# Finding the Flux of a Vector Field

## Homework Statement

A vector field is pointed along the z-axis,
v → = a/(x^2 + y^2)z .

(a) Find the flux of the vector field
through a rectangle in the xy-plane between a < x < b and
c < y < d .

(b) Do the same through a rectangle in the
yz-plane between a < z < b and c < y < d . (Leave your

Φ = EA
E
= F/q

## The Attempt at a Solution

The problem looks similar to this (I used an image of a similar example):

This example here had flux E0⋅A = E0ab

So for this problem, I tried doing the same thing but first finding electric field of vector v: E = F/q = a/(x^2 + y^2)z / q = aq / (x^2 + y^2)z

And then plugging into formula for Φ:
Φ = EA = aq / (x^2 + y^2)z ⋅ xy

I know I messed up somewhere, especially because the question asks to have an integral for an answer but I'm not sure which formula to use or how to use it.

#### Attachments

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kuruman
Homework Helper
Gold Member
The expression to use is not Φ = EA, because the field is not constant over the area. The integral to use is Φ = ∫E⋅dA. Because the field and the directed area are both along the z-axis, you can write Φ = ∫E dA =∫dx ∫ E dy. Do you know how to do double integrals?

Delta2 and Civil_Disobedient
The expression to use is not Φ = EA, because the field is not constant over the area. The integral to use is Φ = ∫E⋅dA. Because the field and the directed area are both along the z-axis, you can write Φ = ∫E dA =∫dx ∫ E dy. Do you know how to do double integrals?

I don't, haven't covered double integrals. Will I be able to do this without double integrals? How would I find electric field in this case? I know area of the rectangle is just length times width, within the x and y bounds. Would those be the bounds of the integral? But there are two sets of boundaries, one for x and one for y

v → = a/(x^2 + y^2)z .

Is that a unit vector z? And is it in the numerator?

Is that a unit vector z? And is it in the numerator?

Yes, and the z isn't in the numerator

Yes, and the z isn't in the numerator

Actually, I don’t know why I asked. It has to be in the numerator. I don’t even know what it would mean if it was in the denominator.

Delta2
kuruman
Homework Helper
Gold Member
I don't, haven't covered double integrals. Will I be able to do this without double integrals? How would I find electric field in this case? I know area of the rectangle is just length times width, within the x and y bounds. Would those be the bounds of the integral? But there are two sets of boundaries, one for x and one for y
Imagine a small rectangle of area ##dA=dxdy## at some point ##(x,y)## and on the area. This rectangle is so small that you can write its contribution to the flux as the product of the field at that point times the small area ##d\Phi=E~dxdy##. In other words, you assume that the field is constant over the area if the area is small enough. Now consider a strip of length ##b## running parallel to the ##y##-axis which means that the value of ##x## is fixed at ##x##. To find the flux through it, you need to add continuously the contributions through all the small rectangles that make it up. You have$$d\Phi{(x)}=dx\int_0^b \frac{a}{x^2+y^2}dy$$The value of this integral is a function of ##x## multiplying ##dx## and represents the flux through the strip of length ##b##. To find the flux through the rectangle of of length ##b## and width ##a##, you need to add, also continuously, the contributions from all such strips that make up the rectangle.
$$\Phi=\int_0^a d\Phi{(x)}dx=\int_0^a\left (\int_0^b \frac{a}{x^2+y^2}dy \right)dx$$ Note that the term between parentheses is a function of ##x## only that you get by doing the integral between them. That, in a nutshell, is how you do a double integral. It's nothing mysterious, just two continuous additions instead of only one.
Actually, I don’t know why I asked. It has to be in the numerator. I don’t even know what it would mean if it was in the denominator.
I fully agree.

Civil_Disobedient, Delta2 and BvU
BvU
Homework Helper
I 'liked' the tutorial approach by @kuruman and won't retract that. However, it turns out this can be interpreted as giving the full answer -- depending on whether the
applies to the whole exercise or only to part b ...

@Civil_Disobedient ?

kuruman
Homework Helper
Gold Member
I 'liked' the tutorial approach by @kuruman and won't retract that. However, it turns out this can be interpreted as giving the full answer -- depending on whether the
applies to the whole exercise or only to part b ...

@Civil_Disobedient ?
I do not believe I gave away the full answer for either part (a) or (b). My "tutorial" is related to the figure posted by OP, which is a rectangle in the xy-plane. That rectangle has one corner at the origin while the one described in part (a) of the question has its equivalent corner at point (a,c). Furthermore, I did not post the result of the double integration which is what part (a) requires. As for part (b), the rectangle described in the question is in the yz plane, it involves a completely different integral and I have said nothing about setting that up. Anyway, thank you @BvU, for continuing to "like" the post. I tried to provide the rationale behind setting up a double integral.

Civil_Disobedient
BvU
Homework Helper
No offence intended...

We haven't heard from @Civil_Disobedient whether b) only or a) and b) have to be answered in the form of an integral ?

Whatever, I'd like to see @Civil_Disobedient provide the outcome for part b) in the form of an evaluated integral as well ... in short: what's the result ?

kuruman
kuruman
Homework Helper
Gold Member
No offence intended...

We haven't heard from @Civil_Disobedient whether b) only or a) and b) have to be answered in the form of an integral ?

Whatever, I'd like to see @Civil_Disobedient provide the outcome for part b) in the form of an evaluated integral as well ... in short: what's the result ?
I agree with all of the above.

Imagine a small rectangle of area ##dA=dxdy## at some point ##(x,y)## and on the area. This rectangle is so small that you can write its contribution to the flux as the product of the field at that point times the small area ##d\Phi=E~dxdy##. In other words, you assume that the field is constant over the area if the area is small enough. Now consider a strip of length ##b## running parallel to the ##y##-axis which means that the value of ##x## is fixed at ##x##. To find the flux through it, you need to add continuously the contributions through all the small rectangles that make it up. You have$$d\Phi{(x)}=dx\int_0^b \frac{a}{x^2+y^2}dy$$The value of this integral is a function of ##x## multiplying ##dx## and represents the flux through the strip of length ##b##. To find the flux through the rectangle of of length ##b## and width ##a##, you need to add, also continuously, the contributions from all such strips that make up the rectangle.
$$\Phi=\int_0^a d\Phi{(x)}dx=\int_0^a\left (\int_0^b \frac{a}{x^2+y^2}dy \right)dx$$ Note that the term between parentheses is a function of ##x## only that you get by doing the integral between them. That, in a nutshell, is how you do a double integral. It's nothing mysterious, just two continuous additions instead of only one.

I fully agree.

The integral here makes sense, but shouldn't the boundaries of each integral correspond to the integrals of the sides of the rectangle? i.e bounds of x axis for part a should be between a and b and the integral for the y axis be between c and d?

The integral form applies to both part a and b I believe

BvU