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

[Abhishekdas]

Anyway what i did after taking the elements is pretty simple...

i have dA and i have E...So now i just got an expression for E.dA integrated it taking limits of y from 0 to a...And doing that i get (ka^3)/2 and i have no clue why its wrong...See if you get anything different...I am dying to know how i am wrong...this seems sooo correct...

 

[Abhishekdas]

I got to go now...think on this and post your replies... I will get back to you tomorrow...

 

[asz304]

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


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

@ 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...

Wow..both of our answers were right. The computer is the only thing that's not doing its job . For the answer, instead of 1.83*10^-1 Nm^2/C or 1.831*10^-1 Nm^2/C, I used 0.183 Nm^2/C and the computer accepted it.

Thanks guys


EDIT: Just to make sure I understand the concept of the gauss' law, is the flux at the bottom of the surface equal to the flux at the top? and if I wanted to find the flux in the front part of the cube, is it just 0? since E(x,y,z) = Kz j + Ky k.

 

[SammyS]

EDIT: Just to make sure I understand the concept of the gauss' law, is the flux at the bottom of the surface equal to the flux at the top? and if I wanted to find the flux in the front part of the cube, is it just 0? since E(x,y,z) = Kz j + Ky k.

Yes, except for the sign. Since the z-component of E does not depend on z, in this case, the flux at the bottom is the negative of the flux at the top, because dA is in the ‒z direction, ( the ‒k direction).

 

[Abhishekdas]

@ Abhishekdas


EDIT: Just to make sure I understand the concept of the gauss' law, is the flux at the bottom of the surface equal to the flux at the top? and if I wanted to find the flux in the front part of the cube, is it just 0? since E(x,y,z) = Kz j + Ky k.

Hi asz304...

Yes the flux in the bottom of the surface is equal in magnitude to the flux at the top but diff in magnitude...And you in the front part the flux is zero because the area vector(along -i) and field vecor are perpendicular...So dot product is zero...

And about our answers matching i really didnt know how i could be wrong...Anyway good that we got it and hope you understood the solution...