1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Charge Problem, Is this right?

  1. Jul 4, 2007 #1
    1. The problem statement, all variables and given/known data

    13. In. Fig. 21-26, particle 1 of charge +1.0 [itex]\mu[/itex]C and particle 2 of charge -3.0 [itex]\mu[/itex]C are held at separation of L = 10.0 cm on an x axis. If particle 3 of known charge [itex]q_{3}[/itex] is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the
    (a) x and
    (b) y coordinates?

    [​IMG]

    2. Relevant equations

    Coulomb's Law

    Vector Form:

    [tex]
    \vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{r^2}\hat{r}_{12}
    [/tex]

    Scalar Form:

    [tex]
    |\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{r^2}
    [/tex]

    3. The attempt at a solution

    [itex]q_{1} = +1.0 x 10^{-6} C[/itex]
    [itex]q_{2} = -3.0 x 10^{-6} C[/itex]
    [itex]L = 0.100 m[/itex]
    [itex]x = ?[/itex]
    [itex]y = ?[/tex]

    There can only be two possible scenarios,

    Scenario A

    [​IMG]

    and

    Scenario B

    [​IMG]

    The reason that charge [itex]q_{3}[/itex] can not be between charges [itex]q_{1}[/itex] and [itex]q_{2}[/itex] is because the forces would not balance each other but instead would add vectorally in the same direction, and therefore the net force on charge [itex]q_{3}[/itex] would not be zero.

    Beginning with scenario A,

    [tex]
    |\vec{F_{31}}| = |\vec{F_{32}}|
    [/tex]

    [itex]r_{31} = L+x[/itex] and [itex]r_{32} = x[/itex]

    [tex]
    \frac{k_{e}|q_{3}||q_{1}|}{r_{31}^2} = \frac{k_{e}|q_{3}||q_{2}|}{r_{32}^2}
    [/tex]

    [tex]
    \frac{|q_{1}|}{(L+x)^2} = \frac{|q_{2}|}{(x)^2}
    [/tex]

    Solving for x,

    [tex]
    x = \frac{-L}{1\mp\sqrt{\frac{|q_{1}|}{|q_{2}|}}}
    [/tex]

    And evaluating for x with [itex]sig. fig. \equiv 2[/itex],

    x = -0.24 m, -0.06 m

    Continuing with scenario B,

    [tex]
    |\vec{F_{31}}| = |\vec{F_{32}}|
    [/tex]

    [itex]r_{31} = x[/itex] and [itex]r_{32} = L+x[/itex]

    [tex]
    \frac{k_{e}|q_{3}||q_{1}|}{r_{31}^2} = \frac{k_{e}|q_{3}||q_{2}|}{r_{32}^2}
    [/tex]

    [tex]
    \frac{|q_{1}|}{(x)^2} = \frac{|q_{2}|}{(L+x)^2}
    [/tex]

    Solving for x,

    [tex]
    x = \frac{\pm L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{1\mp\sqrt{\frac{|q_{1}|}{|q_{2}|}}}
    [/tex]

    Now simplifying by factoring out a [itex]\mp[/itex] out of the denominator,

    [tex]
    x = \frac{\pm L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{\mp\left(\mp1+\sqrt{\frac{|q_{1}|}{|q_{2}|}}\right)}
    [/tex]

    Now noting that the following is true,

    [tex]
    \frac{\pm}{\mp} = \frac{+}{-} or \frac{-}{+} = -
    [/tex]

    Our equation simplifies to

    [tex]
    x = \frac{-L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{\left(\mp1+\sqrt{\frac{|q_{1}|}{|q_{2}|}}\right)}
    [/tex]

    [tex]
    x = \frac{-L\sqrt{\frac{|q_{1}|}{|q_{2}|}}}{\sqrt{\frac{|q_{1}|}{|q_{2}|}}\mp1}
    [/tex]

    And evaluating for x with [itex]sig. fig. \equiv 2[/itex],

    x = +0.14 m, -0.04 m

    Therefore, of scenarios A and B there are four possible answers,

    A: x = -0.24 m, -0.06 m
    B: x = +0.14 m, -0.04 m

    The book lists, x = 0.14 m as the answer, but that can’t be the only answer as that only fits with scenario B.

