Electric Field inside a cylinder

1. Oct 9, 2007

orb123

1. The problem statement, all variables and given/known data

A long cylindrical insulator has a uniform charge density of 0.94 µC/m³, and a radius of 7cm (R). Find the electric field inside the insulator at a adistance of 4cm (r). Answer in units of N/C.

2. Relevant equations

Variables:

q= charge in the gaussian surface
R= Radius of cylinder
r= radius of gaussian surface of cylinder ; r<R
L= Lenght of gaussian surface of cylinder

Equations:

EA=q/ε0
V= (pi)R²L
V'= (pi)r²L
q=pV' volume of the gaussian surface of the cylinder
Q=pV; Q= charge, p=charge density
A=2(pi)RL = Area of cylinder (only the curved surface)

3. The attempt at a solution

Ok, I'm solving the gaussian surface of the cylinder. I know q<Q, and q=pV' = p{(pi)r²L}.
Then, I know p=Q/V by definition, so q=Q{(pi)r²L}/V.

Pluggin it into the first equation:

EA = Q{(pi)r²L}/{V(ε0)}
E{2(pi)RL}=Q{(pi)r²L}/{V(ε0)}

Solving:

E= Qr²/{2V(ε0)R} ; Then replacing V

E= Qr²/{2(pi)R²L(ε0)R} (??)

Why do I still have L there? I was not given a value for that, and it was supposed to be eliminated during solving the equations. And even if L wasn't there, the final solution would not be in units of N/C. What am I doing wrong?

Thanks!

Last edited: Oct 9, 2007
2. Oct 10, 2007

Staff: Mentor

Express the charge within your Gaussian surface as $\rho V = \rho \pi r^2 L$.

3. Oct 10, 2007

orb123

I think I already did that. I replaced V' in q=pV' with (pi)r²L, but I think what I did after that was wrong, but I don't know why.

Thanks for your help!

4. Oct 10, 2007

Staff: Mentor

Then why do I see Qs everywhere?

5. Oct 11, 2007

orb123

Yeah, thanks!

I also realized I was solving the Area wrong. Instead of A=2(pi)RL, should be A=2(pi)rL since I am analyzing the area of the gaussian surface of the cylinder, not the area of the entire cylinder.

Thanks again.

6. Jul 13, 2009

tua82912

Hey Doc Al, this is Aleksey, I'm doing same problem with different values for variables

so: p = Q/V
where r< R

and I get the final answer as exactly pr/2ε0, but when I plug in my answer into the answer post in one of the programs we use in school for HW - it's wrong. What am I doing wrong?

7. Jul 13, 2009

Staff: Mentor

Beats me. That answer looks correct.

8. Jul 13, 2009

tua82912

thanks, at least know on the right track from a Doc. Do you have a masters in Physics or you just like physics in general as an art and mentorship?

9. Jul 13, 2009

tua82912

I got it, my answer was right! Thanks so much for the opinion, i kept writing the formula correctly pr/2ε0 but was plugging into my calc r^2 all the time instead of r.

10. Feb 11, 2010

jhexp

How was the equation Pr/2ε0 derived?

I have a similar problem and the formula appears to be correct, but I'd like to know how you get to it

I know E = Q/ε0 = (PV)/(Aε0) = (P*pi*r^2)/(ε0(2*pi*r^2+2*pi*r*L))

I separated the equation into two and got P*L/(2*ε0)+P*r/(2*ε0). Why is it that only the latter part is the correct equation to use?

By the way, my problem involves a straight and long cylindrical charge cloud. We're given the radius and charge density, but no length.

11. Feb 12, 2010

Staff: Mentor

When calculating the flux through your Gaussian surface, only the curved side of the cylinder counts since the field is radial. The flux through the end pieces is zero since the field is perpendicular to those surfaces, so those areas don't count.