]\lambda L=\int\rho dV=L\rho\int2\pi r dr \\ \lambda L =L\rho \pi r^2\\ \rho=\frac{\lambda}{\pi r^2}[/tex]
I have attached the picture explaining how we got the formula for the volume.
On the picture, I marked the rectangle. You can of this rectangle as the base, and the height would be the circumference of the cylinder.

Ver imagen Demiurgos

Answer:

Given

dq=density*dv

q=lamda*I

Taking double integration

density=lamda/2*pi*r^2

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