Electric Field of Infinite Plate: Why Use Cylindrical Gaussian Surface?

[IKonquer]

I'm trying to show that the electric field for an infinitely large plate is

<br /> <br /> E = \frac{\sigma}{2\epsilon_o}<br /> <br />

I was wondering why you can only use a cylindrical gaussian surface and not a cube?

Thanks

 

[collinsmark]

No, you've got the right idea in the first place!

For an infinitely large, flat plate, you do use a cube for the Gaussian surface.

A cylindrical Gaussian surface is used for objects with cylindrical symmetry such as an infinitely long wire, an infinitely long cylinder or infinitely long cylindrical shell.

A spherical Gaussian surface is used for objects with spherical symmetry, such as a point charge, solid sphere, or hollow shell.

A Gaussian "cube" is used for a infinitely large plane (with or without any thickness) because it preserves the symmetry.

 

[SammyS]

You CAN use a cylinder, as long as the round surface is parallel with the field.

After all, a cube IS a cylinder, --- one having a square base and height equal in length to one side of a base.

 

[collinsmark]

You CAN use a cylinder, as long as the round surface is parallel with the field.

You're right! I forgot about that. You can use a cylinder just fine, as long as the flat sides are parallel to the flat, charged plane.

I always use a cube though. It works just as well. Both are fine choices.

 

[IKonquer]

I used the following equations with a cylindrical Gaussian surface:

<br /> <br /> \sigma = \frac{Q}{A}<br /> <br /> \hspace{10 mm}<br /> <br /> \phi = EA cos\theta<br /> <br /> \hspace{10 mm}<br /> <br /> A_{cylinder} = 2\pi rh<br /> <br /> \hspace{10 mm}<br /> <br /> E = \frac{F}{q}<br /> <br />

At the end when solving for E I got

<br /> <br /> E = \frac{\sigma}{2\epsilon_{0}}<br /> <br />

When I am using a cube, what should the area be? Would it be the surface area?

 

[collinsmark]

I used the following equations with a cylindrical Gaussian surface:

<br /> <br /> \sigma = \frac{Q}{A}<br /> <br /> \hspace{10 mm}<br /> <br /> \phi = EA cos\theta<br /> <br /> \hspace{10 mm}<br /> <br /> A_{cylinder} = 2\pi rh<br /> <br /> \hspace{10 mm}<br /> <br /> E = \frac{F}{q}<br /> <br />

Be careful about the area of the cylinder. Your formula is not quite right. [Edit: actually it is right for part of the cylinder. My point is that your equation is not complete. See below.]

In this case you need to break it up into 3 parts.

• The area of the curved surface, that lies parallel to the electric field (it's normal vector is perpendicular to the field -- that turns out to be critically important -- its surface area vector dA of any small part of it is always perpendicular to the field).
• The "top" endcap (normal vector of the area is parallel to the field).
• The "bottom endcap (normal vector of the area is also parallel to the field).
  
  Then you need to take each section and calculate the vector dot product \vec E \cdot \vec A. Don't forget what the vector dot product means. What's the dot product of two parallel vectors? What's the dot product of two perpendicular vectors?
  
  (Alternately, and perhaps better mathematically, you can take the vector dot product before the integration \oint \vec E \cdot \vec{dA} but you'll still need to break up that closed integral into three separate integrals, one for the curved surface, and one for each of the two endcaps. The final result is the same.)

When I am using a cube, what should the area be? Would it be the surface area?

For a cube you need to break up the closed surface into six open surfaces, one for each side (or break up the closed integral into 6 integrals. Either way will work). But play close attention to whether each surface's normal vector is parallel or perpendicular to the field. It makes a big difference when taking the vector dot product.

The area of square with a side of length a is a². You'll need to do this 6 times. The critical part is making note of the direction of the surface vectors.

 

[IKonquer]

1. The area of the curved surface, that lies parallel to the electric field (it's normal vector is perpendicular to the field -- that turns out to be critically important -- its surface area vector dA of any small part of it is always perpendicular to the field).
2. The "top" endcap (normal vector of the area is parallel to the field).
3. The "bottom endcap (normal vector of the area is also parallel to the field).

I understand your points 2 and 3, but I don't quite grasp why the normal vector is perpendicular to the field in point 1. I still see all the normal vectors being parallel to the electric field.

