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!

Calculating extra electrons

  1. Apr 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A 2.9 x 10^-3 kg latex sphere is moving up, at a constant velocity of 0.013 mm/s, between two large plates. The plates are 1.500 cm apart, with a potential difference of 3.00 x 10^2 V between them. How many extra electrons does the sphere have?


    2. Relevant equations
    N = q/e
    q = (mgr)/[tex]\Delta[/tex]V



    3. The attempt at a solution
    Attempt 1:
    m = 2.9 x 10^-3 kg
    v = 0.013 mm/s = 1.3 x 10^-5 m/s
    r = 1.500 cm = 0.015 m
    [tex]\Delta[/tex]V = 3.00 x 10^2 V

    q = (mgr)/[tex]\Delta[/tex]V
    q = [ (2.9 x 10^-3 kg)(9.80 m/s^2)(0.015 m) ]/ 3.00 x 10^2 V
    q = 1.421 x 10^-16 C
    N = q/e
    N = (1.421 x 10^-16 C )/(1.602 x 10^-16 C)
    N = 8.9 x 10^2 extra electrons

    Attempt 2:

    [tex]\Delta[/tex]Ee = [tex]\Delta[/tex]Ek
    q[tex]\Delta[/tex]V = mv^2/2
    q = mv^2/2[tex]\Delta[/tex]V
    q = [ (2.9 x 10^-3 kg)(1.3 x 10^-5 m/s)^2 ]/ [ 2(3.00 x 10^2 V) ]
    q = 8.168 x 10^-26 C
    N = q/e
    N = (8.168 x 10^-26 C )/(1.602 x 10^-16 C)
    N = 5.1 x 10^-7 extra electrons

    Not too sure if what I have done so far is right or not. But any help will be appreciated.
     
  2. jcsd
  3. Apr 16, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi ProtoXtype,

    Your approach in attempt 1 looks okay to me; however I believe you have a few errors. When you solved for q you say you got 1.421 x 10^-16; this exponent -16 here does not look right; I think you just misread your calculator.

    A couple of lines below that you wrote the charge on an electron as 1.6 x 10^-16, but it looks like maybe you used the correct value in your calculation.


    For your attempt 2, the first equation is not correct as it leaves out the change in gravitational potential energy. If you put that in and also set the change in kinetic energy = 0 (since the object is moving at constant speed) you'll find yourself with the expression you used in attempt 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculating extra electrons
Loading...