Question

An electric field given by \(\overrightarrow{E}= 9.19i - 7.91(y^2 + 1.84)j\) pierces the Gaussian cube of the figure, where the cube is 0.244 m on a side. (E is in newtons per coulomb and y is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube?

Answer 1

I know,

\(\Large{\Phi=\int E \cdot dA}\)

Right?

Answer 2

Yes
You can do this the hard way and calculate the flux through each surface, finding \[\iint\vec E\cdot d\vec A\]and find an expression for the dot product at each point, or you can be clever and answer part (e) first

Answer 3

I hate to interrupt, but for whatever reason, our teacher isn't having us use double integration. That being said, teach me master :P
I want the most simplistic way possible.

Answer 4

right, so that would be to calculate part (e) first
tell me, does the fact that the chosen surface affect the answer to part (e) at all?

Answer 5

time out, what is y doing mixed in with i and j ?

Answer 6

My thought exactly. That is how the problem was given, I will put up exactly what is on my screen momentarily.

Answer 7

ok I think I get it... my idea will not work here

Answer 8

Here we are.

Answer 9

ok so we will have to do each side individually
what is the dot product of \(\vec E\) with \(\vec A\) for the top of the cube?

Answer 10

A is just area, which is 0.0595m^2
E is as listed above, and what would we do with dot products? I am horrible with vectors.

Answer 11

which direction is the area vector pointing in?

Answer 12

In the y direction? I assume.

Answer 13

when dealing with vectors we call them \(\hat i,\hat j,\hat k\)
which of those represent the +y direction?

Answer 14

\(\hat j\)

Answer 15

right, so we want the dot product\[\iint \vec E\cdot d\vec A=\vec E\cdot\vec A=[9.19\hat i+7.91(y^2+1.84)\hat j]\cdot0.244^2\hat j\]

Answer 16

Okay, gimme just a sec.

Answer 17

\(0.47093(y^2+1.84)\hat j\)

Is this even remotely correct?

Answer 18

gotta check the calculator (by which I mean google), one sec

Answer 19

yep, that's right :)
and what is y for the top of the cube?

Answer 20

oh, and \(\hat i\cdot\hat i=1\), so you shouldn't have any vectors left in your expression

Answer 21

Would it be .244 m for the y? and lose the \(\hat j\)?

Answer 22

lose the \(\hat i\), and "yes" to your first question

Answer 23

So we put y=.244 into,

\(\large{0.47093(y^2+1.84)}\)

Answer 24

sorry, you were right, lose the \(hat j)\
but you dropped the - sign
and yes, that is the right substitution

Answer 25

Where did the - sign go?

Answer 26

apparently I forgot to put it as well\[\vec E\cdot\vec A=[9.19\hat i-7.91(y^2+1.84)\hat j]\cdot0.244^2\hat j\\=9.19\cdot0.244^2(\hat i\cdot\hat j)-7.91\cdot0.244^2(y^2+1.84)(\hat j\cdot\hat j)\\=9.19\cdot0.244^2(0)-7.91\cdot0.244^2(y^2+1.84)(1)\\=-7.91\cdot0.244^2(y^2+1.84)\]

Answer 27

Well, that is certainly interesting, never even noticed it.

Very cool, and so you just plug in the y value and you get the flux for the top of the cube?

Answer 28

yep :)

Answer 29

Okay, now for the others, would the y just be zero because they don't move in the y direction?

Answer 30

well, at the bottom of the cube y=0, if that is what you mean
but the flux through the bottom is not 0

Answer 31

use the exact same formula, only now y=0, and the area vector is pointing down

Answer 32

That is what I meant, y=0 for the bottom. What would it be for the other sides? And if it points down, it would just be \(-0.244^2 \hat j\) correct?

Answer 33

perfect!
onto the other sides

Answer 34

Okay, I got 0.866511 for the bottom. Is that correct?
And where would we begin with the sides?

Answer 35

same question
1) what is the area vector for the left side of the cube
2) what is \(\vec E\cdot\vec A\) ?
I'll check your answer for the bottom, one sec

Answer 36

answer for the bottom is right, butyou have way too many sig figs

Answer 37

Would the left side be in the z or \(\hat k\) direction?

Answer 38

I know, I need to trim down... It would be .867 for the bottom.

Answer 39

the axis-vector mapping is\[x\to\hat i\\y\to\hat j\\z\to\hat k\]

Answer 40

z is not a direction

Answer 41

but what you would be tempted to call the "z direction" would be the k direction

Answer 42

Okay, along the z-axis is in the \(\hat k\) direction. Is that more correct?

Answer 43

I don't think so; the picture has the z-axis as the vector coming towards us...
they want the left face of the cube. If I point left in the drawing, along which axis am I pointing?

Answer 44

x axis, correct?

Answer 45

correct, in the positive or negative direction?

Answer 46

Negative.

Answer 47

correct, so which unit vector do we use for the direction of \(\vec A\) ?

Answer 48

\(-.244^2 \hat i\)

Right?

Answer 49

right
since all the areas are the same \(0.244^2\) in magnitude, I would just have said "it points in the \(-\hat i\) direction" for short

Answer 50

now just do \(\vec E\cdot\vec A\) again

Answer 51

-0.547 ?

Answer 52

prolly, let me check

Answer 53

yep

Answer 54

Okay, and for the right side, would it just cancel out when finding the net flux?
Jumping a little ahead I know.

Answer 55

yes, your intuition is exactly right

Answer 56

Very cool. And so for the "back" it would be \(\large-\hat z\)

Answer 57

\(-\hat k\), yes

Answer 58

I'm an idiot, of course it is k... okay, so it would be zero?

Answer 59

right, since \(\hat i\cdot\hat k=\hat j\cdot\hat k=0\)

Answer 60

and your \(\vec E\) field has no \(\hat k\) component

Answer 61

Okay, and it would be zero in the positive \(\hat k\) direction as well.

Answer 62

right
@austinL

Answer 63

I find net flux through the cube to be –0.028 V.m

Answer 64

so as you said, the right and left cancel, and the front and back have flux of zero
so calculating the sum of just the flux of top and bottom should give Vincent's result (I'm assuming he got it right, I didn't check)

Answer 65

It is what I got as well :D

Thank you so much! I really appreciate you taking the time to make sure I understand this!

Answer 66

Happy to help, and thanks for participating in actually learning :)

Answer 67

If I have some more questions on electrostatics, could you help? I may hit another roadblock lol

Answer 68

of course, can't promise I'll be here, but feel free to tag me :)
Vincent also knows his stuff (probably better than I do) so hit him up, too.
Good luck!

Answer 69

I double checked by working out the inner charge too, and it confirms your result. :)

Answer 70

Very cool. I will post a new question shortly I am sure. Thanks again guys.