Guass' Law (Conducting Cylinder)

[mrshappy0]

Homework Statement

Two coaxial cylindrical conductors are shown in perspective and cross-section above. The inner cylinder has radius a = 2 cm, length L = 10 m and carries a total charge of Qinner = + 8 nC (1 nC = 10-9 C). The outer cylinder has an inner radius b = 6 cm, outer radius c = 7 cm, length L = 10 m and carries a total charge of Qouter = - 16 nC (1 nC = 10-9 C). What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r = 4 cm from the origin and makes an angle of 30o with the x-axis?

Homework Equations

E∫dA=q_enclosed/ε

The Attempt at a Solution

Okay so I found the charge per unit length (λ) which is L_guassσ. λ=0.8 nC. Then I performed the integral and solved for E to get E=q^enclo/(2*Pi*ε*r*L). I wasn't sure how to get the x component though but figured it must be just E mag * Cos[theta] but this is wrong. Can anyone help me? I hope I showed enough work to get a response. I can double check my units and everything but I am pretty sure I converted everything to the proper units.

 

[vela]

Homework Statement

Two coaxial cylindrical conductors are shown in perspective and cross-section above. The inner cylinder has radius a = 2 cm, length L = 10 m and carries a total charge of Qinner = + 8 nC (1 nC = 10-9 C). The outer cylinder has an inner radius b = 6 cm, outer radius c = 7 cm, length L = 10 m and carries a total charge of Qouter = - 16 nC (1 nC = 10-9 C). What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r = 4 cm from the origin and makes an angle of 30o with the x-axis?

Homework Equations

E∫dA=q_enclosed/ε

The Attempt at a Solution

Okay so I found the charge per unit length (λ) which is L_guassσ. λ=0.8 nC.

If $\lambda$ is charge per unit length, it should have units of C/m.

Then I performed the integral and solved for E to get E=q^enclo/(2*Pi*ε*r*L).

What values did you use in the formula?

I wasn't sure how to get the x component though but figured it must be just E mag * Cos[theta] but this is wrong. Can anyone help me? I hope I showed enough work to get a response. I can double check my units and everything but I am pretty sure I converted everything to the proper units.

Please post the figure the problem is referring to.

 

[mrshappy0]

Opps I meant q_enclosed =0.8nC

r=0.04m
1/(ε*Pi*2)=8.85*10^-12 N*m^2/C^2
L=10m

 

[vela]

1/(ε*Pi*2)=8.85*10^-12 N*m^2/C^2

Is that what you really meant? $\epsilon_0 = 8.85\times10^{-12}~\text{N m}^2\text{/C}^2$ by itself.

 

[mrshappy0]

1/(8.85*10^-12 (Pi)*2)=1.79836*10^10

 

[vela]

Your work looks okay to me. What answer did you get?

 

[mrshappy0]

35.8N/C but this is wrong...

 

[vela]

Did you multiply by cos 30 since you only want the x-component?

 

[mrshappy0]

Yes and still I had the wrong answer.

 

[mrshappy0]

Well I multiplied 35.8 N/C by Cos(30deg)

 

[vela]

What result did you get?

 

[mrshappy0]

31.0028 n/c

 

[vela]

It might just be rounding error or the number of significant figures. I get 31.1 N/C to three sig figs.

 

[mrshappy0]

I actually entered the formula directly into the online homework thing. I just tried 31.1 N/C and still didn't work. I really dislike the way this homework is set up. Waste of time!

 

[gneill]

The OP needs to check his value for linear charge density; it could be that he's using a value that's 10x too small.

 

[vela]

Yup, that's it. I grabbed 0.8 nC from one of your earlier posts, but the original problem said 8 nC.

 

[mrshappy0]

Yeah thanks, I ended up figuring it out that it should be 8 not .8. This is very frusterating because the online homework assignment said that is was 0.8. So I wasted a lot of time due to that.