# Electrostatic Force

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 $\mu$C, $x_{1}$ = 3.5 cm, $y_{1}$ = 0.50 cm, and $q_{2}$ = -4.0 $\mu$C, $x_{2}$ = -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 $\mu$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:

$$\vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}$$

Scalar Form:

$$|\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{12}}^2}$$

## The Attempt at a Solution

$q_{1}$ = +3.0 $\mu$C

$x_{1}$ = 3.5 cm

$y_{1}$ = 0.50 cm

$q_{2}$ = -4.0 $\mu$C

$x_{2}$ = -2.0 cm

$y_{2}$ = 1.5 cm

$q_{3}$ = +4.0 $\mu$C

$x_{3}$ = ?

$y_{3}$ = ?

(a)

$$|\vec{F}_{21}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{21}}^2}$$

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

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

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

$$sig. fig. \equiv 2$$

$$|\vec{F}_{21}| = 35{{.}}N$$

(b)

$$tan\theta = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

$$\theta = arctan\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$

$$sig. fig. \equiv 2$$

$\theta$ = -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.

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

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

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

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

$$\frac{k_{e}|q_{3}||q_{1}|}{{r_{31}}^2} = \frac{k_{e}|q_{3}||q_{2}|}{{r_{32}}^2}$$

$$\frac{|q_{1}|}{|q_{2}|} = \frac{{r_{31}}^2}{{r_{32}}^2}$$

$$\frac{|q_{1}|}{|q_{2}|} = \frac{(\sqrt{(x_{3}-x_{1})^2 + (y_{3}-y_{1})^2})^2}{(\sqrt{(x_{3}-x_{2})^2+ (y_{3}-y_{2})^2})^2}$$

$$\frac{|q_{1}|}{|q_{2}|} = \frac{(x_{3}-x_{1})^2 + (y_{3}-y_{1})^2}{(x_{3}-x_{2})^2+ (y_{3}-y_{2})^2}$$

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,
---------------------------------------------------------------------------------
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,

$q_{1}q_{2} < 0$ $\therefore$ $q_{1}q_{2} \equiv -$

$$|q_{1}| < |q_{2}|, |\vec{r}_{31}| < |\vec{r}_{32}|, |\vec{r}_{32}| > L$$

$$|q_{1}| = |q_{2}|$$, No equilibrium exists on x-axis.

$$|q_{1}| > |q_{2}|, |\vec{r}_{31}| > |\vec{r}_{32}|, |\vec{r}_{32}| > L$$
---------------------------------------------------------------------------------

$$|q_{1}| < |q_{2}|$$

Then,

$$|\vec{r}_{31}| < |\vec{r}_{32}|$$

Where,

$$|\vec{r}_{32}| > L$$

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,

$$\frac{|q_{1}|}{{r_{31}}^2} = \frac{|q_{2}|}{{r_{32}}^2}$$

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

Then,

$$\frac{|q_{1}|}{{(c)}^2} = \frac{|q_{2}|}{{(r_{12} + c)}^2}$$

Then, through some algebra the following result is arrived,

$$c = \frac{r_{12}}{\pm \sqrt{\frac{|q_{2}|}{|q_{1}|}-1}}$$

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

Then,

$$r_{12} = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$$

Substituting,

$$c = \frac{\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}}{\pm \sqrt{\frac{|q_{2}|}{|q_{1}|}-1}}$$

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

Last edited:

Dick
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:
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,

$$r_{13} = c$$

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,

$$(x_{3}, y_{3})?$$

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

$$\vec{r}_{13} = x_{3}\hat{i}-y_{3}\hat{j}$$

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