Calculating Electric Flux through a Square in the xy-Plane

[ehabmozart]

Homework Statement

In a region of space there is an electirc field E that is in the z-direction and that has magnitude E=(964 N/(c*m))x. FInd the flux for this field through a square in the xy-plane at z=0 and with side length .350 m. One side of the square is along the + x-axis and another side is along the +y-axis

Homework Equations

Electric Flux = ∫ E. da

The Attempt at a Solution

I\ve take da to be constant as A since the field strikes all the area. Then I integrated E which is in terms of x from 0 to 0.35... I am still getting the wrong answer

 

[collinsmark]

Homework Statement

In a region of space there is an electirc field E that is in the z-direction and that has magnitude E=(964 N/(c*m))x. FInd the flux for this field through a square in the xy-plane at z=0 and with side length .350 m. One side of the square is along the + x-axis and another side is along the +y-axis

Homework Equations

Electric Flux = ∫ E. da

The Attempt at a Solution

I\ve take da to be constant as A since the field strikes all the area.

I'm not sure what you mean by that.

Then I integrated E which is in terms of x from 0 to 0.35... I am still getting the wrong answer

E(x) is a function of x. So you're going to need a dx in there somewhere when you evaluate the integral $\int E(x) \ dx.$ One can't integrate over x without the differential dx being in the integrand. (This sort of relates to the fundamental theorem of calculus.)

[Edit: And one can't just throw a dx in there all willy-nilly either. Rather one must express dA in terms of dx (such that it satisfies the original $\int \vec E \cdot \vec {dA}$)].

Here's a hint if it helps. $dA = dy \ dx.$ Now evaluate

$$\int_y \int_x E(x) \ dy \ dx.$$

[Another edit: In this particular problem, $\vec E$ is always parallel to $\vec{dA}$, both always pointing along the z axis. So in this particular problem, $\vec E \cdot \vec{dA}$ reduces to the simple $E \ dA$, and we don't need to worry about cosines resulting from the dot product. But in more complicated, future problems this won't always be the case.]

 

[ehabmozart]

I understand, but i need an approach without double integral. we haven't used this tool so far. i am sure that there must be an alternative way. Thanks anyway

 

[collinsmark]

I understand, but i need an approach without double integral. we haven't used this tool so far. i am sure that there must be an alternative way. Thanks anyway

Yes, you can do it without the double integral. Just express dA in terms of dx.

In this case, think of dA as being the area of an infinitesimally thin, rectangular strip, with length 0.350 m (along the y-axis), and width dx (along the x-axis).

 

[ehabmozart]

Yesssss! This is exactly the answer. But please can you be more elaborate. I mean I need this in details because I can't get the idea... Thanks a lot!

 

[collinsmark]

Yesssss! This is exactly the answer. But please can you be more elaborate. I mean I need this in details because I can't get the idea... Thanks a lot!

I'll do my best to give you a run-down.

We know that for a uniform electric field passing through a flat area the electric flux $\Phi_E$ is
$$\Phi_E = \vec E \cdot \vec A$$
where the direction of $\vec A$ points normal to the surface.

In general, $\vec E \cdot \vec A$ = $EA \cos \theta$, where $\theta$ is the angle between the surface's normal and the direction of $\vec E$. But since they are always parallel for this particular problem, we're not going to worry about $\theta$ for the rest of this post.

Okay, so in summary, if you have uniform electric field passing directly through a flat area (parallel to the surface's normal, i.e. perpendicular to the surface), the flux is
$$\Phi_E = EA$$
where A is the area of the surface.

But what if E is not uniform and is a function of x? Well, we could take the "average" value of E and multiply that times the area. But that wouldn't be a good approximation in most situations.

Maybe we can do a better approximation. We could divide the area in half, into two smaller areas. We could take the average E in one area and multiply it times the smaller area, and also take the average E of the second area, and multiply that times the smaller area. Then we can add the two fluxes together. That would give us a better approximation.

Better yet we could divide the area into 1000 smaller areas. In each of the smaller areas we take the average E in that particular area, and multiply it by the total area/1000. Then we add all the 1000 fluxes together to get the total flux.

Or we could use calculus. We divide the area into an infinite number of infinitesimally small areas, dA. Then we multiply each area by E at that particular location. And we integrate over all locations of concern. And that's the answer.

For a given narrow strip of area, (0.350 [m])dx, the value of E is essentially uniform within that strip (because the strip is so narrow). So we just simply multiply the value of E within that strip times the area of the strip. Then we just do that an infinite number of times along all the locations of the total area, and add up an infinite amount of resulting fluxes. That's what calculus is about.

(It's really the same thing as finding the area under a curve.)

 

[collinsmark]

Allow me to make a correction to my last post (post #6).

I said, "Well, we could take the 'average' value of E and multiply that times the area." Using the term "average" here is a poor choice of wording. As it turns out, one would need to do the calculus first, in order to find "average" value of E. That sort of defeats the whole point of what we're really trying to do.

What I should have said was "Well, we could take the value of E at an arbitrary location within the area, and multiply that times the area."

So whenever I say "average value of E" in post #6, just replace that with "value of E at an arbitrary location within the segment." That's really more in the spirit of what we want to do.

 

[forager]

disregard.. just figured it out and edited my question