Flux Through A Sphere Surrounding A Parallelepiped

[madianned]

Homework Statement

The rectangular parallelepiped with sides $a > b > c$ is filled with charge of constant density $\rho$. A sphere of radius $2a$ is constructed with its center at the origin. Find the flux through the surface of this sphere

Homework Equations

$$ \oint E \cdot da = \frac{q}{\epsilon_0} $$

$$ q = \int \rho dv $$

The Attempt at a Solution

I know that to find the flux I need to find q using the equation above, but for the volume I integrate over, would I need to use the sphere, or the parallelepiped? I'd suspect the parallelepiped, because that's what holds the charge, but then why would they have given me the radius of the sphere surrounding it? I'm probably overthinking it (just confusing myself), but I figured I'd ask regardless.

 

[NFuller]

I'd suspect the parallelepiped, because that's what holds the charge,

Since you are looking for the flux through the sphere, you need the charge contained within that.

Also how is the parallelepiped oriented? Where is it with respect to the sphere?

 

[madianned]

...how is the parallelepiped oriented? Where is it with respect to the sphere?

The parallelepiped has the origin at one corner and edges along the positive directions of the rectangular axes, and the sphere is centered at the origin.

So if I want the charge inside I'd have to use the volume of the parallelepiped as such;
$$ Q_e = \int_v \rho dv = \rho (abc) $$
right?

 

[NFuller]

Yes since the entire parallelepiped is inside the sphere.

 

[madianned]

Okay, awesome! Thank you for your help!