Gauss's Law for Thin-Walled Cylinders

[exitwound]

Homework Statement

Homework Equations

$$\Phi = \int{d\vec E \cdot d\vec A}$$
$$\Phi = \frac{q_{enc}}{\epsilon_0}$$

The Attempt at a Solution

I'm having a problem with this problem. I don't really understand the givens.

If we're dealing with thin-walled cylinders, we're dealing with Area, not with lengths as if we had a line of charge. Why am I given Q/L as sigma instead of Q\A?

 

[Doc Al]

What matters when applying Gauss's law is the total charge contained within the Gaussian surface. For this problem, what would you choose as your Gaussian surfaces?

 

[exitwound]

Cylinders on the common axis so that i have an electric field penetrating the shell of the cylinders (as opposed to the top and bottom which have zero electric field penetration).

EA=Q/e0 would become
E(2piRL)=Q/e0
E=Q/2piRLe0

Oh at this point, could I substitue in the Q/L into the equation here making it

E=sigma/2piRe0

 

[ideasrule]

Oh at this point, could I substitue in the Q/L into the equation here making it

E=sigma/2piRe0

Yup, you've got it. Just plug in the numbers and you'll be done.

 

[Doc Al]

Cylinders on the common axis so that i have an electric field penetrating the shell of the cylinders (as opposed to the top and bottom which have zero electric field penetration).

EA=Q/e0 would become
E(2piRL)=Q/e0
E=Q/2piRLe0

Good.

Oh at this point, could I substitue in the Q/L into the equation here making it

E=sigma/2piRe0

Q/L would be lambda the charge per unit length, not sigma which is the charge per unit area.

 

[exitwound]

Oh, yes, lambda. My fault.

Thanks!

 

[exitwound]

For the second part of this, I'd create a Guassian cylindrical shell in the same way.

Can I use:

$$E_{inner}=\frac{\lambda_{inner}}{2\pi R \epsilon_0}$$

$$E_{outer}=\frac{\lambda_{outer}}{2\pi R \epsilon_0}$$

and add them together? I know that the electric field created by them will be:


$$E_{inner}=\frac{5E-6}{2\pi(.86)(8.85E-12)}=1.05E-4$$

$$E_{outer}=\frac{-8.5E-6}{2\pi(.86)(8.85E-12)}=-1.78E5$$

E_inner has an electric field pointing outward at +1.05E-4 N/C. And E_outer is producing an electric field pushing the opposite direction of -1.78E5 N/C. Can I say that the total electric field at r=.86 is -1.78E5 N/C?

 

[Doc Al]

For the second part of this, I'd create a Guassian cylindrical shell in the same way.

Can I use:

$$E_{inner}=\frac{\lambda_{inner}}{2\pi R \epsilon_0}$$

$$E_{outer}=\frac{\lambda_{outer}}{2\pi R \epsilon_0}$$

and add them together?

Definitely.

You can even just use Lambda_net = Lambda_inner + Lambda_outer.

I know that the electric field created by them will be:


$$E_{inner}=\frac{5E-6}{2\pi(.86)(8.85E-12)}=1.05E-4$$

$$E_{outer}=\frac{-8.5E-6}{2\pi(.86)(8.85E-12)}=-1.78E5$$

I didn't confirm your arithmetic, but the radius needed is 8.6cm = 0.086m.

You definitely have the right idea to use superposition to find the net field.

 

[exitwound]

Thanks for the .086 error. Problem worked out fine.