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Finding the velocity of a proton at the surface of a conductor

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data

    A charged conducting sphere with radius 30cm is having a charge of -50nC. A proton is released from rest at R= 10m. Calculate the velocity of the proton when it reaches the surface of the conductor. What happens to the velocity of the proton when the charge on the sphere is changed to +50nC?

    2. Relevant equations

    V = kq / r
    W = qΔv = 1/2mv2

    3. The attempt at a solution

    First, I connected the potentials for the sphere and the proton:

    Vsphere = k(-50nC) / 0.3 = -1500
    Vproton = k(+q) / 10 = 1.44e-10

    I'm pretty sure this part is where I'm wrong. But I added them up and used -1500 for ΔV.
    Then I plugged in +q into W=qΔv equation, solved for v, and got 547722m/s as the answer.
    I think I'm close, because it says the correct answer is 52e4 m/s, but I'm not sure where I've messed up.

    For the second part where charge on the sphere is +50nC, do I just plug this number into Vsphere and do the same thing to get the velocity?

    Any help is appreciated, thanks.
     
  2. jcsd
  3. Jul 15, 2014 #2
    I'm not completely following your process here -- are you using a conservation of energy approach, involving the kinetic energy of the proton?

    It's a good start using potentials to calculate potential energy, but I'm not sure why you have a potential for both the sphere and the proton. To do a conservation of energy strategy here, you would want to calculate the change in the proton's potential energy as it moves toward the sphere, so that would involve the proton moving through different potentials due to the sphere. So what do you mean when you have ##V_{proton}##?

    And when say you solved for v from the ##W = q\Delta V##, do you mean you also used a kinetic energy relationship? It looks like you mistook the ##\Delta V## for velocity, but I sure hope that isn't the case!
     
  4. Jul 15, 2014 #3
    yes.

    I set qΔV and 1/2mv2 equal to each other, and I used the potential I got for the sphere for ΔV.


    I'm really not sure how to go about doing this problem... but am I on the right track with "qΔV = 1/2mv2" to find the velocity? How do I get ΔV?
     
  5. Jul 15, 2014 #4
    Yes, that is the correct equation to use, it's just your calculation of ##\Delta V## that's off. Think of it instead in terms of:$$\Delta U = \Delta K$$where ##\Delta U## is the change in potential energy of the proton. Using the relationship ##\Delta U = q\Delta V##, you can see that the change in potential energy is proportional to the change in potential that the proton moves through. So think of the proton being at one potential created by the sphere at R = 10m, and then at another potential (still due to the sphere) at R = 0.30m. This will be your ##\Delta V##. And here you'll only use the equation you used for ##V_{sphere}##, since it's the proton moving through the sphere's electric field, not the other way around (although you could claim that the proton moving toward the sphere is equivalent to the sphere moving toward the proton, but let's keep things simple).
     
  6. Jul 15, 2014 #5
    Are you saying find the potential at R=0.30m and at R=10m with -50nC as q for both of them to get delta V? And multiply by q again? Sorry, I'm a little confused.
     
  7. Jul 15, 2014 #6
    Yes, that's right. The idea is that you find the potential due to the sphere at different points in space, using
    $$V = k\frac{q}{r}$$
    where ##q## here is the charge of the sphere. Then once you find the potential difference between two points in space (R = 10m and R = 0.30m), you use the formula
    $$\Delta U = q\Delta V$$
    where ##q## here is now the charge of the proton.

    In a nutshell, the proton moves across a potential difference, and the energy lost/gained equals the potential times the charge.
     
  8. Jul 15, 2014 #7
    Hm, I got 539444m/s... I guess I'm getting close?

    So I got -45 (at R=10) and -1500 (at R=0.3), I did -45-(-1500) and got 1455 for delta V. Then I multiplied by 1.6e-19 to get delta U, which equals delta K. Did I mess up again?
     
  9. Jul 15, 2014 #8

    NascentOxygen

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    Staff: Mentor

    I wonder does the mass of the proton increase slightly? Its speed passes 0.1% of the speed of light.
     
  10. Jul 15, 2014 #9
    I'm not quite sure. The process is right, but I got a similar answer. I sure hope special relativity isn't involved, but even so we're off by a whole order of magnitude according to the answer -- it's on the order of 104, and our answers are on the order of 105. Both are so close to ~5.3 as a base though, so could the answer perhaps have a typo? Are you sure you have all the givens correct? I will double check the answer again.
     
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