1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric potential problem

  1. Feb 12, 2006 #1
    On planet Tehar, the free fall acceleration is the same as that of Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A 4.05 kg ball having a charge of 4.65 [tex] \mu C [/tex] is thrown upward at a speed of 22.2 m/s, and it hits the ground after an interval of 1.86 s. What is the potential difference between the starting point and the top point of the trajectory? Answer in units of kV.

    First I drew a free body diagram and found that the only forces on the ball were the force of the charge and gravity.
    So qE + mg = ma
    where a is constant.
    Then I figured out the height of the ball by adding the final and initial velocities and dividing by 2 and multiplying by the time/2.
    22.0+ 0 /2 *1.86/2
    h= 10.323
    Then I used conservation of energy
    KE_o + PE_0 +EPE_o = KE_f + PE_f + EPE_f
    1/2mv^2_o +0 +0 = 1/2 mv^2_f + mgh +qV
    So V= 1/q ((1/2)mv^2_0 -mgh)
    V= 1/4.65 x 10^-6 ((1/2)4.05(22.2^2) - 4.05 (9.8)( 10.323))
    V=1.26x 10^8
    V= 1.26 x 10^4 kV
    This isn't right.. can someone tell me what's wrong?
     
  2. jcsd
  3. Feb 13, 2006 #2

    andrevdh

    User Avatar
    Homework Helper

    Well what I did is first calculated the gravitational acceleration of the planet from the data. That enables you to calculate the electric field strength from your first equation. That with the height gave me
    [tex]V=Ed=1.26 \times 10^8\ V[/tex]
     
  4. Feb 13, 2006 #3
    I got that, but the answer is supposed to be in kV. So that would make it 1.26 x 10^4, right?
     
  5. Feb 13, 2006 #4
    Is this the right approach or am I doing it completely wrong?
     
  6. Feb 15, 2006 #5

    andrevdh

    User Avatar
    Homework Helper

    Sorry I do not understand the system, since sometimes it notifies me via e-mail when someone posted in the thread, but it did not do it in this case. kilo is [itex]10^3[/itex] therefore [itex]10^5[/itex] will be left after taking away 3 from 8.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric potential problem
Loading...