Only perpendicular field lines contribute to flux. Why?

Question

The flux due to an electric field is given by $E\cdot A$, which means only the electric field lines that are perpendicular to the area A count (same for the perpendicular component of the field, if it exists), while calculating flux. Is there a way to think of it rather intuitively?

Answer 1

That's just not flux due to an electric field - that's just any flux in general.

Forget about electric fields for now.

Imagine you insert a rectangular hoop into a river, and you want to see how much water flows through the hoop per second - we'll define the amount of water that goes through the hoop per second as the "flux" of the hoop.

If you insert the hoop parallel to the direction of the flowing water, no water will go through the hoop - the water will flow above and below it.

If you insert the hoop perpendicular to the flowing water, then you'll be maximizing the amount of water that goes through the hoop per second.

Now say you insert the hoop perpendicularly to the flowing water, and slowly start to rotate it. As it becomes more and more parallel to the flowing water, less water will flow through the hoop per second.

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How much less?

Well, let's think about it. Say the water is flowing at a velocity $V_w$...

But wait a sec... what does it even mean for the water to be flowing at a velocity $V_w$?!!!!

It means this: if $Vw$ is $10 \frac{meters}{second}$, a water particle will move 10 meters in one second.

But it also means this:

If we were to consider the flowing river as made up of a bunch of rectangular cross-section of water (imagine 2D sheets of water inserted one after the other in the river, moving in the direction of the flowing river), each sheet of water would move 10 meters in one second.

It'll be helpful to think of the river as made up of rectangular sheets. You'll soon see why.

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Let's say the area of a single sheet of water is $A$. And let's say $V_w$ is actually 10 meters per second.

We now want to see how much "space" that sheet of water moves through in one second.

We usually measure the amount of "space" something moves through with volume.

If that sheet of water were to move 10 meters, it would've moved through a space with volume $10 *A$. If it were to move 20 meters, it would've moved through space with volume $20*A$. In both cases, it would've moved through a volume shaped like a rectangular prism (since remember, we're considering a rectangular sheet of water moving through space). The base of the prism would have the dimensions of the sheet of water, and the height would be the distance that the sheet moved through. The volume of that prism is the amount of space the sheet of water moved through.

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Let's go back to the flowing river.

Remember the flowing river is being pictured as being made up of a bunch of 2D sheets of water, one inserted after the other, moving in the direction of the river's flow.

We want to figure out the volume of water that flows past a location in the river in one second. Let's call the location in the river $P$.

Let $t=0$ be the start of that second, and $t=1$ be the end of that second.

At $t=0$ there's a sheet of water at $P$. Let's call this sheet $S_1$. Behind it there's another sheet of water, $S_2$, and behind $S_2$ there's $S_3$, and so on and so on.

After one second, at $t=1$, $S_1$ will have moved through a volume of space equal to $A*10$, since it was moving at 10 meters per second in the direction of the river's flow, and its area was $A$.

However, (and this is the key insight), every sheet of water behind it was moving at the same velocity, and they all followed $S_1$ in the same direction! $S_2$ is still behind $S_1$, and $S_3$ is still behind $S_2$, and so on and so on. The volume of water that $S_1$ moved through is filled up by water - filled up by the sheets that were behind $S_1$.

And every sheet of water that's filling up the volume that $S_1$ moved through MUST'VE PASSED THROUGH $P$!!!! Meaning that the amount of water that passed through $P$ is equal to the volume of water that $S_1$ moved through in one second - $A*10$!!!

So, the ammount of water flowing past $P$ per second is equal to the area of each cross-section of water, multiplied by the water's velocity: $V_w*A$.

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Now, let's go back to the hoop. We insert the hoop at $P$. The "flux" through the hoop will be defined as the amount of water that goes through it per second.

For now, picture the hoop being inserted perpendicularly to the flowing water - perpendicularly to the direction of the river. Still picture the river as being made up of sheets of water.

However, imagine the sheets of water as being bigger than area of the hoop. When the water flows, sections of the water-sheets that pass above and below the hoop won't flow through it, and sections of the water-sheets that flow to the right and left of the hoop won't flow through it. Only the section of the water sheets that fit within the hoop's area will flow through it - all the other sections of the water sheets will flow around the hoop, and not go through it.

So, the volume of water that flows through the hoop per second won't be equals to the area of the water sheets multiplied by their velocity, but the area of the hoop multiplied by the water's velocity!

Water that flows through hoop per second = $V_w * A_{hoop}$

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This is the final step. Let's now tilt the hoop so that it's no longer inserted perpendicularly to the water.

The sheets of water are still flowing in the same direction. However, notice something:

[insert picture]

From the perspective of the sheets, the height of the hoop decreased when we tilted it.

[Insert picture]

The sheets can only move forwards - in the direction of the river's flow. They can't also tilt their velocities so that their new velocities are perpendicular to the hoop - they must move in the same direction they were moving previously.

Now, there is more of the water sheets "above" and "below" the hoop, and less of them that can actually go through the hoop.

[Insert picture here]

If the hoop was entirely parallel to the river's flow, ALL sheets of water would be either above or below the hoop, and none of the sheets would go through it.

[Insert picture here]

And now that the hoop is tilted, only a section of the water sheets will be able to go through the hoop. What will the area of this section be? Here's a hint:

[Insert picture here]

(I'll finish this tonight and add the pictures...I gtg!!!!!)

Answer 2

Electric flux is simply the number of field lines passing though the given area. If it is passing through it then it would have a perpendicular component for sure!

You said "we count only those lines which are passing through and have perpendicular component too" . But we can simply say that "we count only those lines which are passing through"

Since we are taking about number of fields, the flux we obtain by this method would be relative only. Now Electric field, $E$ which tells relative density of field lines changes with changing angle of area vector, we need to change the value of $E$ to get flux. The new density at any angle $\theta$ relative to density $E$, when field line make 0° angle with area vector is $E cos \theta$ now we multiply this by small area vector $A$ and get $\phi = E cos\theta A$.

We can change the meaning of the expression $E cos\theta A$ but it would be mathematically correct. We can even write it as $E • A$. It is mathematically correct but can lead to different point of view of electric flux.

Adding more details:

Part 1: We can say that we are taking density of fields $E cos\theta$ and then multiplying it by area to get number of field lines.
The flux becomes $E cos\theta A$.

Part 2: Or we can say that we are taking component of Area perpendicular to electric field(relative density of field lines) $E$ or component of area vector parallel to $E$. The component area is $A cos\theta$. Since this Area is perpendicular to field lines we can use same relative density $E$ for flux. Flux again becomes $E A cos\theta$.

Taking component of field lines means using $E cos\theta$ which I told you in part 1.

So what we actually calculate for flux is only number of field lines passing through given area and not their component.

Take example:

There two Area given $S_1$ and $S_2$. These two are put in same uniform field at different angles. We set a density of field lines and let say $S_1$ vector is parallel to field lines and $S_2$ makes some angle with field lines. In both cases 50 field lines passes through area. We can say that flux is same in both cases as number of field lines passing is same. We don't need to look at angle or area for this.