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Gauss's Law and electric flux

  • Thread starter rambo5330
  • Start date
84
0
Can someone please help me work through this problem I've spent over an hour on this trying to figure out what to do.. heres the question

A nonuniform electric field is given by the expression E = ay^i + bz^j + cx^k,
where a, b, and c are constants. Determine the electric flux through a rectangular
surface in the xy plane, extending from x = 0 to x = w and from y = 0 to y = h.

this question can be viewed better here http://web.uvic.ca/~jalexndr/week 3 problems.pdf (question #54)


i basically use gauss's law and get it down to something like

= C[tex]\int[/tex] (x dA )


my method here was two take the dot product of the electric field and dA which is said to be perpendicular to the surface in the x y plane therefore it will act along the z axis
so this dot product comes out equalling cxdA where c is a constant.... where do i go from here.. the answer is given as (1/2 chw^2)
 
14
0
You have the right answer. To evaluate your expression, use dA = dxdy, so now you have [tex] \int\int{x*dxdy}[/tex] with x going from 0 to w, and y going from 0 to h. If you go through the steps, you will get the same answer.
 
84
0
Oh excellent, so judging by what you wrote to continue past where I left off it involves a double integral? if this is the case I have not learned the double integral yet this semester which makes more sense why I as so stuck ..
 
14
0
well, this case does not have variables x or y in the limits. do the definite integrals separately and just multiple the results together. so it'll be like this [tex]c*\int_{0}^{w}xdx*\int_{0}^{h}dy[/tex]
 
1,132
0
Your surface is in XY plane
flux due to field in i and j derection is 0

only k left, which i assume is easy!!!
 

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