 

[collinsmark]

1. The area of the curved surface, that lies parallel to the electric field (it's normal vector is perpendicular to the field -- that turns out to be critically important -- its surface area vector dA of any small part of it is always perpendicular to the field).
2. The "top" endcap (normal vector of the area is parallel to the field).
3. The "bottom endcap (normal vector of the area is also parallel to the field).

I understand your points 2 and 3, but I don't quite grasp why the normal vector is perpendicular to the field in point 1. I still see all the normal vectors being parallel to the electric field.

Suppose that a charged plane lies along the x-y plane. The normal vector of the charged plane points along +z for positions above the plane and -z for positions below the plane.

Now let's consider a cylindrical Gaussian surface, passing through the charged planes, such that the endcaps are oriented parallel to the charged surface. The normal vectors of each the endcaps will point in the same direction as the plane's normal vectors.

But now consider the curved part of the cylinder. Chop it up into small sections. The normal vector dA of any given section will point somewhere along the x-axis, y-axis, or combination thereof. The actual direction depends upon which section you're working with at a given time, but its normal always points somewhere along the x-y plane (technically parallel to the x-y plane [i.e. perpendicular to the x-y plane's surface normal vector]). Anything that points anywhere along the x or y direction (or combination thereof) is always perpendicular to the z-axis.

 

[IKonquer]

I think I'm almost there. When you say, "But now consider the curved part of the cylinder. Chop it up into small sections," how are you chopping the cylinder up? Is dA supposed to be a circular cross section of the curved part or am I slicing a baguette in half?

 

[collinsmark]

I think I'm almost there. When you say, "But now consider the curved part of the cylinder. Chop it up into small sections," how are you chopping the cylinder up? Is dA supposed to be a circular cross section of the curved part or am I slicing a baguette in half?

Slice the curved area into strips such that the cuts go from positive z to negative z. Each strip will have height (along the z-direction) that was the original height of the cylinder. The width of each strip will be small. But you can now treat each strip as a 2-dimensional flat surface. What is the normal surface vector of each of these flat strips? Of course, the surface normal vector of each of these strips will be in a different direction, depending on the strip in question. But maybe a better question is, "for any given strip, is its surface normal vector parallel or perpendicular to the z-axis?"

 

[IKonquer]

I see. Taking strips parallel to the z-direction will give a strip that will have normal vectors perpendicular to the z-axis and the electric field. But why is it incorrect to take strips dA that are perpendicular to the z-axis?

Thanks for such quick responses!

 

[collinsmark]

I see. Taking strips parallel to the z-direction will give a strip that will have normal vectors perpendicular to the z-axis and the electric field. But why is it incorrect to take strips dA that are perpendicular to the z-axis?

I'm not quite sure I'm following you.

The curved part of the cylinder's area is is the part of the cylindrical that ends up with surface vectors dA being perpendicular to the electric field E

It's the end-caps that have surface vectors parallel to E (by "surface vectors" I'm mean the vector that is normal to the surface -- that's what's implied in Gauss' law). In other words, it is the area of the end-caps that end up being the important part of this problem. They are the only parts of the cylinder that give rise to a non-zero dot product. The surface vector(s) of the curved part of the cylinder gives rise to 0, when taking the dot product. The surface area of the curved part of the cylinder makes no difference to the final result.

In an earlier post you wrote that the area of a cylinder is A_cylinder = 2πrh. But that's the area of the curved part of the cylinder only -- which ignores the area of the end-caps. But the area of the end-caps is what's important, not the curved part.

So what's the area of the end-caps? (And remember, there are two of them. )

 

[IKonquer]

***In an earlier post you wrote that the area of a cylinder is Acylinder = 2πrh. But that's the area of the curved part of the cylinder only -- which ignores the area of the end-caps. But the area of the end-caps is what's important, not the curved part. ***

So would the area just be 2\pi r^{2} ? If so, then I follow the idea that the normal vectors to the caps are parallel to the the electric field.

"The curved part of the cylinder's area is is the part of the cylindrical that ends up with surface vectors dA being perpendicular to the electric field E"

This part still bugs me even though you tried to explain it to me earlier. Isn't dA just extremely small areas? If so, when I take dA to the areas parallel to the xy-plane, shouldn't the electric field be perpendicular to them?

