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TITLE: Hard Electric Charge Problem.

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.

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]

[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] = ?

[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]

[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]

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

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

## 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]

**and**__(c)__**(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: