How Is Electric Flux Calculated Through a Nonuniform Field in a Box?

[asz304]

Homework Statement

A cubic cardboard box of side a = 0.440 m is placed so that its edges are parallel to the coordinate axes, as shown in the figure. There is NO net electric charge inside the box, but the space in and around the box is filled with a nonuniform electric field of the following form: E(x,y,z) = Kz j + Ky k, where K= 4.30 N/C.m is a constant. What is the electric flux through the top face of the box? (The top face of the box is the face where z=a. Remember that we define positive flux pointing out of the box.)

http://img834.imageshack.us/content_round.php?page=done&l=img834/2066/prob07acube.gif&via=mupload"


"[URL=[PLAIN]http://img834.imageshack.us/i/prob07acube.gif/[/URL]

Homework Equations

Flux = Electric Field*Area

The Attempt at a Solution

I think I should start getting the integral of an area since E is not constant..but for which area? cube or one square (surface), and how do I do integrals with areas.

Thanks

 

[Abhishekdas]

hi asz304...

Is the answer ka³/2? which is equal to .3662912?

If the answer matches i am going to start helping you out with the sum...So reply fast...

 

[sachin123]

I thought answer would be 0.Flux is dot product of E and Area.they are perpendicular to each other for the top surface.
IF it were for the side face,we would have to divide the side face into slits parallel to z axis and integrate,right?
Is Gauss law applicable for uniform Electric fields?
Because the total flux through the cube here is not equal to 0 right?

 

[asz304]

I thought answer would be 0.Flux is dot product of E and Area.they are perpendicular to each other for the top surface.
IF it were for the side face,we would have to divide the side face into slits parallel to z axis and integrate,right?
Is Gauss law applicable for uniform Electric fields?
Because the total flux through the cube here is not equal to 0 right?

Thats what I answered on my first try, but it turned out to be wrong.


@ Abhishekdas

I tried your answer but it doesn't work. How did you do it anyway?

 

[Piyu]

Pretty sure gauss law is applicable here.
A gaussian(arbitrary or imaginary) surface of the cube will have zero net flux out of all its faces.

My working for this(im just trying here) is integrate to K*y*a dy from 0 to a.

This is because flux is E . d A and the upper face experiences the k component of the field. and dA = dxdy = ady since dx is constant.

 

[SammyS]

Gauss's Law isn't much good for this problem. asz304 only needs to know the flux through the top square, not through all 6 sides of the closed surface (cube).

What is z at the top of the cube?

What is E when z = 0.440 m ?? or z=a if you prefer.

The electric flux through the top is: \int_{Top}\vec{E}(z=a)\cdot d\vec{A}\,,\text{ where } d\vec{A}=\hat{k}\,dx\,dy

 

[asz304]

E(top) = Kz j

Is this right? I tried substituting the values for that, and I tried something different for getting the integral of E(top) but still got the wrong answers for both tries.

 

[Piyu]

Ah my bad, i came across wrongly, i really meant that gauss's law can be used since all u need is an imaginary surface.

asz304, mind showing ur working for the integration and the answer?

maybe we could analyse from there

 

[sachin123]

There is no component of E in the z direction.The flux should be zero right?
Can someone explain how it isn't?

And can someone try to find the net flux through the whole cube?I am getting a non zero value for flux!

 

[Piyu]

Check ur initial equation for E field. take in mind i j k refers to x y z direction respectively. This means that there is no E in the X direction since its basically 0 i.

In the z direction Ky k just means that the e field in the z direction can be described ky

 

[sachin123]

Oh sorry,I didn't see it properly
its ka^3 over 2 then,i guess

 

[asz304]

This is what I did but I still get the wrong answers...

a)\int EA = \int (ka)(a^2) = k*a^3/3 = 1.22*10^-1 Nm^2/C


b)\int EA = \int(ka)(a^2) = k*a^3/2 = 1.83*10^-1 Nm^2/C


What am I missing here?

