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 30 degrees with the x-axis?
Ex =?

Question

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 30 degrees with the x-axis?
Ex =?

anonymous · Answer

I did all the work and got the equation $$E=q \div(2\pi*E _{0}*R*L) $$. Now to find the x compent I multiplied by cos(30) and got 31.1N/C and the answer is wrong still. Any ideas?

anonymous · Answer

[url= http://postimage.org/image/d16pkwrc5/ ][img] http://s18.postimage.org/d16pkwrc5/coaxial_1.jpg [/img][/url]

anonymous · Answer

^is an image of the problem.

anonymous · Answer

The problem doesn't ask for accuracy to any specific digits but rounding to 31.1 or not doesn't make a difference.

anonymous · Answer

Hey turing, I know for a fact that the q enclosed = 0.8nC. What could i have possibly done wrong? I know for a fact that that is the correct formula. I am just baffled.

turingtest · Answer

I'm getting a different answer, but I may have made a unit mistake.
Your formula is right though, yeah...

anonymous · Answer

1/(2Pi*vacuum perm.)=1.79*10^10N*m^2/C^2

R=0.04m
L=10m
q=0.8*10^-9C

turingtest · Answer

$$E=\frac{q_i}{2\pi\epsilon_0rh}\cos\theta=\frac{8\times10^{-9}}{2\pi(8.85\times10^{-12})(0.04)(10)}\cos(30^{\circ})\approx311$$why do you have 0.8nC ? 
I thought it was 8nC.

anonymous · Answer

hmm well on the online hw problem it walks you through this one and I got a calculated 0.8nC which it siad was correct.

turingtest · Answer

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)
^is what you wrote above

anonymous · Answer

but clearly that was the problem. cause 311 is the answer. How did you get 8nC?

turingtest · Answer

from you...

anonymous · Answer

hmm wierd. i meant to type 0.8nC. I looked over the problem and the computed value is 0.8nC

turingtest · Answer

lol, I don't know what to tell you about that :/

anonymous · Answer

hah thanks i guess