Cylinder rod charge density (Gauss Law)

[Arman777]

Homework Statement

there's a cylinder rod inner radius a outer raidus b.we want to find electric field between a and b,like point r (or radius r) a<r<b.

Homework Equations

Gauss Law

The Attempt at a Solution

[/B]I am trying to find Q enclosed but something make me confused.I am taking an integral and I found $E=\frac {Q} {2π(b^2-a^2)ε_0}$ is this correct ?

 

[TSny]

Your result for E doesn't look correct.

The problem doesn't give any information about the length of the cylindrical rod or how the charge is distributed inside the rod.

When using Gauss' law, clearly state the geometry of the Gaussian surface and show your steps in arriving at the result.

 

[Arman777]

Its length is infinite and charge distribution is uniform.I found now $E=/frac {pr} {2e_0}$

 

[Arman777]

I was thinking it was wrong too cause it was looking so weird.

 

[Arman777]

If last result is true I ll be happy

 

[haruspex]

Its length is infinite and charge distribution is uniform.I found now $E=/frac {pr} {2e_0}$

That makes no sense. Did you mean $E=\frac {\rho r} {2e_0}$? $E=\frac {\pi r} {2e_0}$? Or...?

 

[Arman777]

That makes no sense. Did you mean $E=\frac {\rho r} {2e_0}$? $E=\frac {\pi r} {2e_0}$? Or...?

oh sorry I typed from my phone . ok İts $E=\frac {\rho r} {2e_0}$

 

[haruspex]

oh sorry I typed from my phone . ok İts $E=\frac {\rho r} {2e_0}$

I think that would right for a=0. Clearly it must depend on a though (e.g. consider r=a).

 

[Arman777]

I ll share the whole question to make things clear.Its confusing like this

 

[Arman777]

Cylindrical Shell: Consider an infinite cylindrical shell with inner radius a, outer radius b and uniform volume charge density ρ. Consider a point at a distance r from the central axis. Compute the electric field at this point
(a) when the point is inside the cavity of the cylinder (r < a),
(b) when the point is within the shell (a < r < b),
(c) and when the point is outside (b < r).
(d) Sketch the E vs r graph. (Here E is the radial component of the electric field.)

 

[Arman777]

 

[Arman777]

I ll share my steps

 

[Arman777]

I changed my approach.

Lets take a length $L$ horizontal so Total Flux will be $E(2πrL)$

$Q_{enc}=\int ρ \, dv$
$dv=π2rLdr$
$ρ$ is constant so
$Q_enc=ρ\int_f^g π2rL \, dr$
What should be $f$ and $g$ ?
$f=a , g=r$ ?

 

[haruspex]

$f=a , g=r$ ?

Yes.

 

[Arman777]

ok thanks.