# Electric Fields of Cylinders and Cylindrical Shells

1. May 22, 2014

### tristanm

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

2. Relevant equations
∫EdA = $\frac{QEnclosed}{εo}$
σ=Q*A
A=2πrL
ρ=$\frac{Qinsulator}{εo(c2-b2)L}$

3. 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 $\frac{Qenclosed}{εo2πrL}$
I then put in the values in meters, μCs and got -7.1901E4N/C

2. E of the insulator is equal to $\frac{Qenclosed}{εo}$ which is equal to $\frac{ρπ(r2-b2)L}{2εor}$
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?

2. May 22, 2014

### haruspex

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

3. May 23, 2014

### tristanm

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

What about the actual physics question?

4. May 25, 2014

### haruspex

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.

5. May 30, 2014

### tristanm

Sorry about replying so late. I did in fact figure it out! Thank you for the help