Electric Fields of Cylinders and Cylindrical Shells

AI Thread Summary
The discussion focuses on using Gauss' Law to calculate the electric field at specific radii from the z-axis and determining the surface charge density on a metal cylinder. For a radius of 5.0 cm, the electric field was calculated to be -7.1901E4 N/C, while at 8.0 cm, the field was found to be -3.1176E4 N/C after considering contributions from both the insulator and the metal. There was some confusion regarding the calculation process, particularly in ensuring all terms were included and correctly formatted in LaTeX. The surface charge density was calculated using the formula σ=Q*A, resulting in -2.7527E-5. Overall, the participants engaged in clarifying the calculations and addressing formatting issues.
tristanm
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Homework Statement


1. Use Gauss' Law to calculate the electric field at a radius of 5.0cm from the z-axis
2. Use Gauss' Law to calculate the electric field at a radius of 8.0cm from the z-axis
3. What is the surface charge density σmetal on the outer surface of the metal cylinder?


Homework Equations


∫EdA = \frac{Q<sub>Enclosed</sub>}{ε<sub>o</sub>}
σ=Q*A
A=2πrL
ρ=\frac{Q<sub>insulator</sub>}{ε<sub>o</sub>(c<sup>2</sup>-b<sup>2</sup>)L}

lwA7Yot.png


The Attempt at a Solution


1. Using Gauss' law, I took E and A out of the integral as they are both in the same direction, moved A over to the RHS of the equation, and subbed in 2πrL to give \frac{Q<sub>enclosed</sub>}{ε<sub>o</sub>2πrL}
I then put in the values in meters, μCs and got -7.1901E4N/C

2. E of the insulator is equal to \frac{Q<sub>enclosed</sub>}{ε<sub>o</sub>} which is equal to \frac{ρπ(r<sup>2</sup>-b<sup>2</sup>)L}{2ε<sub>o</sub>r}
which gives the result of 4.0724E4N/C
Adding this value to Emetal I get -3.1176E4N/C

I'm not sure whether or not this is the correct procedure to calculate the electric field inside the shell's material, however for part 1. I know that inside the shell it has no electric field.

3. Is this just a matter of using σ=Q*A where Q is the charge of the metal and A is (2πr2 + 2πrl)?
I did this and got -2.7527E-5


Aside: Why are my [/itex] formatting not working?
 
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tristanm said:
Why are my [/itex] formatting not working?

Because you put non-LaTeX controls (sub, /sub) inside. Use _{} etc.:
\frac{Q_{Enclosed}}{ε_o}
 
haruspex said:
Because you put non-LaTeX controls (sub, /sub) inside. Use _{} etc.:
\frac{Q_{Enclosed}}{ε_o}

That makes sense. I can't seem to edit my post for some reason, but no matter.

What about the actual physics question?
 
tristanm said:
That makes sense. I can't seem to edit my post for some reason, but no matter.

What about the actual physics question?

Sorry for the delay - sporadic internet access.
For 1, the formula looks ok - haven't checked the numbers.
For 2, you seem to be missing some terms in the denominator (like those in your equation for 1). You've also missed out some algebraic steps in getting to the numerical result. If you post all your algebra I'm happy to check that.
 
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Sorry about replying so late. I did in fact figure it out! Thank you for the help
 
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