How Do You Calculate Electric Flux Through a Nonuniform Field?

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To calculate electric flux through a nonuniform electric field described by E = ay^i + bz^j + cx^k, the dot product of the electric field and the differential area element dA is taken, focusing on the z-axis component. The expression simplifies to cxdA, leading to the need for a double integral over the rectangular surface in the xy-plane. The correct approach involves evaluating the double integral as c * ∫(0 to w) x dx * ∫(0 to h) dy, since the limits are constant. The flux contributions from the i and j components are zero, confirming that only the k component is relevant. Ultimately, this method yields the expected result of (1/2 chw^2).
rambo5330
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Can someone please help me work through this problem I've spent over an hour on this trying to figure out what to do.. here's 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%203%20problems.pdf (question #54)


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

= C\int (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)
 
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You have the right answer. To evaluate your expression, use dA = dxdy, so now you have \int\int{x*dxdy} 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.
 
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 ..
 
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 c*\int_{0}^{w}xdx*\int_{0}^{h}dy
 
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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