1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Question about Electric field.

  1. Feb 24, 2014 #1
    [SOLVED] Question about Electric field.

    1. The problem statement, all variables and given/known data
    A point charge [itex]q_{1}=15.00\mu C[/itex] is held fixed in space. From a horizontal distance of [itex]3.00cm[/itex] , a small sphere with mass [itex]4.00*10^{-3}kg[/itex] and charge [itex]q_{2}=+2.00\mu[/itex]C is fired toward the fixed charge with an initial speed of [itex]38.0m/s[/itex] . Gravity can be neglected.
    What is the acceleration of the sphere at the instant when its speed is [itex]20.0m/s[/itex] ?
    Express your answer with the appropriate units.

    2. Relevant equations

    [itex]a = \frac{k*q_{1}*q_{2}}{m*r^{2}}[/itex]

    3. The attempt at a solution

    [itex]a = \frac{9*10^{9}*15*10^{-6}*2*10^{-6}}{4*10^{-3}*r^{2}}[/itex]

    [itex]a = 67.5 * \frac{1}{r^{2}}[/itex]

    By integration,

    [itex]v = -67.5 * \frac{1}{r} + C[/itex]

    Initial conditions: [itex]v = 38 m/s, r = 0.03m [/itex]

    [itex]38 = -67.5 * \frac{1}{0.03} + C[/itex]

    [itex]C = 2288[/itex]

    [itex]v = -67.5 * \frac{1}{r} + 2288[/itex]

    When [itex]v = 20m/s[/itex],

    [itex]20 = -67.5 * \frac{1}{r} + 2288[/itex]

    [itex]r = 0.029761904[/itex]

    [itex]a = 67.5 * \frac{1}{0.029761904^{2}}[/itex]

    [itex]a ≈ 76200 m/s^{2}[/itex]

    Since both charges are +ve,

    [itex]a = -76200 m/s^{2}[/itex]

    However, the solution is not correct.

    May anyone pointing out the errors? Thank you very much.
    Last edited: Feb 24, 2014
  2. jcsd
  3. Feb 24, 2014 #2


    User Avatar

    Staff: Mentor

    Hi miniake, Welcome to Physics Forums.

    I think the problem lies with your assumption that you can integrate the acceleration with respect to distance to yield velocity. Velocity is the integral of acceleration with respect to time.

    Have you considered using a conservation of energy approach to find the separation for v = 20.0 m/s ?
  4. Feb 24, 2014 #3
    Hello miniake

    What is the correct answer ?

    Edit :Conservation of energy is the correct way to approach the problem.
    Last edited: Feb 24, 2014
  5. Feb 24, 2014 #4
    Thank you gneill.

    It works, using the formula [itex]U_{1}+KE_{1} = U_{2}+KE_{2}[/itex].

    Thanks again.

    Using the above approach will work.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted