Electrostatic Force

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TITLE: Hard Electric Charge Problem.

Homework Statement



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.

Homework 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]

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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
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:
  • #3
170
0
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.

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:
  • #4
Dick
Science Advisor
Homework Helper
26,263
619
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.
 

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