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 Force/ Electrical Fields

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    How many electrons must be removed from each two of 5.81kg copper spheres to make the electric force of repulsion between them equal to the magnitude to the gravitational attraction between them

    2. Relevant equations

    F: Ke (q1)(q2)/r^2

    3. The attempt at a solution
    ( 8.99x10^9)( 9.11x10^-31)(1.67x 10^-27)/ 2 (5.81)
    = 3.97x 10^ -47
     
  2. jcsd
  3. Feb 6, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Maybe you should write out the variables for the equations and simplify before throwing the numbers in?

    Besides I'm not familiar with what a 10-47 of an electron looks like.
     
  4. Feb 7, 2009 #3
    i'm not sure wat to do? probably F=ma then F=Ke(q1)(q2)/r sqed
     
  5. Feb 7, 2009 #4
    Gravitational force has its own expression,
    http://www.physicsclassroom.com/Class/circles/u6l3c.cfm
    If you're not familiar with it already.

    And I suppose for the question that you neglect the mass being removed when you remove electrons? It's a very small mass, but still a mass.

    Anyway, the way I understand the question is that the same amount of electrons is being removed from the spheres, which means one thing, which simplifies everything; the spheres will have the same charge.

    You've got the expression for gravitational force and electric force. Relate them in the way the question asks you to.
     
  6. Feb 7, 2009 #5
    (m)(a)=F then G=(m1)(m2)/d sqed????? or F: Ke (q1)(q2)/r^2 or how would i use these?
     
  7. Feb 7, 2009 #6

    LowlyPion

    User Avatar
    Homework Helper

    As Hannisch suggests consider the gravitational force of attraction between two 5.8 kg masses.

    That must equal the force of repulsion at the same distance - at whatever r because they are both proportional to 1/r2.

    Set them equal and discard the terms that cancel. Then solve.
     
  8. Feb 7, 2009 #7
    so do i start with F= 6.673x 10^-11 * 5.81/ d sq, but d sq cancels, but then (5.81) a= F, therefore 3.87x 10^-10 = F= (5.81) a, ?
     
  9. Feb 7, 2009 #8

    LowlyPion

    User Avatar
    Homework Helper

    No.

    GMm/r2 = kqq/r2

    The masses are equal. The charges equal.

    q = m*(G/k)1/2
     
  10. Feb 7, 2009 #9
    therefore is i solve for q, then (5.81)(( 6.673x 10^11/8.99x10^9)) ^1/2
     
  11. Feb 7, 2009 #10
    so that would be 1.74 electrons?
     
  12. Feb 7, 2009 #11

    LowlyPion

    User Avatar
    Homework Helper

    I'm not sure how you got something bigger than 1 times 5 to equal 1.74.
     
  13. Feb 7, 2009 #12
    wait if i plug in the equation u gave me then i get 2.15 x 10^-20
     
  14. Feb 7, 2009 #13

    LowlyPion

    User Avatar
    Homework Helper

    I haven't the slightest idea what you calculated.

    But I am fairly certain you didn't calculate the square root correctly.

    As an example (4*10-6)1/2 will yield 2*10-3
     
  15. Feb 7, 2009 #14
    (5.81)(( 6.673x 10^11/8.99x10^9)) ^1/2 ----------> 6.673x 10^11/8.99x10^9)) ^1/2=8.61x 10^-11 then times 5.81 equals 5x 10^-10
     
  16. Feb 7, 2009 #15

    LowlyPion

    User Avatar
    Homework Helper

    G is 10-11 but you got the right number.

    But what are your units?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electric Force/ Electrical Fields
Loading...