Electric Field due to a disk of charge. (Problem in derivation)

Question

This might be a really silly question, but I don't understand it.

In finding the electric field due to a thin disk of charge, we use the known result of the field due to a ring of charge and then integrate the relation over the complete radius. But I had a problem in the derivation, as follows:

We assume a ring at a distance $r$ , and of an infinitesimal thickness $dr$ from the center of the disk. Then the very next step in every book I've referred is $dA = 2 \pi rdr$. This is what I don't understand. Shouldn't the area be $$dA = \pi [ (r+dr)^2 - r^2] ~ ?$$ Please help me out here.

Answer 1

$$A=\pi r^2$$ $$\frac{dA}{dr}=\pi\cdot2r$$ $$dA=2\pi rdr$$

Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$

You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit while taking infinitesimal rings.

Answer 2

You're exactly right. Try expanding your expression out.

$$\pi [(r + dr)^2-r^2] = \pi [r^2 + 2 r dr + dr^2] = \pi [2rdr + dr^2]$$

Since it is an infinitesimal, $dr^2 = 0$. Indeed it is defined this way. When we do integration problems like the one you describe, we always consider a small element (like a ring of width $dr$) but then eventually take the limit as $dr \to 0$. In such situations $dr^2$ will always be negligible compared to $dr$.

Answer 3

Imagine that you take the thin strip of width $dr$ , cut it (say at $\theta=0$) and stretch it into a straight line. You would then have a rectangle (almost) of width $dr$ and length $2\pi r$, with approximate area $2\pi r^2$. As $dr$ tends to zero you can drop the "approximately" - this is a basic trick in calculus.

Answer 4

dA=π[(r+dr)²−r²]
=π[ r² + 2 r dr + dr² − r² ]
=π[ 2 r dr + dr² ]

We drop the dr² term and are left with dA=2πrdr