1. The problem statement, all variables and given/known data A 2kg charge of +3C moves through a potential difference of 6V and begins at rest. What is its final velocity? 2. Relevant equations m = 2kg q = +3C dV = 6V Vi = 0 m/s Vf = ? F = m*a (Vf)2 = (Vi)2 + 2(a)(d) dV = (ke)(q)/(r) 1 N/C is equal to 1 V/m 3. The attempt at a solution Okay I am not sure if this was some mistake given by our professor. This is one of the review questions at the end of his presentation. These are to be done as extra practice. We didn't go over anything like this in class so I finagled my way to what I believe is the answer. I used the propert 1 N/C = 1 V/m to move over and get 1 N = (V)(C)/(m) so F (newtons) = (V)(C)/(m) now from here I said **** I don't have a distance. What equation could I use to find it (r = distance is assumed here) Using the dV = ke(q)/r I solved for r = d = 4.5x10^9 m Plugging into my finagled force formula F = (6N*m/C)(3C) / (4.5x10^9m) --> F = 4.0x10^-9 N a = F / m --> a = (4.0x10^-9kg*m/s^2) / (2kg) --> a = 2.0x10^-9 m/s^2 putting this back into my kinematic equation (from christ..3 years ago) I got Vf = sqrt [2(2.0x10^9m/s^2)(4.5x10^9m)] --> sqrt [18m^2 / s^2] Vf = 4.24 m/s Now I know this is an outrageous stretch here. And chances are that I am a fool and wasted my time trying to solve an incomplete / misplaced problem. However being that I calculated all of this using proper units and they all cancelled out accordingly I think I am right. Can anyone please chime in on whether or not this is right and if not how it could (if at all) be calculated differently. This is more for my own personal sanity and pride because like I said we didn't go over this type of problem and I don't expect it on the test. Thanks in advance.