Let us revisit our last integral.
\(\displaystyle \int k\frac{dQ}{r}\)
Here \(\displaystyle r\) is the distance from a point on the flat ring to a point on the \(\displaystyle x\)-axis (not along the radius of the ring, except at \(\displaystyle x = 0\)). I'll make a sketch to show you the idea.
If the
blue color is a point on the flat ring, then \(\displaystyle r\) is the purple distance \(\displaystyle (h)\). I should have called it \(\displaystyle h\) (hypotenuse) instead.
So, the integral should be:
\(\displaystyle \int k\frac{dQ}{h}\)
From the sketch, \(\displaystyle h = \sqrt{x^2 + r^2}\), where \(\displaystyle r\) is the distance from the center of the ring to the
blue point
Then, the integral becomes:
\(\displaystyle \int k\frac{dQ}{\sqrt{x^2 + r^2}}\)
\(\displaystyle \sigma = \frac{Q}{A} = \frac{dQ}{dA} = \frac{dQ}{2\pi r \ dr}\)
\(\displaystyle dQ = \sigma 2\pi r \ dr\)
Then,
\(\displaystyle \int k\frac{dQ}{\sqrt{x^2 + r^2}} = \int_{R_1}^{R_2} k\frac{\sigma 2\pi r \ dr}{\sqrt{x^2 + r^2}}\)
We are ready to integrate, and that's what we'll do in the next post.
