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Homework Help: Coulomb's Law particle charge

  1. Jan 13, 2006 #1
    A particle with charge 5[tex]\mu[/tex]C is located on the x-axis at the point -6 cm, and a second particle with charge -8 [tex]\mu[/tex]C is placed on the x-axis at 10 cm. The Coulomb constant is 8.9875 x 10^9 Nm^2/C^2. What is the magnitude of the total electrostatic force on a third particle with charge 9[tex]\mu[/tex]C placed on the x-axis at -2 cm. Answer in units of N.
    First I converted everything.
    -8[tex]\mu[/tex]C= -8 x 10^-6 C
    5 [tex]\mu[/tex]C= 5 x 10^-6 C
    9 [tex]\mu[/tex]C= 9 x 10^-6 C
    -6 cm= -.06 m
    -2 cm= -.02 m
    10 cm=.10 m
    then I drew a free body diagram for the forces on the third particle and found that both forces point to the right, since the first particle has a positive charge, and the second has a negative charge.
    Then I used Coulomb's Law,
    For the first charge,
    [tex] F= 1/4\pi \epsilon * (5 x 10^-6 * 9 x 10^-6) / (.04)^2 [/tex]
    Simplifying this gave me [tex] 1/4\pi \epsilon * 2.81 x 10^-8 [/tex]
    Plugging in 8.9875 x 10^9 for E and solving gave me 2.48 x 10-19.
    Then I did the same thing for the next charge.
    [tex] F= 1/4\pi \epsilon * (-8 x 10^-6 * 9 x 10^-6)/ (.12)^2 [/tex]
    so [tex] 1/4\pi \epsilon * 5 x 10^-9 [/tex]
    Plugging in for E and solving gave me 4.43 x 10^-20.
    I then added these together, and got 2.923 x 10^-19 N, which wasn't right.
    Can someone please help me? I don't understand electrostatics at all!
  2. jcsd
  3. Jan 13, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I haven't checked your arithmetic, but I suspect the problem is here:
    Realize that the Coulomb constant [tex]k = 1/(4\pi \epsilon_0)[/tex].

    Realize that that number is k, not [tex]\epsilon_0[/tex].

    Other than that, your solution looks OK.
  4. Jan 13, 2006 #3
    Thanks so much Doc Al. Now I'm finally getting some right answers on my homework. :cool:
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