Cylinder rod charge density (Gauss Law)

1. Mar 11, 2017

Arman777

1. The problem statement, all variables and given/known data
Theres 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.
2. Relevant equations
Gauss Law
3. The attempt at a solution
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 ?

2. Mar 11, 2017

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.

3. Mar 11, 2017

Arman777

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

4. Mar 11, 2017

Arman777

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

5. Mar 11, 2017

Arman777

If last result is true I ll be happy

6. Mar 11, 2017

haruspex

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

7. Mar 11, 2017

Arman777

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

8. Mar 11, 2017

haruspex

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

9. Mar 11, 2017

Arman777

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

10. Mar 11, 2017

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.)

11. Mar 11, 2017

Arman777

12. Mar 11, 2017

Arman777

I ll share my steps

13. Mar 11, 2017

Arman777

I changed my approach.

Lets take a lenght $L$ 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 $g$ ?
$f=a , g=r$ ?

14. Mar 12, 2017

haruspex

Yes.

15. Mar 12, 2017

ok thanks.