Electric Field due to Charged Washer

[yango_17]

Homework Statement

A thin flat washer lies in the x − y plane, with the center at the coordinate origin. The washer has an inner radius R1 and an outer radius R2 (so it looks like a disk of radius R2 with a concentric circular cut-out of radius R1). The surface of the washer is uniformly charged with a surface charge density σ.
1. (a) What is the electric field (as a vector) at the distance z along the z-axis (which coincides with the axis of symmetry of the washer)?[Hint: subdivide the washer into infinitely thin concentric rings.]

Homework Equations

E=q/r^2 (electric field due to a point charge)

The Attempt at a Solution

I know that you have to integrate in order to determine the total magnitude of the entire washer on the point, but I'm not sure if I'm going about setting up the integration the right way. Currently, I have dE= k(δds)/(R2^2-R1^2), where δ represents the surface charge density of the washer. Any hints as to where to go from here would be appreciated.

 

[LittleMrsMonkey]

That's not a correct formula.First of all, what is ds?You are dealing with a surface and the surface charge density is ##σ= \frac {Q}{A}##.
Also,why are ##R_1## and ##R_2## in the formula for dE?What would the interval of the integration be?

 

[RUber]

I could be slightly off on this one, so bear with me.

For a fixed Z, the distance from a point on the disk (x,y,~0) to the point (0,0,Z) is
##d = \sqrt{x^2 + y^2 + Z^2} = \sqrt{r^2 + Z^2}##
Using charge density sigma, you should come up with an electric field that looks like:
##E = \frac {\sigma}{4\pi \epsilon d^2 } \hat d ## from a single point.
The portion of this field pointed in the ##\hat z## direction is ##\frac{Z}{d}##
Clearly, for large Z, this will tend toward being able to replace d with Z, and simplify quickly to some factor of sigma time the area divided by Z^2.

Assuming Z is not large enough to make that simplification, but large enough to negate the thickness of the washer, you want to integrate over all the points in the disk.
Changing to cylindrical coordinates seems best here.
##\int\int f(x,y,z) dx dy \to \int\int f(r, \theta, z) r dr d\theta dz.##
Which gives
##\int_0^{2\pi} \int_{R_1}^{R_2} \frac {\sigma}{4\pi \epsilon d^2 }\frac{Z}{d} r \, dr d\theta. ##
It seems like this is a simple enough integral to work out if you use the substitution ##u = d^{-1}##

Does this seem like a viable method? I noticed that my E field equation has extra scaling factors in it compared to yours...the math is the same either way down to a constant multiple.

 

[LittleMrsMonkey]

I can tell you there is no need for double integrals because this kind of problem is in the curriculum of an exam I'm taking in a few days,and I won't be learning double integrals until spring 2016.
I suggest taking a look at the thread right below this one,which is on oractically the same problem.

 

[RUber]

I can tell you there is no need for double integrals because this kind of problem is in the curriculum of an exam I'm taking in a few days,and I won't be learning double integrals until spring 2016.
I suggest taking a look at the thread right below this one,which is on oractically the same problem.

Right. The theta integral turns into a multiplication by 2pi, since the function has no theta dependence. That's simply what I see to be standard notation for integrating over an area.

 

[LittleMrsMonkey]

Right. The theta integral turns into a multiplication by 2pi, since the function has no theta dependence. That's simply what I see to be standard notation for integrating over an area.

Again,I don't know anything about these integrals so I'll take your word for it.

 

[yango_17]

I saw cylindrical coordinates and got very intimidated. I'm not sure if RUber's method is useful, as our professor doesn't really expect us to use double integrals. However, I do understand what you said about the distance Z and your electric field equation. Still trying to work out what the integral for "infinitely thin concentric rings" would look like.

 

[RUber]

Ignore the double integral and cylindrical coordinates, the result is the same. It becomes a single integral in terms of radius over the washer multiplied by a constant.
The function you are integrating will be in terms of the function d(r) I mentioned in the first post.