# Electric field of a disk

1. Sep 7, 2013

### tuggler

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 $$2r\pi dr$$.

Last edited: Sep 7, 2013
2. Sep 7, 2013

### 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