# Cylinder rod charge density (Gauss Law)

Gold Member

## Homework Statement

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.

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 ?

Related Introductory Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
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.

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

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

Gold Member
If last result is true I ll be happy

haruspex
Homework Helper
Gold Member
Its lenght 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....?

Gold Member
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
Homework Helper
Gold Member
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).

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

Gold Member
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.)

Gold Member

Gold Member
I ll share my steps

Gold Member
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## ?

haruspex