# Electric force/velocity

by kristi.lynn
Tags: electric, force or velocity
 P: 13 It seems like this should be really easy but I've been stuck for awhile... A small particle has charge -5.00 uC and mass 2.00E-4 kg. It moves from point a, where the electric potential is Va=+200V, to point b, where the electric potential is Vb=+800V. The electric force is the only force acting on the particle. The particle has speed 5.00 m/s at point a. What is its speed at point b? Here's what I've tried. I did Ka + Ua = Kb + Ub so that's (.5)(.200g)(5.00m/s)^2 + (-5.00E-6C)(200V) = (.5)(.200g)(vb)^2 + (-5.00E-6C)(800V) And that gave me vb of 5.00 m/s so I figure that lead me in a circle bc I know that's not the answer bc the answer is in the back of the book... So then I tried: V=kq/r and did -600V[Va-Vb]=9E9Nm^2/C^2 x -5.00E-6C / r and got r=75m so I assumed that's the distance between a and b. So then I used that with E=V(ab)/d and got E=-8N/C and so I tried to do something like E=kq/r^2 to find q but obviously, that only will give me q=-5.00E-6C so I'm making lots of little circles but I can't seem to figure out the charge of these points.. Any suggestions on what to do with those potentials (+200 and +800) I don't know where to go from here.. Kristi-Lynn
Mentor
P: 40,263
 Quote by kristi.lynn Here's what I've tried. I did Ka + Ua = Kb + Ub so that's (.5)(.200g)(5.00m/s)^2 + (-5.00E-6C)(200V) = (.5)(.200g)(vb)^2 + (-5.00E-6C)(800V)
Looks good to me, except that you changed to grams for some reason. Stay with SI units.

 And that gave me vb of 5.00 m/s so I figure that lead me in a circle bc I know that's not the answer bc the answer is in the back of the book...
How did you get vb = 5.0 m/s??? That's what you started with. Do it over. (Simplify before doing any arithmetic.)
 PF Patron Sci Advisor Emeritus P: 10,400 Use an energy argument. Do you know what an electron-volt is? An electron-volt is a unit of energy that is gained by an electron moving across a potential difference. The mass of the electron is irrelevant, and so is the distance it travels. All that matters is its charge and the potential difference it crosses. This can be seen by just looking at the dimensions of the quantities involved. Work (energy) is defined as force times distance, K = F d. The force on a charge of q Coulombs in an electric field E is F = q E. Substituting, K = q E d. E can be expressed as newtons per coulomb (or as volts per meter). We'll choose to represent it as volts per meter: K = q (V/d) d The distance cancels, and you're left with work (energy) = charge * potential difference. That's all there is to it. Calculate how much energy is gained (or lost) by the particle moving through that potential difference, and add it to the particle's initial kinetic energy, found with K = 0.5 m v_i^2. Then find its velocity again with K = 0.5 m v_f^2. - Warren
P: 13

## Electric force/velocity

Ok i got it...I didn't know to leave it in kg...but that made the difference... Thanks yet again physics man!
 PF Patron Sci Advisor Emeritus P: 10,400 Scratch that, I see that you already were making an energy argument. - Warren
 P: 13 Thanks just the same :) I've finished yet another physics assignment! Well except for part c of one of them... why isn't a balloon charged with 1200V potential difference (relative to me) dangerous? I know that it isn't, but I don't think they're looking for "bc it's a balloon and a balloon rubbed on a carpet won't kill me, it'll stick to a wall..." But I'm not sure why that's only a small amount when a 6v battery can probably at least give you a hurting shock.. any ideas?
 PF Patron Sci Advisor Emeritus P: 10,400 The 1200V balloon is not dangerous because you don't really have to remove (or add) much charge to get that kind of potential. The balloon is not a constant source of electrons at 1200V, which would be dangerous. Instead, it's a source of only a few electrons. Potentials don't kill you -- the movement of electrons, with the subsequent generation of heat, kills you. If you touched the balloon, your body would add (or remove) only a tiny bit of charge to neutralize its charge. The movement of such a small amount of charge is not enough to cause any significant damage. - Warren
 P: 13 Awesome thanks! :)

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