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!

Electrostatic Force

  1. Jul 9, 2007 #1
    TITLE: Hard Electric Charge Problem.

    1. The problem statement, all variables and given/known data

    15. The charges and coordinates of two charged particles held fixed in an xy plane are [itex]q_{1}[/itex] = +3.0 [itex]\mu[/itex]C, [itex]x_{1}[/itex] = 3.5 cm, [itex]y_{1}[/itex] = 0.50 cm, and [itex]q_{2}[/itex] = -4.0 [itex]\mu[/itex]C, [itex]x_{2}[/itex] = -2.0 cm, [itex] y_{2}[/itex] = 1.5 cm. Find the
    (a) magnitude and
    (b) direction of the electrostatic force on particle 2 due to particle 1. At what
    (c) x and
    (d) y coordinates should a third particle of charge [itex]q_{3}[/itex] = +4.0 [itex]\mu[/itex]C be placed such that the net electrostatic force on particle 3 due to particle 1 and 2 is zero.

    2. Relevant equations

    Coulomb's Law:

    Vector Form:

    [tex]
    \vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}
    [/tex]

    Scalar Form:

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

    3. The attempt at a solution

    [itex]q_{1}[/itex] = +3.0 [itex]\mu[/itex]C

    [itex]x_{1}[/itex] = 3.5 cm

    [itex]y_{1}[/itex] = 0.50 cm

    [itex]q_{2}[/itex] = -4.0 [itex]\mu[/itex]C

    [itex]x_{2}[/itex] = -2.0 cm

    [itex] y_{2}[/itex] = 1.5 cm

    [itex]q_{3}[/itex] = +4.0 [itex]\mu[/itex]C

    [itex]x_{3}[/itex] = ?

    [itex] y_{3}[/itex] = ?

    (a)

    [tex]
    |\vec{F}_{21}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{21}}^2}
    [/tex]

    [itex]r_{21} = \sqrt{(x_{2}-x_{1})^2+((y_{2}-y_{1})^2}[/itex]

    [tex]
    |\vec{F}_{21}| = \frac{k_{e}|q_{1}||q_{2}|}{(\sqrt{(x_{2}-x_{1})^2+((y_{2}-y_{1})^2})^2}
    [/tex]

    [tex]
    |\vec{F}_{21}| = \frac{k_{e}|q_{1}||q_{2}|}{(x_{2}-x_{1})^2+((y_{2}-y_{1})^2}
    [/tex]

    [tex]
    sig. fig. \equiv 2
    [/tex]

    [tex]
    |\vec{F}_{21}| = 35{\textcolor[rgb]{1.00,1.00,1.00}{.}}N
    [/tex]

    (b)

    [tex]
    tan\theta = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}
    [/tex]

    [tex]
    \theta = arctan\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)
    [/tex]

    [tex]
    sig. fig. \equiv 2
    [/tex]

    [itex]\theta[/itex] = -10 degrees

    [itex]\theta[/itex] = 3.5 x 10 degrees or [itex]6.1 rad.[/itex]

    (c) and (d)

    Ok, here is where I am stuck. I can’t find the [itex](x_{3}, y_{3})[/itex] such that the net force on [itex]q_{3}[/itex] will be zero.

    So here is how I approach these parts.

    [tex]\Sigma \vec{F}_{3} = 0[/tex]

    [tex]0 = \vec{F}_{31} + \vec{F}_{32}[/tex]

    [tex]-\vec{F}_{31} = \vec{F}_{32}[/tex]

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

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

    [tex]
    \frac{|q_{1}|}{|q_{2}|} = \frac{(\sqrt{(x_{3}-x_{1})^2 + (y_{3}-y_{1})^2})^2}{(\sqrt{(x_{3}-x_{2})^2+ (y_{3}-y_{2})^2})^2}
    [/tex]

    [tex]
    \frac{|q_{1}|}{|q_{2}|} = \frac{(x_{3}-x_{1})^2 + (y_{3}-y_{1})^2}{(x_{3}-x_{2})^2+ (y_{3}-y_{2})^2}
    [/tex]

    Two unknowns, one equation…, so then I figured that [itex]q_{3}[/itex] can only be placed on the line of force through [itex]q_{1}[/itex] and [itex]q_{2}[/itex].

    The line of force is the imaginary axis (line) through [itex]q_{1}[/itex] and [itex]q_{2}[/itex].

    Therefore, we essentially have a simple single axis problem where we need to find where on the same axis a charge [itex]q_{3}[/itex] can be placed such that the force on it due to [itex]q_{1}[/itex] and [itex]q_{2}[/itex] is zero.

