Can Coulomb's Law Help Solve Where Forces Cancel Out?

[PFStudent]

Homework Statement

http://img238.imageshack.us/img238/3648/hrwfop7echpt2128br6.jpg

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:

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

Magnitude Form:

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

The Attempt at a Solution

Yea, this is a pretty hard problem.

Here is how I tackled it:

First off, I let the angle $\theta$ beginning on the positive axis and moving counter-clockwise equal $\theta_{1}$ and let the bottom angle beginning on the positive axis and moving clockwise equal $\theta_{2}$.

The reason I did this was to preserve consistency when measuring angles,

Therefore,

$$\theta_{1} = \theta$$

$$\theta_{2} = 2\pi - \theta$$

$$q_{1} = q_{2} = -e$$

$$q_{3} = q_{4} = -q$$

$$q \leq 5e$$

$q_{1} \equiv$ fixed in place

$q_{2} \equiv$ free to move

$q_{3} \equiv$ fixed in place

$q_{4} \equiv$ fixed in place

$$q_{1}(x_{1}, y_{1}) = q_{1}(-2R, 0)$$
$$q_{2}(x_{2}, y_{2}) = q_{2}(-R, 0)$$
$$q_{3}(x_{3}, y_{3}) = q_{3}\left(0, Rtan\left(\theta\right)\right)$$
$$q_{4}(x_{4}, y_{4}) = q_{4}\left(0, Rtan\left(2\pi - \theta\right)\right)$$

The problem asks for the three smallest values of theta for which electron 2 will be held in place, therefore,

$$\Sigma\vec{F}_{2} = 0$$

$$0 = \vec{F}_{21} + \vec{F}_{23} + \vec{F}_{24}$$

Breaking into x-components,

$$0 = \vec{F}_{21}_{x} + \vec{F}_{23}_{x} + \vec{F}_{24}_{x}$$

$$-\vec{F}_{21}_{x} = \vec{F}_{23}_{x} + \vec{F}_{24}_{x}$$

$$|{F}_{21}_{x}| = |{F}_{23}_{x} + {F}_{24}_{x}|$$

$$F_{21}_{x} = |\vec{F}_{21}|cos\theta_{21}, \theta_{21} = 0$$
$$F_{21}_{x} = |\vec{F}_{21}|$$

$$F_{23}_{x} = |\vec{F}_{23}|cos\theta_{23}, \theta_{23} = \theta + \pi$$
$$F_{23}_{x} = |\vec{F}_{23}|cos\left(\theta + \pi\right)$$

$$F_{24}_{x} = |\vec{F}_{24}|cos\theta_{24}, \theta_{24} = \pi - \theta$$
$$F_{21}_{x} = |\vec{F}_{21}|cos\left(\pi - \theta)$$

$$\left|\left(|\vec{F}_{21}|\right)\right| = \left|\left(|\vec{F}_{23}|cos\left(\theta + \pi\right)\right) + \left(|\vec{F}_{21}|cos\left(\pi - \theta\right)\right)\right|$$

$$\left|\left(\frac{k_{e}|q_{2}||q_{1}|}{{r_{21}}^2}\right)\right| = \left|\left(\frac{k_{e}|q_{2}||q_{3}|}{{r_{23}}^2}\right)cos\left(\theta + \pi\right) + \left(\frac{k_{e}|q_{2}||q_{4}|}{{r_{24}}^2}\right)cos\left(\pi - \theta\right)\right|$$

$$\left|\frac{k_{e}|q_{2}||q_{1}|}{{r_{21}}^2}\right| = \left|\frac{k_{e}|q_{2}||q_{3}|}{{r_{23}}^2}cos\left(\theta + \pi\right) + \frac{k_{e}|q_{2}||q_{4}|}{{r_{24}}^2}cos\left(\pi - \theta\right)\right|$$

Canceling out: $|k_{e}|$ and $||q_{2}||$; from both sides.

$$\left|\frac{|q_{1}|}{{r_{21}}^2}\right| = \left|\frac{|q_{3}|}{{r_{23}}^2}cos\left(\theta + \pi\right) + \frac{|q_{4}|}{{r_{24}}^2}cos\left(\pi - \theta\right)\right|$$

