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Voltage Needed for a Proton to Arrive at Sphere at A Certain Speed

  1. Oct 2, 2009 #1
    A metal target sphere of 30 cm diameter suspended out in space is given a charge of +1.00 nC. Through what voltage must a distant proton be accelerated from rest (by some sort of space weapon) if it is to arrive at the sphere at a speed of 8.2*10^4 m/s? (Neglect any gravitational effects.)

    V= (k0*Q)/R
    W= qe*delta V
    VB-VA= k0*Q(1/rb-1/ra)

    I calculated the voltage for the sphere: V= (8.99*10^9)(1.00*10^-9C)/0.15m= 59.93V
    Then I calculated W: (1.6*10^-19)(59.93V)= 9.5888E-18J
    Now, this is where I'm having trouble. What is the next step? What does speed have to do with any of this? I am completely lost so any help would be greatly appreciated.
  2. jcsd
  3. Oct 2, 2009 #2


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    My EM is rusty and I don't remember how you do potential for the surface of a charge . . . the approach I would take is to say
    energy from V = KE at the sphere + work done pushing charges together

    That last term would be W = k*q*e/R where q is the charge on the sphere and R its radius.
  4. Oct 3, 2009 #3
    Sorry, but you lost me with that. KE of what sphere? A proton? And what is e? In my first post, qe represents an electron volt according to my textbook. What does your e represent? If KE is for the proton, I would use KE= 1/2mv^2 right? Any other suggestions?
  5. Oct 3, 2009 #4


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    e is the elementary charge of the proton. e = 1.6 x 10^-19 C.
    The energy given to the proton by the Voltage must give it the energy that it needs to do work moving through the sphere's electric field plus the kinetic energy it has when it gets to the sphere.
    The only unknown in that equation is the V so you can find it that way.
    Yes 1/2mv^2 is the last term.
  6. Oct 3, 2009 #5
    Ok, I only have three trials left so just to be clear:
    V= 1/2(1.67*10^-27kg)(82000m/s^2)+(8.99*10^9)(59.93V)(1.6*10^-19C)/0.15m
  7. Oct 3, 2009 #6


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    I don't think so.
    Energy from V = KE + Work
    eV = .5*m*v^2 + kqe/R
    V = .5*m*v^2/e + kq/R divided both sides by e
    = 1/2(1.67*10^-27kg)(82000m/s^2)/(1.6x10^-19) + (8.99*10^9)(10^-9C)/0.15m
  8. Oct 3, 2009 #7
    I got 59.93V, but that's incorrect. Maybe this will help. In my textbook the question is broken up into two parts.

    35) A metal target sphere of 20cm diameter suspended out in space is given a charge of +1.00nC. How much work is done on a proton in taking it from very far away (essentially infinity) to the surface of the of the sphere? Answer: 1.4*10^-17J

    37) With problem 35 in mind, through what voltage must the proton be accelerated from the rest (by some sort of space weapon) if it is to arrive at the sphere at a speed of 8.5*10^4m/s? Neglect any gravitational effect. Answer: 1.3*10^2V

    My online hmwk combines the questions into one. I just don't see how they are related in this particular case.
    Last edited: Oct 3, 2009
  9. Oct 3, 2009 #8


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    Yikes, I got 59.93V the first time, too! It was because the "m/s^2" confused me. I thought only the seconds were squared, but the 82000 should be squared, too! Best not to write units in intermediate calcs.

    I checked our formula with the speed 85000 and diameter 20 cm and got 128, which is pretty close to the 1.3*10^2V answer. That certainly confirms that the formula is correct.
  10. Oct 3, 2009 #9
    Ok, it goes something like this, the proton had an initial energy of E.

    After it loses the W you calculated above, 9.5888E-18J it has kinetic energy remaining, 1/2*mv^2

    So, E-W=KE

    This means that initially, your proton had energy E=KE+W. You know KE because you know the final velocity and you know W which you calculated above. Next, this E is actually q*DeltaV where delta V is the potential difference it has been accelerated through, which you are looking for.

    After all this, I think you should get 95V.
  11. Oct 3, 2009 #10
    Sorry, but I'm still not getting it. What do you mean by the 85000 and the seconds are supposed to be squared too? 85000^2/60^2?

    (1/2)(1.67*10^-27)(85000^2)(1.6*10^-19) + (8.99*10^9)(10^-9)/0.10= 89.9

    And thanks for helping out physicsnoob93, but 95V isn't right either.
  12. Oct 3, 2009 #11


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    I just meant that instead of writing "82000m/s^2" it would have been better to leave out the units and just write "82000^2" to avoid confusion.

    I got 95.0, too. Check the numbers in the first post again. Is it 8.2*10^4 m/s or possibly 8.5 like the book question? Diameter 30 cm? Do they ever have a wrong answer?
  13. Oct 3, 2009 #12
    Those are the correct numbers. I cut and pasted the question directly from my online hmwk and I'm looking at it right now. I don't think they've ever had a wrong answer, but I'll try it with 8.5 like the book. I only have one trial left anyway so I'm just going to wing it; I can live with a 93%, but it would be nice to know how to do the problem because it might be on my test. So thanks for everything! I know you tried really hard to help and I appreciate it. Now that I am officially out of trials and can't get credit for this question, can someone please just show me how to do it for the test?
    Last edited: Oct 3, 2009
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