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Homework Help: Do i use coulomb's law?

  1. Dec 18, 2003 #1
    Problem 11.
    Find the magnitude electric field at a point midway between two charges of 34.4*10^9 C and 78.6*10^-9 C separated by a distance of 58.2 cm. Answer in N/C.
    Note: Do i use coulomb's law? If so, when i multiply the constant to the quiotent is that my answer?

    Problem 14.
    An electron moving through an electric field experiences an acceleration of 6*10^3m/S^2.
    a. Find the magnitude of the electric force acting on the electron. Answer in N.
    Note: Do i use F=mass*acceleration?
    b. What is the magnitude of the electric field strength? Answer in N/C.
    What formula do I use?
     
  2. jcsd
  3. Dec 18, 2003 #2
    E=qF will help u
    F=ma will do
     
  4. Dec 19, 2003 #3

    ShawnD

    User Avatar
    Science Advisor

    Re: physics

    Find the field from 1 then the field from the other then subtract.

    field from first charge:
    [tex]F = \frac{k q_1 q_2}{d^2}[/tex]

    One of the charges isn't there so just divide it out.
    [tex]\frac{F}{q_1} = \frac{k q_2}{d^2}[/tex]

    [tex]\frac{F}{q_1} = \frac{(9x10^9)(34.4x10^-^9)}{0.291^2}[/tex]

    [tex]\frac{F}{q_1} = 3656[/tex] N/C

    field from second charge:
    [tex]\frac{F}{q_1} = \frac{k q_2}{d^2}[/tex]

    [tex]\frac{F}{q_1} = \frac{(9x10^9)(78.6x10^-^9)}{0.291^2}[/tex]

    [tex]\frac{F}{q_1} = 8354[/tex] N/C

    field at that point:
    8354 - 3656 = 4698 N/C




    For question A:
    You know the formula [tex]F = ma[/tex]. You know the mass of an electron and its rate of acceleration. Sub in to find the force.

    For question B:
    The force is given by the field strength x charge. For field I'll just put L since I don't know what it should be.
    F = Lq
    L = F/q

    You solved the force in part A and you know the charge of an electron.
     
    Last edited: Dec 19, 2003
  5. Dec 19, 2003 #4
    Sorry

    I rechecked and for problem 11 it was 34.4*10^-9C.
     
  6. Dec 19, 2003 #5
    so the answer will be 4698 * 10^-9
     
  7. Dec 19, 2003 #6
    Question for problem 14.

    I had a=6*10^3 and m=9.109*10^-31. I subsituted those values for ma in F=ma and got 5.4654*10^-27. When i posted the answer I got it wrong. What did I do wrong?
     
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