 

[IKonquer]

"shouldn't the electric field be perpendicular to them? " Sorry I used perpendicular instead of parallel, which is what I meant to say.

 

[collinsmark]

"shouldn't the electric field be perpendicular to them? " Sorry I used perpendicular instead of parallel, which is what I meant to say.

Let me elaborate on a couple of points for clarity.

Defining the surface vector:
For a given surface, its surface vector dA points in a direction that is normal (i.e. perpendicular) to the surface. For example, a plane that lies on the x-y plane has a surface vector that points along the z-direction (which is either +z or -z depending on how you define it). When using Gauss' law, the surface vectors always point out of the surface. So for this problem, the surface vector of the top end-cap points up (in the +z direction). The surface vector of the bottom end-cap points down. The surface vectors of the curved part of the cylinder point all point somewhere parallel to the x-y plane. Another way of phrasing that last statement is: the surface vectors of the curved part of the cylinder are all perpendicular to the x-y plane's own surface vector.

Vector dot product.
a · b = ab, if a and b are parallel.
a · b = 0, if a and b are perpendicular.

Gauss's law:

\oint_S \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon_0}

So as you can see from Gauss's law, and given the constraints of this particular problem, we're only interested in the surface area of those parts of the Gaussian surface that have surface vectors parallel to the electric field: the end-caps (the curved part makes no difference).

That's also why using a box as the Gaussian surface is just as easy, if not easier.

 

[IKonquer]

"For a given surface, its surface vector dA points in a direction that normal (i.e. perpendicular) to the surface." - Wow, I completely ignored that fact... But I understand now : )

<br /> dA = 4 \pi r \hspace{10 mm} E = \frac{Qr^{2}}{4\pi \epsilon_{0}}<br /> <br />

Since the area normal vectors of the cap at the top and bottom of the cylinder are parallel to the electric field, I need to multiply by two right? And unfortunately, since I have not covered surface integrals, I don't know how to integrate surface integrals....

 

[collinsmark]

"For a given surface, its surface vector dA points in a direction that normal (i.e. perpendicular) to the surface." - Wow, I completely ignored that fact... But I understand now : )

<br /> dA = 4 \pi r \hspace{10 mm} E = \frac{Qr^{2}}{4\pi \epsilon_{0}}<br /> <br />

Uh, be careful with your dA. It's still not correct.

And as a matter of fact, I wouldn't even express it in terms of differentials. You could if you wanted to. But it's just easier to use the total area of each end-cap, since you already know the surface vector is completely parallel to E everywhere. All you need to do for each end-cap is multiply E times A. [Edit: by that I mean ∫E · dA simplifies to just EA. See below for more detail.]

The area of a flat circle is A = πr².

Since the area normal vectors of the cap at the top and bottom of the cylinder are parallel to the electric field, I need to multiply by two right?

Yes, one for each end-cap.

It's true that each end-cap's surface vector points in the opposite direction from the other (one points up and the other points down). But also note that the electric field above the charged plane points up, and the electric field below the charged plane points down.

And unfortunately, since I have not covered surface integrals, I don't know how to integrate surface integrals....

Surface integrals are not too important for this problem. The important thing to note is you've chosen your Gaussian surface such that the electric field E has a constant magnitude over any section of area -- in this case, the area described by either end-cap is just the area of a circle. That, and E is always parallel to the surface vector of that section of surface. ... But let me step back a bit.

Remember what we discussed earlier about the dot product of parallel vectors,

a · b = ab, if a and b are parallel.

Well, we know for a given end-cap that E is parallel to dA, so

\int_S \vec E \cdot \vec{dA} \rightarrow \int_S EdA

[Note how the vector notation goes away, and we are now only dealing with scalars]
and since we know that E is a constant over the whole area, we can pull it out from under the integral.

= E \int_S dA = EA.

Note that you can't use the above simplifications all the time and in all situations. It's only valid because we've picked our Guassian surface such that its surface vector is parallel to E (for the section we're working with) and such that the magnitude of E is constant over that section of surface.

So now I think you're on your way.
Reformulate Guass's law using all the simplifications we've used so far for this problem (remembering to break it up into three parts -- two end-caps and one curved section)