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Charged particle thrown upward please help (almsot there)

  1. Feb 28, 2005 #1
    A particle is uncharged and is thrown vertically upward from ground level with a speed of 25.0 m/s. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 30.0 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge –q. Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

    I figured out h with
    Vf^2 = Vi^2 + 2ah
    (38.55m)

    I also know difference in energy = mv^2/2 - mv^2/2 = 275 J
    I also know that at the top, its positively charged, and that the speed required for -q will be less

    how can i solve the rest?
     
  2. jcsd
  3. Feb 28, 2005 #2

    I hate to spoil it for you, but you're not almost there.

    How did you get these numbers? (Show your work so we can see where you need help.)

    First, use the uncharged situation to figure out the total energy (as a multiple of the mass). This tells you the gravitational potential energy at the top (as a multiple of the mass), and note that the GRAVITATIONAL potential at the top is the same for all 3 trials.

    Then, the initial speed of the second throw lets you figure out how much ADDITIONAL energy was required to oppose the electrical field in getting the positively charged ball to the same height h.

    Finally, in the third throw, recognize that the effect of the electric field on the negatively charged ball will be exactly the opposite of its effect on the positively charged ball, and from that you can figure out how much (less) kinetic energy was needed to throw the ball the 3rd time. From that, you can solve for the speed.


    I don't know what your point was here. The first time (at the bottom AND at the top), it has no charge. The second time (at the bottom AND at the top), it has a positive charge. And so on...
     
  4. Mar 1, 2005 #3
    The electric potential at the height h exceeds the electric potential at ground level
    potential at top > potential at bottom, potential at top is more positive then at bottom

    I figured out h with
    Vf^2 = Vi^2 + 2ah
    30^2 = 0 + 2(-9.81)h
    h =45.87m (changed it)

    energy required to get it to this height is mv1^2/2

    the 2nd time, a is not -9.81 because the particle is charged, and Fe effects it.
    energy required to get it to 45.87m is mv2^2/2

    the difference in energies (relates to the strength of the electric potential) is
    mv1^2/2 - mv2^2/2
    =m(900/2) - m(625/2)
    = 137.5 * m (fixed this too, wtf was i thinking)

    but now im stuck again..
     
  5. Mar 1, 2005 #4
    is wrong. vf in all cases is 0, and vi for the uncharged case is 25 m/s. But actually it doesn't matter: you don't have to solve for h. You just have to know that h is the same for all 3 cases, and that's given.

    There, you have the key to the solution. This tells you that it required an extra (137.5*m)J of energy to get the ball up to height h working against gravity AND the electric field, as compared to the energy required to get there working against gravity alone.

    So, what can you conclude about the energy required if the same electric field is helping?
     
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