Solve Electrostatic Force Problem

[PFStudent]

TITLE: Hard Electric Charge Problem.

Homework Statement

15. The charges and coordinates of two charged particles held fixed in an xy plane are q_{1} = +3.0 \muC, x_{1} = 3.5 cm, y_{1} = 0.50 cm, and q_{2} = -4.0 \muC, x_{2} = -2.0 cm, y_{2} = 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 q_{3} = +4.0 \muC 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:

<br /> \vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}<br />

Scalar Form:

<br /> |\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{12}}^2}<br />

The Attempt at a Solution

q_{1} = +3.0 \muC

x_{1} = 3.5 cm

y_{1} = 0.50 cm

q_{2} = -4.0 \muC

x_{2} = -2.0 cm

y_{2} = 1.5 cm

q_{3} = +4.0 \muC

x_{3} = ?

y_{3} = ?

(a)

<br /> |\vec{F}_{21}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{21}}^2}<br />

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

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

<br /> |\vec{F}_{21}| = \frac{k_{e}|q_{1}||q_{2}|}{(x_{2}-x_{1})^2+((y_{2}-y_{1})^2}<br />

<br /> sig. fig. \equiv 2<br />

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

(b)

<br /> tan\theta = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}<br />

<br /> \theta = arctan\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)<br />

<br /> sig. fig. \equiv 2<br />

\theta = -10 degrees

\theta = 3.5 x 10 degrees or 6.1 rad.

(c) and (d)

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

So here is how I approach these parts.

\Sigma \vec{F}_{3} = 0

0 = \vec{F}_{31} + \vec{F}_{32}

-\vec{F}_{31} = \vec{F}_{32}

|\vec{F}_{31}| = |\vec{F}_{32}|

<br /> \frac{k_{e}|q_{3}||q_{1}|}{{r_{31}}^2} = \frac{k_{e}|q_{3}||q_{2}|}{{r_{32}}^2}<br />

<br /> \frac{|q_{1}|}{|q_{2}|} = \frac{{r_{31}}^2}{{r_{32}}^2}<br />

<br /> \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}<br />

<br /> \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}<br />

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

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

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

In solving this problem, I referred to the following principle,
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Given any two arbitrary un-like sign charges: q_{1} and q_{2}, placed on an x-axis a distance L from each other. Then, the placement (on the x-axis) of a charge q_{3} such that the net force on q_{3} due to: q_{1} and q_{2}, will be zero. Can be given as follows,

q_{1}q_{2} &lt; 0 \therefore q_{1}q_{2} \equiv -

<br /> |q_{1}| &lt; |q_{2}|, |\vec{r}_{31}| &lt; |\vec{r}_{32}|, |\vec{r}_{32}| &gt; L<br />

<br /> |q_{1}| = |q_{2}|<br />, No equilibrium exists on x-axis.

<br /> |q_{1}| &gt; |q_{2}|, |\vec{r}_{31}| &gt; |\vec{r}_{32}|, |\vec{r}_{32}| &gt; L<br />
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<br /> |q_{1}| &lt; |q_{2}|<br />

Then,

<br /> |\vec{r}_{31}| &lt; |\vec{r}_{32}|<br />

Where,

<br /> |\vec{r}_{32}| &gt; L<br />

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

q_{1} and q_{3} = c

Now, going back to the original relationship,

<br /> \frac{|q_{1}|}{{r_{31}}^2} = \frac{|q_{2}|}{{r_{32}}^2}<br />

Now noting that: r_{31} = c and r_{32} = r_{12} + c

Then,

<br /> \frac{|q_{1}|}{{(c)}^2} = \frac{|q_{2}|}{{(r_{12} + c)}^2}<br />

Then, through some algebra the following result is arrived,

<br /> c = \frac{r_{12}}{\pm \sqrt{\frac{|q_{2}|}{|q_{1}|}-1}}<br />

Since, r_{12} \equiv distance r from 1 to 2.

Then,

<br /> r_{12} = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}<br />

Substituting,

<br /> c = \frac{\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}}{\pm \sqrt{\frac{|q_{2}|}{|q_{1}|}-1}}<br />

Letting, (for convenience) sig. fig. \equiv 4,

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: x_{3} and y_{3}?

Any help would be appreciated. :)

Thanks,

-PFStudent

 

[Dick]

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.

 

[PFStudent]

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,

<br /> r_{13} = c<br />

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,

<br /> (x_{3}, y_{3})?<br />

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

<br /> \vec{r}_{13} = x_{3}\hat{i}-y_{3}\hat{j}<br />

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

Thanks for the help, any more information is appreciated.

Thanks,

-PFStudent

 

[Dick]

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.