    Then, if x= 0.14 m is the only answer what justifies it to be so out of all the other solutions, is it because it is the only positive solution?
    What is the other answer for scenario A?
    And the answer for scenario B (x = 0.14m) seems to fit more with scenario A, then why did I not get this answer while solving for solutions in scenario A?

    Are the other solutions for x real solutions to this problem?

    Thanks,
    -PFStudent
     
  2. jcsd
  3. Jul 4, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You solved it perfectly, you just have to correctly interpret what you found.
    Realize that these negative values for x violate your assumption of being to the right of q_2. That means there is no solution for scenario A.

    This should make some intuitive sense. You used good reasoning to eliminate searching for solutions that place q_3 between the other charges. You can also reason that since q_2 > q_1, putting q_3 to the right of q_2 (your scenario A) will always keep the force between q_2 & q_3 greater than the force between q_1 & q_3. Conclusion: There is no way to get a net force of zero on q_3 in scenario A.
     
    Last edited: Jul 4, 2007
  4. Jul 7, 2007 #3
    Hey,

    Thanks for the information Doc Al.

    Ok, so knowing that placing [itex]q_{3}[/itex] between: [itex]q_{1}[/itex] and [itex]q_{2}[/itex], will cause the forces exerted by: [itex]q_{1}[/itex] and [itex]q_{2}[/itex], on [itex]q_{3}[/itex] to be in the same direction.

    In addition, noting that because [itex]|q_{2}| > |q_{1}|[/itex] then placing [itex]q_{3}[/itex] to the right of [itex]q_{2}[/itex] is ruled out because, everywhere (on the x-axis) to right of [itex]q_{2}[/itex] the force, [itex]\vec{F}_{23} > \vec{F}_{13}[/itex]. This is, because placing [itex]q_{3}[/itex] to the right of [itex]q_{2}[/itex] creates a dynamic such that as [itex]q_{3}[/itex] approaches [itex]q_{2}[/itex] the force, [itex]\vec{F}_{23} \rightarrow \infty {\textcolor[rgb]{1.00,1.00,1.00}{.}} as {\textcolor[rgb]{1.00,1.00,1.00}{.}} \vec{F}_{13} \rightarrow \max(\vec{F}_{13})[/itex].

    Then, placing [itex]q_{3}[/itex] to the left of [itex]q_{1}[/itex] is the only possible solution, because if [itex]q_{3}[/itex] is placed at [itex]q_{1}[/itex]’s location and begins moving away (to the left) from [itex]q_{1}[/itex] the force [itex]F_{13}[/itex] approaches zero (from infinity), however because [itex]|q_{1}| < |q_{2}|[/itex], then as [itex]\vec{F}_{13}[/itex] decreases there exists a single point such that [itex]\vec{F}_{13}[/itex] will balance [itex]\vec{F}_{12}[/itex].

    Therefore, q_{3} must be placed to the left of q_{1} on the x-axis in order for q_{3} to experience a net force of zero, due to q_{1} and q_{2}.

    Now going back to my solution,

    Scenario A.

    [​IMG]

    For scenario A, because I now know there is no solution to this scenario for the placement of q_{3}, and because I am calculating a (real and therefore positive) distance, x it (now) makes intuitive sense that negative answers (x = -0.24 m, -0.06 m) for x just mean that there is no solution to this scenario.

    Is this the correct interpretation? Also, since this scenario has no solutions, what significance do the negative solutions mean? None? And in addition, why is it that I even arrived at solutions (negative numbers) as opposed to imaginary solutions? If there really are no solutions to this scenario, should not my solutions have been imaginary rather than negative (even though I understand that both imply there is no real-world solution to this problem)?

    Scenario B.

    [​IMG]

    And now going back to scenario B, because I (now) realize that this is the correct approach, then in interpreting my solutions, x = +0.14 m, -0.04 m. I now understand that only x = 0.14 m, can be the correct solution, because the other solution is negative which implies a non-real distance and therefore cannot be a real solution to this problem.

