Gauss's Law and electric flux

In summary, the person is trying to solve a problem they have been stuck on for hours, and has found a guide on the website. They have used Gauss's law to evaluate the expression, and have found that it involves a double integral. They have also found that the flux is due to the field in the x and y directions, and that the last term is easy to calculate.
  • #1
rambo5330
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.. 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[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)
 
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  • #2
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.
 
  • #3
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 ..
 
  • #4
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]
 
  • #5
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!
 

1. What is Gauss's Law and how is it related to electric flux?

Gauss's Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. In other words, it describes the relationship between the electric field and the charges that create it.

2. How is Gauss's Law mathematically expressed?

Gauss's Law is expressed as ∫E∙dA = Q/ε₀, where ∫E∙dA represents the electric flux through a closed surface, Q represents the enclosed charge, and ε₀ is the permittivity of free space.

3. What is the significance of Gauss's Law in understanding electric fields?

Gauss's Law is significant because it allows us to calculate the electric field at a point without having to directly measure it. It also helps us understand how electric fields are affected by the distribution of charges in space.

4. Can Gauss's Law be applied to any shape or only to simple geometric shapes?

Gauss's Law can be applied to any shape, as long as the electric field and charge distribution are known. However, it is most commonly used with simple geometric shapes, such as spheres, cylinders, and planes.

5. How does Gauss's Law relate to the concept of electric flux?

Gauss's Law is directly related to the concept of electric flux, as it is used to calculate the electric flux through a closed surface. It also helps us understand the direction and magnitude of the electric field lines passing through a given surface.

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