Homework Help: Total flux through small cube

1. Sep 11, 2007

physicsstooge

I have the hw problem below with which I'm totally stumped. Can anyone out there help me out?

A small cube of volume 9.0 cm^3 is 0.30 cm from a metal sphere that has charge 3.00 coulombs. If the cube is empty, what is the total flux through it?

2. Sep 11, 2007

Staff: Mentor

Consider Gauss's law.

3. Sep 11, 2007

Staff: Mentor

By "total flux", do you mean "net flux"? Like how much is leaving the cubical volume, versus how much is entering it? What do you think the answer is?

4. Sep 11, 2007

physicsstooge

"Consider Gauss's law."

I suppose the setup in this particular case that is confusing me. I'm probably making it harder than what it needs to be. I instinctively want to say it's the E * the Area of the cube, but that's not right.

5. Sep 11, 2007

Staff: Mentor

You can calculate the flux through each one of the 6 sides of the cube that way. Did you re-check the question to see if it is talking about the net flux? What do they mean by "total" flux?

6. Sep 11, 2007

Staff: Mentor

That's pretty close. More precisely, the net flux through the cube is the sum of $\vec{E}\cdot\vec{A}$ over the entire surface of the cube. What does Gauss's law tell you about the net flux through some volume?

7. Sep 11, 2007

physicsstooge

"What does Gauss's law tell you about the net flux through some volume?"

Sum should be zero. I knew I was making it harder than it was. Thanks guys

8. Sep 12, 2007

Staff: Mentor

Right! As long as the cube is empty (no charge inside), the net flux must be zero.

9. Sep 14, 2007

dynamicsolo

I'll just mention that some version of this problem is a favorite (often multiple-choice) exam question *precisely* because students may think that the external field is somehow important. The configuration of that field is often chosen to make calculating the flux through each surface of the cube very complicated. Thus, many students in desperation will select a choice thinking that there's supposed to be some clever way to do the flux integral, *rather* than realizing that they just want to apply Gauss' Law.

Just as a way to "psyche-out" such questions (related to useful approaches in physics problem solving generally), notice that nothing is said about the *orientation* of this cube in the external field of the charged sphere. (There's nothing in the statement of the problem that allows you to *assume* any symmetry.) How could these details *not* matter in finding the net flux? Only if the answer were the one where the external field didn't matter: zero.

Last edited: Sep 14, 2007