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Homework Help: Electric charge

  1. Jan 14, 2006 #1
    Four point charges, each of magnitude 5.5[tex]\mu[/tex]C, are placed at the corners of a square 76.5 cm on a side. The value of Coulomb's constant is 8.9875 x 10^9 Nm^2/C^2. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Answer in units of N.

    First I drew a free body diagram and found that the negative charge would be attracted to all of the other charges. So then I used Coulomb's Law.
    F_14= [tex] 1/4\pi\epsilon_o * (5.5 x 10^-6 *5.5 x 10^-6)/ .765^2. [/tex] I know that E_o = 8.85 x 10^-12, by setting 8.9875 x 10^9= [tex] 1/4\pi \epsilon_o. [/tex]
    Solving this gave me 4.58 x 10^-22.
    This would be the force for 2 of the charges, since all of the charges and distances are equal, except one force is in the x-direction, and the other is in the y-direction.
    Next I solved for the other force, which would be the diagonal from the negative charge. I found the distance between the two charges to be 1.08 m by using the pythagorean theorem.
    I split it up into x and y components and used Coulomb's Law.
    In the x-direction:
    F= [tex] 1/4\pi \epsilon_0 * (5.5 x 10^-6 * 5.5 x 10^-6) / 1.08^2 * (.765/1.08). [/tex] and I got 1.83 x 10^-11.
    Since it is a square, I got the y-direction to be the exact same thing.
    Then I added together the components for both parts to get 1.83 x 10^-11 in both directions. Then I did the [tex] \sqrt (1.83 x 10^-11)^2 +(1.83 x 10^-11)^2 [/tex] for my final force and found it to be 2.59 x 10^-11, which isn't right. Can someone help me? thanks in advance.
     
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  3. Jan 14, 2006 #2

    Astronuc

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    [tex] 1/4\pi \epsilon_0 * (5.5 x 10^-6 * 5.5 x 10^-6) / 1.08^2 * (.765/1.08). [/tex]

    This does not seem right. The distance of the diagonal is 1.08 m, but why multiply by 0.765/1.08?

    Since the two + charges on the adjacent legs are equally apart and perpendicular, the net force will be along the diagonal.
     
  4. Jan 14, 2006 #3
    I multiplied it by .765/1.08 because that is what the sin/cos is equal to. I thought I needed to have that for the x and y directions.
    Do I just take that part out from each of the forces I calculated and add everything together?
    I'm confused on where to go from here.
     
  5. Jan 14, 2006 #4

    lightgrav

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    I got .465 N on my calculator, for the nearby charges.
    You know that the Force by the farther charge is LESS ...

    Re-key your computation!

    the Coulomb constant k = 9E9 Nm^2/C^2 is easier to use than epsilon ...
     
  6. Jan 14, 2006 #5
    Ok I got the .465 for the two nearest charges, and I got .233 for the charge that is furthest away. Do I add these together? Or do I need to break the one that's furthest into components?
     
  7. Jan 14, 2006 #6

    lightgrav

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    Do you want the answer in (F_x , F_y) components, or magnitude?
    Either split the small one into x,y (before adding to the big ones);
    or add the two big ones by Pythagoras (before adding to the small one)
     
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