tag:blogger.com,1999:blog-5834252248412211498.post170874946847518008..comments2023-10-02T19:15:59.038+10:00Comments on BIPH3001-Frontiers in Biophysics 2010: Your Turn 7DMitch Ghttp://www.blogger.com/profile/12435777631824415803noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-5834252248412211498.post-22059365604724036052010-09-08T08:03:36.975+10:002010-09-08T08:03:36.975+10:00I think it makes sense. If you have something smal...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.Matt Grevetthttps://www.blogger.com/profile/04319569394511737669noreply@blogger.comtag:blogger.com,1999:blog-5834252248412211498.post-29349805348882647582010-09-07T13:30:46.033+10:002010-09-07T13:30:46.033+10:00So it is extremely easy to ionise a drop with a ra...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.Heatherhttps://www.blogger.com/profile/13167610447846122237noreply@blogger.comtag:blogger.com,1999:blog-5834252248412211498.post-42363475338098244642010-09-07T01:34:24.624+10:002010-09-07T01:34:24.624+10:00I change my mind. I made the corrections ad did wh...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.<br /><br />Click on the image to see it in more detail.Mitch Ghttps://www.blogger.com/profile/12435777631824415803noreply@blogger.comtag:blogger.com,1999:blog-5834252248412211498.post-64593435556682302802010-09-07T00:49:47.306+10:002010-09-07T00:49:47.306+10:00I posted my entire post as an image, since it had ...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.<br /><br />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^.Mitch Ghttps://www.blogger.com/profile/12435777631824415803noreply@blogger.com