    In solving this problem, I referred to the following principle,
    ---------------------------------------------------------------------------------
    Given any two arbitrary un-like sign charges: [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]

    [tex]
    |q_{1}| < |q_{2}|, |\vec{r}_{31}| < |\vec{r}_{32}|, |\vec{r}_{32}| > L
    [/tex]

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

    [tex]
    |q_{1}| > |q_{2}|, |\vec{r}_{31}| > |\vec{r}_{32}|, |\vec{r}_{32}| > L
    [/tex]
    ---------------------------------------------------------------------------------

    [tex]
    |q_{1}| < |q_{2}|
    [/tex]

    Then,

    [tex]
    |\vec{r}_{31}| < |\vec{r}_{32}|
    [/tex]

    Where,

    [tex]
    |\vec{r}_{32}| > L
    [/tex]

    Therefore, the charge [itex]q_{3}[/itex] must be placed (on the line of force) to the right of [itex]q_{3}[/itex]. We then let the distance between,

    [itex]q_{1}[/itex] and [itex]q_{3}[/itex] = c

    Now, going back to the original relationship,

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

    Now noting that: [itex]r_{31} = c[/itex] and [itex]r_{32} = r_{12} + c[/itex]

    Then,

    [tex]
    \frac{|q_{1}|}{{(c)}^2} = \frac{|q_{2}|}{{(r_{12} + c)}^2}
    [/tex]

    Then, through some algebra the following result is arrived,

    [tex]
    c = \frac{r_{12}}{\pm \sqrt{\frac{|q_{2}|}{|q_{1}|}-1}}
    [/tex]

    Since, [itex]r_{12} \equiv[/itex] distance r from 1 to 2.

    Then,

    [tex]
    r_{12} = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}
    [/tex]

    Substituting,

    [tex]
    c = \frac{\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}}{\pm \sqrt{\frac{|q_{2}|}{|q_{1}|}-1}}
    [/tex]

    Letting, (for convenience) [itex]sig. fig. \equiv 4[/itex],

    c = 0.3614 m, -0.024 m

    Note: Real distance cannot be negative, therefore,

    c = 0.3614 m

    Ok, so I am now stuck, how am I supposed to use this distance, c to find the coordinates: [itex]x_{3}[/itex] and [itex]y_{3}[/itex]?

    Any help would be appreciated. :)

    Thanks,

    -PFStudent
     
    Last edited: Jul 10, 2007
  2. jcsd
  3. Jul 9, 2007 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Your logic is impeccable and the c is correct. To find a position c away from P1 in the direction pointed to by P1-P2, form a vector c*(P1-P2)/sqrt((P1-P2)^2) and add it to P1. Do you see how that works? It's c times a unit vector pointing the correct direction added to the initial position.
     
    Last edited: Jul 9, 2007
  4. Jul 10, 2007 #3
    Hey,

    Thanks for the information, I looked over what you suggested and I couldn't quite, understand what you were trying to say.

    So far, this is as much as I could gather, that given a distance,

    [tex]
    r_{13} = c
    [/tex]

    How is it that I construct a unit vector pointing in the correct direction and add that to the initial position? How do I do that? How does that lead me to to the coordinates,

    [tex]
    (x_{3}, y_{3})?
    [/tex]

    As far as the distance vector for [itex]q_{1}[/itex] to [itex]q_{3}[/itex], isn't that the following,

    [tex]
    \vec{r}_{13} = x_{3}\hat{i}-y_{3}\hat{j}
    [/tex]

    I don't understand how this helps, me solve for [itex](x_{3}, y_{3})[/itex].

    Thanks for the help, any more information is appreciated.

    Thanks,

    -PFStudent
     
    Last edited: Jul 10, 2007
  5. Jul 10, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok. P1=(x1,y1), P2=(x2,y2). P1-P2=(x1-x2,y1-y2) is a vector that points in the direction from P2 towards P1. It has length L=sqrt((x1-x2)^2+(y1-y2)^2) (the distance between P1 and P2). So (P1-P2)/L points in the same direction and has length 1 (its a unit vector). So c*(P1-P2)/L points in the same direction and has length c. So P1+c*(P2-P1)/L lies along the line between P2 and P1 but c units outside of the interval between P1 and P2 and nearest to P1. And I think that's what you want.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Electrostatic Force
  1. Electrostatic force (Replies: 1)

  2. Electrostatic force (Replies: 6)

  3. Electrostatic force (Replies: 13)

  4. Electrostatic Force (Replies: 3)

  5. Electrostatic Forces (Replies: 1)

Loading...