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Application to Coulomb's Law

  1. Aug 30, 2011 #1
    A 5C point charge located at (2m, 3m, 4m) and a 6C point charge located at (4m, 3m, 2m) together exert a net force on a 7C point charge located at (2m, 4m, 3m) is most closely represented by...

    (a) k(-6i + 15j - 10k) N
    (b) k(-8i + 13j - 12k) N
    (c) k(7i + 14j - 11k) N
    (d) k(7i - 14j - 11k) N


    I know that you have to use Coulomb's law however by adding the components i continue to get the wrong answer
     
  2. jcsd
  3. Aug 30, 2011 #2

    PeterO

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    Have you found the ratio of the magnitudes of the forces from each charge - I think I got 15:7, but did it quickly so may be in error

    Have you found separations correctly?

    Currently we don't know what you have found??

    we are not
     
  4. Aug 30, 2011 #3
    honestly my professor did not give me very much direction with this and the math is something i have not taken yet. with the info that i had i thought that i would have to find the force from the 5C charge and the 6C charge on the 7C charge and then add them up. When i do this i get odd answers
     
  5. Aug 30, 2011 #4

    PeterO

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    OK, post here how you got the force of the 5C on the 7C.

    even just the distance from the 5C to the 7C would be a start.
     
  6. Aug 30, 2011 #5

    collinsmark

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    Hello azia,

    Welcome to Physics Forums!

    Here's a few hints. First, take a look at coulombs law for two point charges:

    [tex] \vec F = k \frac{q_1 q_2}{r^2}\hat a_r [/tex]

    Note that F is a vector. But also note that r is not a vector in the above equation. r is a scalar. The scalar distance between the two point charges.

    Here [itex] \hat a_r [/itex] is a unit vector. You coursework might use different notation, such as maybe [itex] \hat r [/itex]. The "hat" on top is typically what distinguishes it as a unit vector in most notations. I'll call it [itex] \hat a_r [/itex] for now, but you should use whatever notation your coursework uses. A unit vector has a magnitude of 1 and a direction. for electrostatics, the direction points from the charge exerting the force to the charge that the force is acting upon (reverse to find the force on the first charge).

    [itex] \hat a_r [/itex] can be written in the notation
    [tex] \hat a_r = (a_x, \ a_y, \ a_z) = a_x \hat \imath + \ a_y \hat \jmath + \ a_z \hat k[/tex]

    Now those hints I mentioned.

    (1) Remember in Coulombs law that r is a scalar. You're going to need to find a separate r for each of two charges that exhibit a force on the third charge. You'll need to find the scalar distances. Well, specifically you'll need to find the square of each of these distances.

    (2) You will also need to find the unit vectors. It isn't too tough. For each case, it's just the position of the 7 C charge minus the position of the another charge, divided by the distance between them. Remember, these unit vectors will be vectors with a magnitude of 1 (and will be unit-less).

    (3) You'll need to do this twice. Once using the 5 C charge and another using the 6 C charge. You'll get two forces.

    (4) These forces are added together to create the final force. Just remember, they are vectors and you need to add the components individually.

    (5) The possible values that you must pick from in the problem statement are highly rounded. they are approximate.

    --------------------------
    Edit:

    By the way, some textbooks/coursework use the mathematical notation:

    [tex] \vec F = k \frac{q_1 q_2}{r^3} \vec r [/tex]

    which works fine too. But you'll still need to find the scalar distance, and the process of solving the problem is mostly the same.

    Use whichever method/notation your coursework uses.

    And by the way, please show your work and maybe we can help you figure out what, if anything, is going wrong.
     
    Last edited: Aug 30, 2011
  7. Aug 30, 2011 #6
    thank you collinsmark!! i was using the wrong formula. instead of using the unit vector i was just using the difference between the 2 vectors. I think once i correct that it should come out okay. Thank you again, i had spent a total of bout 3 hours on this. it's a lot easier than i made it
     
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