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Need phyeics help (electric charge/Forces)

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    K i need help with like 3 phyics problem on electric forces / charges.

    1. One gram of copper has 9.48 10^21 atoms, and each copper atom has 29 electrons.

    (a) How many electrons are contained in 1.08 g of copper?
    _________ electrons

    b. What is the total charge of these electrons?
    __________ C

    2. Calculate the net charge on a substance consisting of a combination of 7.5 10^13 protons and 4.2 10^13 electrons.
    ___________ C

    3. An alpha particle (charge = +2.0e) is sent at high speed toward a gold nucleus (charge = +79e). What is the electric force acting on the alpha particle when the alpha particle is 4.8 10^-14 m from the gold nucleus?
    _______________ N

    2. Relevant equations

    Elctron -1.6 x 10^-19
    Proton: 1.6 x 10^-19
    Coulombs Law: Felectric = Kc (q1*q2)/r^2

    3. The attempt at a solution

    alright iv been having trouble since our teacher did not teach us anything about these barley and the book doesnt help at all.

    1st one i dont know where to start.

    2nd one i tried just simply adding them together and that didnt work and then i tried using pythagorm theorm because iwas maybe thinking that the net force you need to find like the triange.

    3rd one i plugged into the columbs law and i got 6.2x10^38 and i think i got that wrong too?

    Any help please?
  2. jcsd
  3. Mar 8, 2009 #2
    ight well i actully figured out the last two. but i still can seem to get the 1st one. iv tried all that i can think of can some one help me with that please?
  4. Mar 8, 2009 #3


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    For the first one they tell you how many atoms per gram. They tell you how many -e per atom ... where is the problem?
  5. Mar 8, 2009 #4
    the problem is i get the first part of the question, i do 29 * 9.48e21 then multiply it by 1.08 and get my answer 2.962e23

    then how do i find the total charge? do i multiply that by -1.6e-19?
  6. Mar 8, 2009 #5


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