Electric Field of a cube with a rod

[doggydan42]

Homework Statement

The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude $3 \frac{N}{C}$ and directed outward from the axis of the rod. (a) How much charge per unit length exists on the copper rod? (b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?

Homework Equations

For a cylindrical symmetry: $$\phi = 2\pi rLE$$$$\phi = \frac{q}{\epsilon_0}$$$$\lambda = 2\pi r\epsilon_0 E$$
Where $\lambda$ is the linear charge density, $\phi$ is the flux, and L is the length.

I know I am missing an equation, which is the equation I am trying to find.

The Attempt at a Solution

For the first part, I solved for $\lambda$ to be $3.34\times10^{-12}\frac{C}{m}$. I know that this must be used in the second part, but I am unsure of how to include the cube into the equation.

Thank you in advance

 

[haruspex]

but I am unsure of how to include the cube into the equation.

Are you familiar with "Gaussian surfaces"? If the cube is replaced by a rectangular box still length 5cm parallel to the rod but 10cm in the other two directions, would the flux through it be more, the same, or less?

 

[NFuller]

I know that this must be used in the second part, but I am unsure of how to include the cube into the equation.

Does Gauss's law care what the shape of the Gaussian surface is?

 

[alejandromeira]

Recalls the physical interpretation of Gauss's theorem: flow is the closed charge into the surface divided by epsilon. Can you calculate these charge?

Another thing, when the closed electric charge inside the surface is the same, is important the shape of the surface to calculate the total flow?

 

[doggydan42]

Oh. So the charge enclosed would just be $q_{enc}=\lambda L$, where L is the length of the cube. Right? Since if I understand correctly, it doesn't matter what shape it is, as long as the charge enclosed is the same.

So if I solved this correctly, then:
$$\phi=\frac{q_{enc}}{\epsilon_0}=\frac{\lambda L}{\epsilon_0} = \frac{(3.34\times 10^{-12} \frac{C}{m})(0.05 m)}{8.85\times 10^{-12}\frac{C^2}{Nm^2}}
\\ \phi=1.89\times 10^{-2} \frac{Nm}{c}$$

If not what am I missing? Also, how would external charges affect the flux, since the rod extends past the cube?

Thank you.

 

[haruspex]

how would external charges affect the flux, since the rod extends past the cube?

By symmetry, the field lines are all perpendicular to the axis of the rod, so flux from other parts of the rod do not enter the cube. Also, more generally, a line of flux from any charge external to the cube must enter and exit it an equal number of times, so produces no net flux.