 

[sachin123]

Where are your limits of integration?variable?
And what are a and b there?

 

[asz304]

EDIT: integral from b to a where b is 0 and a is 0.440 m.

 

[sachin123]

SO what is the answer?
Your 'b' seems right(though the integral expression isn't).
k*(a^3)/2

 

[asz304]

SO what is the answer?
Your 'b' seems right(though the integral expression isn't).
k*(a^3)/2

\int EA = \int Ea^2 = Ea^3/3? = ka^4/3

 

[sachin123]

It is dimensionally wrong.

 

[asz304]

SO what is the answer?
Your 'b' seems right(though the integral expression isn't).
k*(a^3)/2

How is my integral expression in b) wrong?

 

[Abhishekdas]

Hi...asz304 i see you are online so we can work it out right now...

Ok so tell me what did you say ...i mean i did not get what you said by my answer doesn't work...is that supposed to mean that its wrong...?

 

[asz304]

@ Abhishekdas

Yeah, I tried to input the number and the computer said I gave it the wrong answer, with 3 tries left.

 

[Abhishekdas]

a)\int EA = \int (ka)(a^2) = k*a^3/3 = 1.22*10^-1 Nm^2/C


b)\int EA = \int(ka)(a^2) = k*a^3/2 = 1.83*10^-1 Nm^2/C

I did not get what you have done here...Even after putting the limits which you gave in your next post i don't see any what you have taken as the differential element (dl,dy or whatever)...Can you give me a better picture of what you are trying?

 

[asz304]

I did not get what you have done here...Even after putting the limits which you gave in your next post i don't see any what you have taken as the differential element (dl,dy or whatever)...Can you give me a better picture of what you are trying?

Oh. I think I know why my integral expression is wrong, it's because I forgot to put dA, dA = k dx dy. And I know that the only component that contributes the flux of the top surface is z.

 

[Abhishekdas]

Ok...i will tell you what i did...maybe i am making some mistake which we can work out right now...

Ok so...on the face where the flux is required z = a...So
E=Ka j + Ky k.

Now i need to take small elements to cover the area...So the elements I chose are the dy elements...i take x as constant on the face because the field is independent of x.
So taking x=a and taking elements of dy i get an area element equal to dA=x*dy...

Are you cool till this?

 

[Abhishekdas]

hey asz304 ...
i checked out right now that ka^3/2 = .1831456 i forgot to divide it by 2 earlier so maybe its worth a try in your comp right now...but its up to you...

waiting for your reply...

 

[asz304]

I had that answer before, but the computer still tells me that it's the wrong answer.

 

[Abhishekdas]

ok let's see...
anyway did u get what i said in the previous post about taking the area elements?

 

[asz304]

I didn't get this part:

So taking x=a and taking elements of dy i get an area element equal to dA=x*dy...

 

[Abhishekdas]

See ...the electric field is independent of x...So if i form elements of the area vector i can write it taking the length of each infinitesimally small rectangle as x(=a which is a canstant) and width dy...So can i write a single lement as x.dy=a.dy? Check out the diagram I've attached...Over there the length of the rectangle is a and width is dy...

 

[asz304]

k I got it. So if I were to get the flux of the right side of the cube it would be independent from x and z? and if I took the flux in the front of the cube it would be independent from y and z? or would it just be equal to 0 since E(x,y,z) = Kz j + Ky k?

 

[Abhishekdas]

Not exactly...im saying that the field is independent of x simply because the electric field expression has no x term...It doesn't depend on the axes as you put it...ok?

Whatever be the axes or whatever faces you take the field will depend on those parameters or variables which are there in the expression...and in case one of those parameters turn out to be constant for a given area (or face) it can just make out problem smpler because that component which depends on this constant parameter itself becomes constant(not independent) as in our sum in the given face z=a (a constant) so our field in that region became ka j + ky k leaving only the k componenet variable(depending on y) which is the reason we need to integrate...