I posted my entire post as an image, since it had so many equations in it. I hope that's ok with everyone. Sorry that the size and font are a bit inconsistant. Next time I will write my whole post in a Latex file, turn it into an image, and post that.
Also, I have to appologise, I used E to mean 2 diffent things. E on the left hand side of an equation is the energy, but E elsewhere is the E operator, meaning x10^.
So it is extremely easy to ionise a drop with a radius of 1um but very hard to ionise one with a radius of 1mm. I would assume it to be the other way around as the charge could be distributed across a greater area.
I think it makes sense. If you have something smaller, there is less overall surface to distribute the charge across, thereby making it easier to put a charge to it.
I posted my entire post as an image, since it had so many equations in it. I hope that's ok with everyone. Sorry that the size and font are a bit inconsistant. Next time I will write my whole post in a Latex file, turn it into an image, and post that.
ReplyDeleteAlso, I have to appologise, I used E to mean 2 diffent things. E on the left hand side of an equation is the energy, but E elsewhere is the E operator, meaning x10^.
I change my mind. I made the corrections ad did what I said I would do next time now. I hope that makes things easier to understand.
ReplyDeleteClick on the image to see it in more detail.
So it is extremely easy to ionise a drop with a radius of 1um but very hard to ionise one with a radius of 1mm. I would assume it to be the other way around as the charge could be distributed across a greater area.
ReplyDeleteI think it makes sense. If you have something smaller, there is less overall surface to distribute the charge across, thereby making it easier to put a charge to it.
ReplyDelete