Recent content by pollytheparrot

Thanks for the help man! Really means a lot! You keep doing you!

Where R =d+x and x = Sqrt(R^2 - b^2), so d = R - Sqrt(R^2 - b^2) ?

https://photos-3.dropbox.com/t/2/AADWd9kWZiwMFH1Svwi-bbLvApEmtOjBYuDxYjJxCWD5JQ/12/2668640/jpeg/32x32/1/1436475600/0/2/2015-07-09%2021.04.34.jpg/CODwogEgASACIAMgBCAFIAYgBygBKAIoBw/mATI66H3HO_k1EIE84VJevokK97mR2xUrjAyfX-h5Yk?size_mode=5 Someone else worked it out like this, is this allowed...

So given your triangle, how would I use that to find d? I was just assuming something like this. Don't know how picture got flipped...

It looks kind of legit if you make that line and flip it around and upside down

Surely it can't be that simple? d=√R^2-b^2 ?

Sorry, that is uncertainty and sloppiness in action there. Is this what you meant by making a triangle? If you can find the hypotenuse created by the red line, using some kind of circle magic, you have the side b, and the last side should be d, correct? If this is the right course of action...

That's still an assumption on my part. Between the second and third plates, the only force that should be acting on the particle is the magnetic force, which is F=q*v x B, and that it should move in a circle, and reasoning (albeit crappy explanation it is) thought it should be where it is, is...

Perhaps my second post answered these questions? I'm assuming that the center of the circle is in line with the second (middle) plate

Because my picture was pretty potato quality, here is a new version. Also I found a simplification of R, where it is mv^2/qvB, leading to mv/qB, which is m*sqrt(2qV/m) / qB...

Homework Statement Basically there are 3 plate capacitors, with a distance a between the first two and b between the last two. In between the first two plates is voltage V. A particle of charge q is released from the first plate, and when it leaves the second plate it is no longer...