In addition noting the following,

$$r_{21} = R$$
$$r_{23} = \sqrt{{R}^2+{y_{3}}^2}$$
$$r_{24} = \sqrt{{R}^2+{y_{4}}^2}$$

$${|y_{3}|}^2 = {y_{3}}^2$$
$${|y_{4}|}^2 = {y_{4}}^2$$

$$r_{23} = \sqrt{{R}^2+{y_{3}}^2} = r_{23} = \sqrt{{R}^2+\left({|y_{3}|}^2\right)}$$
$$r_{24} = \sqrt{{R}^2+{y_{4}}^2} = r_{24} = \sqrt{{R}^2+\left({|y_{4}|}^2\right)}$$

Where, $|y_{3}| = |y_{4}|$ and $r_{23} = r_{24}$; so then letting,

$$|y_{3}| = |y_{4}| = y$$

$$r_{23} = r_{24} = r_{2} = \sqrt{{R}^2+{\left(y\right)}^2}$$

And noting: $q_{3} = q_{4} = -q$ and $q_{1} = -e$;

then,

$$\left|\frac{|\left(-e\right)|}{{\left(R\right)}^2}\right| = \left|\frac{|\left(-q\right)|}{{\left(\sqrt{{R}^2+{\left(y\right)}^2}\right)}^2}cos\left(\theta + \pi\right) + \frac{|\left(-q\right)|}{{\left(\sqrt{{R}^2+{\left(y\right)}^2}\right)}^2}cos\left(\pi - \theta\right)\right|$$

Through some algebra the following is arrived,

$$\left|\frac{|e|\left({R}^2+{y}^2\right)}{{R}^2}\right| = \left|q\right| \left|\left(cos\left(\theta + \pi\right) + cos\left(\pi - \theta\right)\right)\right|$$

$e > 0$, then $|e| = e$

Noting that,

$$cos\left(\theta + \pi\right) = cos\left(\pi - \theta)\right)$$

Then let,

$$cos\left(\theta + \pi\right) = cos\left(\pi - \theta)\right) = cos(\phi)$$

$$\left|\frac{\left(e\right)\left({R}^2+{y}^2\right)}{{R}^2}\right| = \left|q\right| \left|\left(cos\left(\phi\right) + cos\left(\phi\right)\right)\right|$$

$$\left|\frac{e\left({R}^2+{y}^2\right)}{{R}^2}\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

$$tan(\theta) = \frac{y_{3}}{R}$$

$$tan(2\pi - \theta) = \frac{y_{4}}{R}$$

$$y_{3} = Rtan(\theta)$$
$$y_{4} = Rtan(2\pi - \theta)$$

$${y_{3}}^2 = {y_{4}}^2 = {y}^2$$

$${\left(tan(\theta)\right)}^2 = {\left(tan(2\pi - \theta)\right)}^2$$

$${tan}^{2}(\theta) = {tan}^{2}(2\pi - \theta)$$

Then, let,

$${tan}^{2}(\theta) = {tan}^{2}(2\pi - \theta) = {tan}^{2}(\gamma)$$

So,

$${y}^2 = {R}^{2}{tan}^{2}(\gamma)$$

Then,

$$\left|\frac{e\left({R}^2+{y}^2\right)}{{R}^2}\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

$$\left|\frac{e\left({R}^2+\left({R}^{2}{tan}^{2}(\gamma)\right)\right)}{{R}^2}\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

$$\left|\frac{e\left({R}^2+{R}^{2}{tan}^{2}(\gamma)\right)}{{R}^2}\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

Ok, I note that I have two unknowns: $q$ and $\phi$; I understand that I have an interval for $q$,

$q \equiv$ on $(-\infty, 5e]$

However, how does this help?