    Finally, then the answer is x = 0.14 m.
    Which correctly interpreted, means that charge q_{3} must be placed 0.14 m on the x-axis to the left of charge q_{1} in order for the net force on q_{3} to be zero due to: q_{1} and q_{2}.

    Then continuing on with part b, it can be shown that placing q_{3) at any other y-coordinate other than zero, will not result in a net force of zero. This is because at any other y-coordinate (other than zero) the forces, [itex]|\vec{F}_{13}_{y}|[/itex] and [itex]|\vec{F}_{23}_{y}|[/itex] would not balance each other, except only at zero. This is because only at y = 0, [itex]|\vec{F}_{13}_{y}| = |\vec{F}_{23}_{y}|= 0[/itex].

    Therefore, although my book gives the coordinates for q_{3} as follows,

    (a) x = 14 cm
    (b) y = 0 cm

    Would not the more correct solution for this problem be,

    (a) x = -0.14 m
    (b) y = 0.0 m

    Therefore, [itex]q_{3}(-0.14{\textcolor[rgb]{1.00,1.00,1.00}{.}}m,{\textcolor[rgb]{1.00,1.00,1.00}{.}}0.0{\textcolor[rgb]{1.00,1.00,1.00}{.}}m)[/itex].

    Since, in the book’s figure they placed the y-axis going through q_{1}?

    Thanks,

    -PFStudent
     
    Last edited: Jul 7, 2007
  5. Jul 7, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, perfectly correct.

    Realize that the negative answers are perfectly valid solutions to your equations. But those equations only correctly model the situation in the region of applicability. For instance, the field from q1 changes sign when you move from the left to the right of it, so the equations used don't really apply in that region.

    This sort of thing is pretty common. Toss a ball at some angle and calculate when it will hit the ground. You'll get two answers: one positive, one negative. Both are valid solutions to the equation, but the equation only models the real world if time goes forward. Make sense?


    All good.

    Yes--you mean y = 0 m for (b) of course. Looks like the book was sloppy.
     
  6. Jul 7, 2007 #5
    Hey,

    Thanks for the explanation Doc Al.

    Ah. That's right, they (the solutions) are valid mathematical solutions that make the equation true. However, because my equations model the real world, not all mathematical solutions will be real-world solutions.

    I never looked at (mechanics) physics equations like that before, but it makes sense. I wonder how I would prove the negative is solution is valid if I let time flow backward though...that sounds like an interesting problem.

    Ok, x = -0.14 m (note the negative sign) is correct? It should be right because of how they placed the y-axis and because they were asking for the x-coordinate.

    Also, if I generalized this problem for any arbitrary un-like sign charge: [itex]q_{1}[/itex] and [itex]q_{2}[/itex], placed on an x-axis a distance L from each other. Then, the placement (on the x-axis) of a charge [itex]q_{3}[/itex] such that the net force on [itex]q_{3}[/itex] due to: [itex]q_{1}[/itex] and [itex]q_{2}[/itex], will be zero. Can be given as follows,

    [itex]q_{1}q_{2} < 0[/itex] [itex]\therefore[/itex] [itex]q_{1}q_{2} \equiv -[/itex]

    [itex]|q_{1}| < |q_{2}|, |\vec_{r}_{31}| < |\vec_{r}_{32}|[/itex]

    [itex]|q_{1}| = |q_{2}|, [/itex] No equilibrium exists on x-axis.

    [itex]|q_{1}| > |q_{2}|, |\vec_{r}_{31}| > |\vec_{r}_{32}|[/itex]

    I pretty sure the above always holds.

    Then, if the above is true, can I further generalize this problem for any arbitrary charges: [itex]q_{1}[/itex] and [itex]q_{2}[/itex]? I was trying this the other night and it got pretty long and tedious.

    Thanks,

    -PFStudent
     
    Last edited: Jul 7, 2007
  7. Jul 7, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Absolutely: x = -0.14 m is correct.

    Good. In addition: [itex]|\vec_{r}_{32}| > L[/itex]

    Good. In addition: [itex]|\vec_{r}_{31}| > L[/itex]

    I don't see why you couldn't generalize it for arbitrary charges, but I doubt it's worth your while. You've got this one nailed; move on to something else. :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electric Charge Problem, Is this right?
Loading...