# Finding Area of Ring Segment to Find Electric Field of Disk

• ChiralSuperfields
In summary, the area of a ring segment can be found by cutting the ring and stretching it out to form a rectangular strip of length 2πr and width dr. This means that the area is equal to 2πrdr, which is the same as the formula for the area of an annulus. When integrating this term, it is not necessary to include the (dr)^2 term as it approaches zero much faster than the 2rdr term. The derivative of the area function with respect to r is 2πr, showing that the higher order terms in dr or Δr will vanish in the limit. This method can also be used to find the volume of a spherical shell.

#### ChiralSuperfields

Homework Statement
A disk of radius R has a uniform surface charge density s. Calculate the electric field at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk
Relevant Equations
Continuous charge distribution formula
Hi!

For this problem,

Why is the area of each ring segment dA equal to (2π)(r)(dr)?

However, according to google the area of a ring segment (Annulus) is,

Many thanks!

Cut the thin ring and stretch it out. You get a rectangular strip of length ##2\pi r## and width ##dr##. What is its area?

The area of the annulus is indeed what google says. Note that
##R^2-r^2=(R+r)(R-r)##
For a thin ring ##R## and ##r## are very close to each other so that ##R-r=dr##. You can then approximate ##R+r\approx 2r## so that ##R^2-r^2\approx 2rdr##.

ChiralSuperfields
kuruman said:
Cut the thin ring and stretch it out. You get a rectangular strip of length ##2\pi r## and width ##dr##. What is its area?
Thanks! So that is thought process that gives the area! However, why is that not equal to the area from the Annulus formula?

From the annulus formula I got A = (π)[(r + dr)^2 - r^2].

Many thanks!

Callumnc1 said:
Thanks! So that is thought process that gives the area! However, why is that not equal to the area from the Annulus formula?

From the annulus formula I got A = (π)[(r + dr)^2 - r^2].

Many thanks!
See my edited post.

ChiralSuperfields
kuruman said:
See my edited post.
Thank you, I have written up and realized I had a new question:

What if we choose not to approximate? How would we integrate the infinitesimal (dr)^2 term?

Many thanks!

Callumnc1 said:
Thank you, I have written up and realized I had a new question:
View attachment 318038
What if we choose not to approximate? How would we integrate the infinitesimal (dr)^2 term?

Many thanks!
Well, you will notice that if you integrate

$$\int dA = 2\pi \int_{r}^{R} r~dr$$

That it is exactly:

$$A = \pi \left( R^2 - r^2 \right)$$

So the term ##(dr)^2## is not contributing anything. In the limit as ##dr \to 0 , 2 r ~dr ## is exact.

ChiralSuperfields
erobz said:
Well, you will notice that if you integrate

$$\int dA = 2\pi \int_{r}^{R} r~dr$$

That it is exactly:

$$A = \pi \left( R^2 - r^2 \right)$$

So the term ##(dr)^2## is not contributing anything. In the limit as ##dr \to 0 , 2 r ~dr ## is exact.

I guess as dr approaches zero then (dr)^2 approach's zero a lot faster than 2r dr because it is squared.

However, just curious had would you integrate the (dr)^2 term anyway?

Many thanks!

Callumnc1 said:
However, just curious had would you integrate the (dr)^2 term anyway?
I don't believe you ever would. It doesn't follow from the definition of the derivative.

$$(A + \Delta A)- A = \pi \left[ \left( r + \Delta r \right)^2 - r^2\right] \implies \Delta A = \pi \left[ 2 r ~\Delta r + (\Delta r)^2 \right]$$

The derivative is given by:

$$\frac{dA}{dr} = \lim_{ \Delta r \to 0 } \frac{\Delta A}{ \Delta r} = \lim_{ \Delta r \to 0 } \frac{ \pi \left[ 2 r ~\Delta r + (\Delta r)^2 \right] }{ \Delta r} = \lim_{ \Delta r \to 0 } \pi [ 2r + \cancel{ (\Delta r)}^0 ] = 2 \pi r$$

So if there are any first order ( or higher ) ##\Delta r ## terms left hanging around they vanish in the limit.

Last edited:
ChiralSuperfields
erobz said:
I don't believe you ever would. It doesn't follow from the definition of the derivative.

$$(A + \Delta A)- A = \pi \left[ \left( r + \Delta r \right)^2 - r^2\right] \implies \Delta A = \pi \left[ 2 r ~\Delta r + (\Delta r)^2 \right]$$

The derivative is given by:

$$\frac{dA}{dr} = \lim_{ \Delta r \to 0 } \frac{\Delta A}{ \Delta r} = \lim_{ \Delta r \to 0 } \frac{ \pi \left[ 2 r ~\Delta r + (\Delta r)^2 \right] }{ \Delta r} = \lim_{ \Delta r \to 0 } \pi [ 2r + \cancel{ (\Delta r)}^0 ] = 2 \pi r$$
Ok thank you erobz!

erobz
If you don't want to see these higher order terms in dr or ##\Delta r##, you can just start from the area of a circle, ##A(r)=\pi r^2## as a function of r. Then the change in the area when the radius increases by dr is the differential ##dA= (\frac{dA}{dr}) dr## = ##2\pi r dr##. This is the area of the ring added to the circle by a change in radius of dr so it is the area of a differential ring. You can use the same method to find the volume of a spherical shell by starting with the volume of a sphere.

ChiralSuperfields
nasu said:
If you don't want to see these higher order terms in dr or ##\Delta r##, you can just start from the area of a circle, ##A(r)=\pi r^2## as a function of r. Then the change in the area when the radius increases by dr is the differential ##dA= (\frac{dA}{dr}) dr## = ##2\pi r dr##. This is the area of the ring added to the circle by a change in radius of dr so it is the area of a differential ring. You can use the same method to find the volume of a spherical shell by starting with the volume of a sphere.
Thank you for sharing with me that really interesting method @nasu !