How Do You Calculate the Electric Field of a Ring?

[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 \Delta q is not a geometric property. And the final expression the book got was \frac{Kx q}{(R^2 + x^2)}^{3/2}.

How come they can integrate with respect to \Delta q?

 

[mfb]

You can integrate over the total charge in the same way you can integrate over the length of the ring, or over the total mass to calculate the moment of inertia. The integration variable does not have to be a length, it can be nearly everything.

 

[Enigman]

Integration is just summation of infinitely small elements of something. Here you are adding up all the Electric field due to all the Δq of the ring at a point along the axis at a distance x. So according to your equation

\sum \Delta E =\sum \frac{K x\Delta q}{(x^2+y^2)^{3/2}}
becomes,
\int \ dE =\int \frac{K x dq}{(x^2+y^2)^{3/2}}
when Δq→0

 

[tuggler]

But 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 can't integrate over \Delta Q because its 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?

 

[mfb]

Integration variables don't have to be geometric quantities.

I don't understand why we had to change it with the field of a line but not with a disk?

You don't have to in both cases.

 

[tuggler]

I am not understanding, then why did my book try and trick me?

 

[tuggler]

They did those two problems different with the same reasoning I mentioned above. And they did the same thing with an electric field of a disk as they did with the line of charge by replacing \Delta Q with the density over the surface area 2r\pi dr.

 

[mfb]

I am not understanding, then why did my book try and trick me?

I don't think it did.
If there are multiple ways to solve a problem, you have to choose one. There is no "right" or "wrong" choice.