Electric field inside and outside concentric cylinders

[mr_coffee]

Hello everyone. I'm so close to getting this problem done but i can't get the last part right!
THe problem is the following:
Two long, charged, thin-walled, concentric cylinders have radii of 3.0 and 6.0 cm. The charge per unit length is 4.9 10- 6 C/m on the inner shell and -8.5 10-6 C/m on the outer shell.

(a) Find the magnitude and direction of the electric field at radial distance r = 4.8 cm from the common central axis. (Take radially outward to be positive.)

I found this by using: E = $$\delta$$/2PIEor;
E = (4.9E-6)/(2PIEo(.048m) = 1835826.8 N/C

(b) Find the magnitude and direction of the electric field at r = 8.5 cm, using the same sign convention.

I can't seem to figure this part out. I tr4eid doing the following:
I added up the E field @ 3.0cm and @ 6.0cm to get 389644.9 but didn't know what do it with it from there. So i tried, just using the same method as above in part (a) but insteed of using .048m i used .085m and i got -1798361 N/C which was wrong. Any help would be great!

 

[cliowa]

What is your knowledge in this field?

It seems very easy to me to just solve this problem by using Gauss' Law. Ever heard of that?

Regards...Cliowa

 

[mr_coffee]

FLux = Q/Eo ? I already tried that and FLux = EA. How would you approach it using Gauss's Law?

 

[Doc Al]

Solve part b exactly as you solved part a. The only change is that you must use the total charge per unit length contained within the radius. (Add the charge from both cylinders.)

 

[mr_coffee]

the problem I'm having is, how do you figure out the charge, when all your given is the Electrical field? Also am I just finding the charge @ 3cm and the charge @ 6cm, then adding them together, then using that charge in the equation -> E = /2PIEor;
But in that equation there is no charge variable.

 

[Doc Al]

the problem I'm having is, how do you figure out the charge, when all your given is the Electrical field?

You are given the charge per unit length on each cylinder, not the electric field; the electric field is what you are asked to find.

Also am I just finding the charge @ 3cm and the charge @ 6cm, then adding them together, then using that charge in the equation -> E = /2PIEor;
But in that equation there is no charge variable.

Yes, add the charges. That equation has no charge because you left it out! (Refer to your first post.)

 

[mr_coffee]

Awesome, thanks agan Doc! worked great!