I could also solve for $|q|$, and recognize that $|q| = |q(\phi)|$, and attempt to find values of $\phi$ that would minimize or maximize $|q(\phi)|$ through differentiation with respect to $\phi$, however I don’t quite think that is the right approach… :-/

I already tried working out the y-components, and ended up with the following,

$$\Sigma\vec{F}_{2} = 0$$

$$0 = \vec{F}_{21} + \vec{F}_{23} + \vec{F}_{24}$$

$$0 = \vec{F}_{21}_{y} + \vec{F}_{23}_{y} + \vec{F}_{24}_{y}$$

$$-\vec{F}_{21}_{y} = \vec{F}_{23}_{y} + \vec{F}_{24}_{y}$$

$$-\vec{F}_{21}_{y} = \vec{F}_{23}_{y} + \vec{F}_{24}_{y}$$

Note, $\vec{F}_{21}_{y} = 0$,

$$-\left(0\right) = \vec{F}_{23}_{y} + \vec{F}_{24}_{y}$$

$$-\vec{F}_{23}_{y} = \vec{F}_{24}_{y}$$

$$|{F}_{23}_{y}| = |{F}_{24}_{y}|$$

Working through the above with substitutions and algebra leads to the following,

$$sin\left(\theta+\pi\right) = sin\left(\pi-\theta\right)$$

Which is another dead end…

Yea, so this problem is really hard and I’m really stuck.

Any help is appreciated,

-PFStudent

 

[Doc Al]

Your formal approach sure makes it look harder than it is. Try this. From symmetry, you should know that only the horizontal component of the forces from the two ions will count towards the net force on electron #2. So just set up the equation for the net force: it will be a function of y and q. Just cycle through the valid values for q, solve for y in each case (if possible), then translate that back to theta.

 

[PFStudent]

Hey,

Doc Al, thanks for the information.

Ok, going back to this point,

$$\left|\frac{e\left({R}^2+{R}^{2}{tan}^{2}\gamma\right)}{{R}^2}\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

I realize that the ${R}^{2}$ can be canceled out on the left side reducing the equation to,

$$\left|e\left(1+{tan}^{2}\gamma\right)\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

From here I note the following Trigonometric Identity,

$${sec}^{2}\theta-{tan}^{2}\theta = 1$$

So with the appropriate substitution,

$$\left|e\left(\left({sec}^{2}\gamma\right)\right)\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

$$\left|e\left({sec}^{2}\gamma\right)\right| = \left|q\right| \left|2cos\left(\phi\right)\right|$$

$$\left|\frac{e}{{cos}^{2}\gamma}\right| = \left|q\right| \left|2cos\phi\right|$$

From the earlier substitution I made,

$${tan}^{2}{(\theta)} = {tan}^{2}{(2\pi - \theta)} = {tan}^{2}{(\gamma)}$$

I then assume*,

$${cos}^{2}{(\theta)} = {cos}^{2}{(2\pi - \theta)} = {cos}^{2}{(\gamma)}$$

From here I proceed to evaluate ${cos}^{2}{\gamma}$ for both possible values,

$${cos}^{2}{\gamma} = {cos}^{2}{\theta} = {cos}^{2}{\left(2\pi - \theta\right)}$$

$${cos}^{2}{\theta} = \left(cos\theta\right) \left(cos\theta\right)$$

$${cos}^{2}{\left(2\pi - \theta\right)} = \left({cos}{\left(2\pi - \theta\right)}\right) \left({cos}{\left(2\pi - \theta\right)}\right)$$

From here noting the following trigonometric identity,

$$cos\left(\alpha\pm\beta\right) = cos\alpha{\textcolor[rgb]{1.00,1.00,1.00}{.}}cos\beta \mp sin\alpha{\textcolor[rgb]{1.00,1.00,1.00}{.}}sin\beta$$

Then,

$${cos}{\left(2\pi - \theta\right)} = cos2\pi{\textcolor[rgb]{1.00,1.00,1.00}{.}}cos\theta+sin2\pi{\textcolor[rgb]{1.00,1.00,1.00}{.}}sin\theta$$

$${cos}{\left(2\pi - \theta\right)} = \left(1\right)cos\theta+\left(0\right)sin\theta$$

$${cos}{\left(2\pi - \theta\right)} = cos\theta$$

Therefore,

$${cos}^{2}{\gamma} = {cos}^{2}{\theta} = {cos}^{2}{\left(2\pi - \theta\right)}$$

