Question

Electrostatics question (really need subdivisions in physics)

Answer 1

For ease of viewing I have an attached image file.

Answer 2

That's a pretty easy one :)
Close the surface by 5 identical squares and apply Gauss law.

Answer 3

I'm not sure that I follow. It said that it is similar to thinking of it as one side of a square. So, by that thought wouldn't it be similar to like finding the top of the last problem?
And the whole, even less numbers thing is horrible.

Answer 4

No, one side of a CUBE.

Answer 5

Have you learnt Gauss law?

Answer 6

\(\Large{\Phi = \frac{q}{\epsilon_{0}}}\)

Right?

Answer 7

Yep!

Answer 8

Okay, forgive my ignorance... but how would we apply this law?

Answer 9

So whenever you have a closed surface, the net flux is easy to express with respect to overall charge within the enclosed volume.

Answer 10

What is the overall flux through the cube?

Answer 11

What is the flux through any of its sides?

Answer 12

\(\Large{\Phi=\frac{+q}{8.85\times10^{-12}}}\)

Answer 13

brb

Answer 14

@Isaiah.Feynman
All-right, but the problem is that E is not uniform on the surface, so working out the integral is pretty difficult.

Answer 15

Vincent, did what I put look okay?

Answer 16

If you mean flux through the cube is \(q/\epsilon_0\) , this is okay.

Answer 17

"But the problem simplifies to E dot A."
Unfortunately, this is a case where it does not simplify so easily.
You have to go round the problem and use Gauss law to avoid long calculations.

Answer 18

How do we apply this?

Answer 19

"How do we apply this?"
I do not understand your question :(
You have already applied Gauss law.

Answer 20

I mean, in terms of finding the answer. How do we work about getting to the answer?

Answer 21

@TuringTest
You have any thoughts?

Answer 22

This is almost the problem I thought the other one was. As Vincent said, to avoid painful calculations here you want to use a conceptual trick.
Think of this: if the charge were surrounded by a sphere, could you tell me the net flux through it?

Answer 23

The six sides of the cube are symmetrical with respect to the charge in the middle.

Answer 24

That is just Gauss' Law, correct. And my problem is, is this just a charge of +e?

Answer 25

I guess we can assume the charge is +e, yeah

Answer 26

That is where I am confused. There are only a few things that we actually know. Charge now, evidently... and \(\epsilon_{0}\)

Answer 27

the answer will be in terms of mostly variables, so don't cry about the numbers
I wouldn't even plug in for the charge or epsilon

Answer 28

"Charge now, evidently... and ϵ0"

Yes, that's all you need. d does not appear in the result.

Answer 29

so once more, in terms of variables alone, what would be the total flux if the charge were surrounded by a sphere?

Answer 30

\(\Large{\Phi=\frac{q}{\epsilon_{0}}}\)

Correct?

Answer 31

correct, now would this be any different if we changed the shape of the gaussian surface?

Answer 32

(so long as it still encloses q completely)

Answer 33

It shouldn't right?

Answer 34

Any arbitrary gaussian surface I believe.

Answer 35

right, we just like to choose a sphere for purposes of making the integral symmetric and easy
now given that you know what the total flux through our hypothetical surface is, lets say that surface is not a sphere, but a cube, with q at the center

as Vincent said: "The six sides of the cube are symmetrical with respect to the charge in the middle."

Hence you know the total flux through the cube, and you have a symmetry about how much flux goes through each side of the cube...

Answer 36

So would you have,

\(\Large{\Phi_{1~side}=(\frac{1}{6})\cdot(\frac{+q}{\epsilon_{0}})}\)

Answer 37

yup

Answer 38

lesson: symmetry makes scary problems simple

Answer 39

Love it :D
That is a huge problem of mine, over complication... I just need to slow down I think.

Answer 40

That problem is a classic, and once you get into actually doing the integrals, you might want to go about doing it the long way and integrating across the entire surface, just to make sure it checks out ;)

Answer 41

Thank you both very much! I appreciate it!
Just need to keep it simple and slow down.

Answer 42

Oh and for the record, I got 3.12E-9 for the answer.