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Coloumb's Law Part II

  1. Nov 25, 2003 #1
    HEHE me again.. i was going to post this Q in the same thread as the last, but i didn't think you guys would see it...

    here's the Q... A point charge of 3.8x10^-6 is placed 0.20 m to the right of a charge of -2.0x10^-6. What is the force on a third charge of 2.3x10^-6 if it is placed where the third charge would experience a net force of zero.
  2. jcsd
  3. Nov 25, 2003 #2


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    I think you copied the problem incorrectly. If the third charge is placed so the force on it is zero, then the force on it is zero.

    - Warren
  4. Nov 25, 2003 #3
    oops here...

    A point charge of 3.8x10^-6 is placed 0.20 m to the right of a charge of -2.0x10^-6. What is the force on a third charge of 2.3x10^-6 if it is placed

    d) Where would the third charge experience a net force of zero?

    i got the answers for a), b) and c).. just need help on d)

    they changed the question basically to WHERE would the third object be placed to get a force where they equal each other so that the object doesn't move...
  5. Nov 25, 2003 #4


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    Do you understand (and can you explain to me) the principle of superposition?

    Can you explain to me what happens when a particle experiences two forces in opposite directions?

    - Warren
  6. Nov 25, 2003 #5
    no.. can't say that i do....
  7. Nov 25, 2003 #6


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    Okay. :smile: Let's start there.

    The first thing you need to know about electric fields is that they just add right on top of one another.

    Imagine that you have one particle all by itself -- a proton, say. It creates an electric field all around itself that tends to pull negative particles toward it. You know how to calculate the force that would be experienced by a particle in that field -- it's just Coulomb's law.

    Now, imagine that you add a second particle (of any arbitrary charge). What happens? Well, the field due to the first particle stays exactly the same -- it is completely unaffected by the new charge.

    The new charge, however, exerts a field of its own. It may be in opposition, or in concert, with the field of the first.

    Imagine that the fields created by both particles are like the transparencies you'd use on an overhead projector. You can put down a tranparency with the field due to the first particle drawn on it. Then you can stack another transparency with the field due to the second particle right on top. The total field is just the sum of the two individual fields. This is called the principle of superposition.

    So we've established that the total force on test particle (the third, movable charge in this problem) is divided up into two parts: some due to the first particle, and some due to the second.

    As you know from playing tug-o-war, if you pull on an object from both directions with equal force, the object won't move. This is what has to happen in this problem -- the first particle has to pull just as hard as the second, in the opposite direction.

    Let's assume that the x-axis is horizontal, and goes through both particles. Let's assume that the first charge, [tex]\inline{q_1 = 3.8 \cdot 10^{-6} C}[/tex] is at the origin of this coordinate system, while the second charge, [tex]\inline{q_2 = -2.0 \cdot 10^{-6} C}[/tex] is at the position [tex]\inline{x_2 = 0.20 m}[/tex].

    What's the total force experienced by the third, test particle?

    It's the sum of two copies of Coulomb's law:

    [tex]F_\textrm{total} = \frac{k q_1 q_3}{r_1^2} + \frac{k q_2 q_3}{r_2^2} = 0[/tex]

    where [tex]\inline{r_1}[/tex] is the distance between the first and third charges, and [tex]\inline{r_2}[/tex] is the distance between the second and third charges.

    All you need to do is solve for either [tex]\inline{r_1}[/tex] or [tex]\inline{r_2}[/tex]. Does this make sense?

    - Warren
  8. Nov 25, 2003 #7
    yeah i think i got it... thx again...
  9. Jan 19, 2004 #8
    slight variation?

    hi i have a slight variation to the same problem and was wondering how it might differently be solved:

    A positive charge of +5.4 µC is located at the origin, and a -2.0 µC charge is placed 17.0 cm away on the +x axis. At what location on the x-axis can a third charge be placed so that it experiences no net force?

    i understand superposition and the application of coulomb's law but i do not understand how to do this problem when the third charge is an unknown.

    any help would be suuper.
  10. Jan 19, 2004 #9
    Assume it to be q and solve hence
  11. Jan 19, 2004 #10
    no luck

    tried my good man, but no dice.

    any other suggestions?
  12. Jan 19, 2004 #11
    nevermind, im just a big dummy

    thanks for your help guy.
  13. Jan 20, 2004 #12
    While were on C's LAw

    Seeing as how this post is on E fields etc, I have a question about 2 charged Particles placed .3m apart and a third point is placed equidistant away from each point(3 points for an equilateral triangle) a 3rd charged particle is placed at Point P.
    I can get my E fields but I am left with 2 Vectors Therefore, I have to figure out the F at P4,I don't know how to multiply vectors, I have only added them to this point.
    FYI Q1+39 microC, Q2-26 microC Q3-36 microC. So know what and can an electric field be expressed as a vector.

    Thanks"feeling negatively Charged"
  14. Jan 20, 2004 #13
    You can start it as new post

    And what is P4 There is no multiplications involved as far as i know

    Any way

    [tex]\vec{E}=\frac{kq \vec{r}}{r^3}[/tex] where r has the direction from the charge to the point where field is to evaluated

    U should take the sign of charge as well
    Last edited: Jan 20, 2004
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