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Field of a uniformly charged disk

by ehabmozart
Tags: charged, disk, field, uniformly
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ehabmozart
#1
Feb19-13, 09:37 AM
P: 194
I have a question about the electric field of a uniformly charged disk with radius a. I'll move point by point till i reach the part i really can't get. First of all, the surface area is made of infinite number of RINGS! So, we basically integrate the charge of the rings from r="0" to r="a". Since we considered a ring to be made of infinite points, we said that a point carries a charge dq of the whole Q. dq= Lambda ds. By analogy, the Q of one ring in the surface is a fraction of the big Q of the surface area which is equal dq= sigma * dA. (I assumed this part by analogy and not by understanding the concept) .. Anyway, it is given that dA itself is 2pi r dr. How can we get this?? I mean, mathematically A= pi a^2.. So dA= 2pi a da. This is from the mathematical side. From the physical side, i really can't get the point the rings with thickness dr. I mean from where did this dr come. I was told that if you get a regtangle with a width dr, and wrap it, you will get a ring with some thickness.. I really can't visualize this. This is my whole problem which is how to get dA which I originally can't know how did we reach to dA itself.

Please clarify and i would be thankful to whoever gives me a hand to the full argument.
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16180339887
#2
Feb19-13, 12:09 PM
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P: 38
Quote Quote by ehabmozart View Post
Anyway, it is given that dA itself is 2pi r dr. How can we get this?? I mean, mathematically A= pi a^2.. So dA= 2pi a da. This is from the mathematical side. From the physical side, i really can't get the point the rings with thickness dr.
If you were integrating just a line of charge from [itex]0[/itex] to [itex]a[/itex] then you are adding up infinitesimal lengths [itex]dr[/itex]. As you said earlier in your description, a disc of charge is like many rings, with infinitesimal width [itex]dr[/itex], stacked up from [itex]0[/itex] to [itex]a[/itex]. Thea area of such a ring would just be the perimeter of it, [itex]2\pi r[/itex] multiplied by the infinitesimal width [itex]dr[/itex], and so to get the total area, we sum them all up.
ehabmozart
#3
Feb19-13, 12:25 PM
P: 194
From where did we add the width dr. ??

jtbell
#4
Feb19-13, 01:21 PM
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Field of a uniformly charged disk

Imagine a single ring with radius 2πr and width dr. Cut it somewhere and straighten it out so you now have a long thin rectangle with length 2πr and width dr. What is the area of that rectangle?
ehabmozart
#5
Feb19-13, 02:02 PM
P: 194
That's fine.. I mean from where did we arise with the width if we were originally dealing with infinite number of rings which we integrated before without using thickness..
jtbell
#6
Feb19-13, 02:55 PM
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Quote Quote by ehabmozart View Post
we were originally dealing with infinite number of rings which we integrated before without using thickness..
I don't understand what you mean here. Are you saying you did an integral that didn't have dr in it?
ehabmozart
#7
Feb19-13, 03:59 PM
P: 194
Exactly... When we did the field due to a ring, we didn't consider its thickness.. However, when we dealt with the disk, we included thickness! Why?
jtbell
#8
Feb19-13, 05:40 PM
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Ah, you did the field due to a ring all by itself... now I see. But when you did it then, I bet you assumed the ring has a given linear charge density (often called λ), in coulombs/meter. Then the total charge is λ times the "length" (circumference) of the ring.

When you do a disk, you usually assume a given surface charge density (often called σ) in coulombs/meter2. Then the total charge of the disk is σ times the area of the disk, and the total charge on a ring that is part of the disk is σ times the area of the ring.
cwilkins
#9
Feb20-13, 03:53 AM
P: 25
From a physical point of view it does make sense to multiply by [itex]dr[/itex]. You are adding rings of a small thickness [itex]dr[/itex] to get the whole area, and subsequently the whole charge on the disk.

As to how you can find the correct [itex]dA[/itex] in general, there is a surefire way to produce it for any common surface. Brace yourself, it might look daunting and laden with unholy quantities, but I assure you that if you understand this it will be forever helpful in many physics courses. (I'm also assuming you have had some exposure to vector calculus.)

You start out by picking a curvilinear 3d basis which lends naturally to your cross-section of interest (in your problem, it would be spherical or cylindrical coordinates, since you can get a disk by taking a slice of either a sphere or a cylinder).

Next, you identify the line element [itex]d\textbf{l}[/itex] for that coordinate system. In general, the line element can be written as [itex]d\textbf{l} = dl_{x_1}\hat{\textbf{x}_1}+dl_{x_2}\hat{\textbf{x}_2}+dl_{x_3}\hat{\tex tbf{x}_3}[/itex], where [itex]\{\hat{\textbf{x}_1}, \hat{\textbf{x}_2}, \hat{\textbf{x}_3}\}[/itex] are the basis vectors for that space. For example, the line element in spherical coordinates is [itex]d\textbf{l} = dr \hat{\textbf{r}} + r d\theta \hat{\theta} + r \sin\theta d\phi \hat{\phi} = dl_{r}\hat{\textbf{r}}+dl_\theta\hat{\theta}+dl_\phi\hat{\phi}[/itex]. (If if it isn't clear where these came from, try imagining how much arc length changes in each spherical direction as you go from [itex]r[/itex] to [itex]r + dr[/itex], [itex]\theta[/itex] to [itex]\theta + d\theta[/itex], and [itex]\phi[/itex] to [itex]\phi + d\phi[/itex].)

Once you've got the line element, you're basically done. By setting one of the coordinate bases as a fixed constant, we simply get [itex]dA = dl_{x_1}dl_{x_2}[/itex], where [itex]x_3[/itex] is the basis coordinate you held constant. This equation holds because the coordinate system is curvilinear--that is, you can get the area of something by summing a bunch of squares over the surface who have area equal to [itex]dl_{x_1}dl_{x_2}[/itex].

Here's the actual example relevant to your question: the circular disk formed by letting [itex]x_3 = \theta = \frac{\pi}{2}[/itex] (that is, it's in the xy-plane). Then [itex]dA = dl_r dl_\phi = (dr)(r\sin\theta d\phi)|_{\theta = \frac{\pi}{2}} = r \sin(\frac{\pi}{2}) dr d\phi = r dr d\phi[/itex]. Now, since you know it's a full disk, integrating over [itex]\phi[/itex] from 0 to [itex]2\pi[/itex] leaves you with the expression from your problem: [itex]2\pi r dr[/itex]. That's where it came from.

I hope this is helpful, and I am happy to clarify questions you might have. (And I apologize for any drudgery suffered in trying to understand the post.)


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