Why Does Integration Differ Between Electric Fields of a Disk and a Line Charge?

[tuggler]

Homework Statement

I am suppose to find an expression for the electric field of a ring.

Homework Equations

E =\frac{Kq}{r^2}

The Attempt at a Solution

I calculated my results and I reached up to this:

\frac{Kx\Delta q}{(R^2 + x^2)}^{3/2} where R = radius, x = distance, K = constant, q = charge.

And then I looked at my book and noticed they integrated with respect to \Delta q which got me confused because when I calculated the electric field due to a line charge \Delta q it wasn't considered a geometric property. The final expression the book gave \frac{Kx q}{(R^2 + x^2)}^{3/2}.

How come they can integrate with respect to \Delta q in this case but not a line charge?

For example, when I was measuring the electric field of a line of charge I got the expression \sum \frac{d \Delta Q}{(y_1^2 + d^2)^{3/2}} but with that expression I couldn't integrate over \Delta Q because the book said it is not a geometric quantity so I had to replace \Delta Q with \Delta Q = Q/L \Delta y. I don't understand why we had to change it with the field of a line but not with a disk?

The book did the same thing with an electric field of a ring as they did with the line of charge by replacing \Delta Q with the density over the surface area 2r\pi dr.

 

[LeonhardEu]

I suppose you talk about a disk ( on your first statement you talk about "ring"). For the ring you have the field of a line so you multiply your linear density with λdx = λrdθ. Now you have a surface with density σ. So dq = σda = σ(2πrdr).
Draw a diagram and you will see the validity of that. look the area that represents 2πrdr