Becomes,

$${cos}^{2}{\gamma} = {cos}^{2}{\theta} = {cos}^{2}{\left(\theta\right)}$$

$${cos}^{2}{\gamma} = {cos}^{2}{\theta}$$

So,

$$\left|\frac{e}{{cos}^{2}\gamma}\right| = \left|q\right| \left|2cos\phi\right|$$

is now,

$$\left|\frac{e}{\left({cos}^{2}{\theta}}\right)\right| = \left|q\right| \left|2cos\phi\right|$$

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|q\right| \left|2cos\phi\right|$$

Now, $cos\phi$ can be reduced in the same manner,

Refer to the earlier substitution,

$$cos\left(\theta + \pi\right) = cos\left(\pi - \theta\right) = cos(\phi)$$

Referring to the earlier trigonometric identity,

$$cos\left(\alpha\pm\beta\right) = cos\alpha{\textcolor[rgb]{1.00,1.00,1.00}{.}}cos\beta \mp sin\alpha{\textcolor[rgb]{1.00,1.00,1.00}{.}}sin\beta$$

Each trigonometric function can then be reduced as follows,

-----------------------------------------------

$$cos\left(\theta + \pi\right) = cos\theta{\textcolor[rgb]{1.00,1.00,1.00}{.}}cos\pi - sin\theta{\textcolor[rgb]{1.00,1.00,1.00}{.}}sin\pi$$

$$cos\left(\theta + \pi\right) = cos\theta\left(-1\right)-sin\theta\left(0\right)$$

$$cos\left(\theta + \pi\right) = -cos\theta$$

$$cos\left(\pi - \theta\right) = cos\theta{\textcolor[rgb]{1.00,1.00,1.00}{.}}cos\pi + sin\theta{\textcolor[rgb]{1.00,1.00,1.00}{.}}sin\pi$$

$$cos\left(\pi - \theta\right) = cos\theta\left(-1\right) + sin\theta\left(0\right)$$

$$cos\left(\pi - \theta\right) = -cos\theta$$

-----------------------------------------------

Reducing the previous equation,

$$cos\left(\theta + \pi\right) = cos\left(\pi - \theta\right) = cos(\phi)$$

to,

$$\left(-cos\theta\right) = \left(-cos\theta\right) = cos(\phi)$$

$$cos(\phi) = -cos\theta$$

Reducing the earlier equation,

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|q\right| \left|2cos\phi\right|$$

to,

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|q\right| \left|2\left(-cos\theta\right)\right|$$

Simplifying,

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|q\right| \left|-2cos\theta\right|$$

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|q\right| \left|2cos\theta\right|$$

From here refer to the following equation,

-----------------------------------------------

$$q = n_{e}e, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}n_{e} = \pm1, \pm 2, \pm 3,...,$$

e $\equiv$ elementary charge

$$e = 1.60217646{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}10^{-19}{\textcolor[rgb]{1.00,1.00,1.00}{.}}C$$

-----------------------------------------------

So let,

$$q = n_{e}e$$

Then,

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|q\right| \left|2cos\theta\right|$$

Becomes,

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|\left(n_{e}e\right)\right| \left|2cos\theta\right|$$

$$\left|\frac{e}{{cos}^{2}{\theta}}\right| = \left|n_{e}e\right| \left|2cos\theta\right|$$

Then through some algebra,

$$\left|\frac{1}{2n_{e}}\right| = \left|{cos}^{3}{\theta}\right|$$

I note from earlier that,

-----------------------------------------------

$$q \leq 5e$$

and defined on the interval,

q $\equiv$ $(-\infty, 5e]$

-----------------------------------------------

Then from,

$$q = n_{e}e$$

Substitute in to,

$$q \leq 5e$$

Yielding,

$$\left(n_{e}e\right) \leq 5e$$

Simplifying to,

$$n_{e} \leq 5$$

From here note that $n_{e}$ is defined on the interval,

$n_{e}$ $\equiv$ $(-\infty, 5]$

Here is the problem, how can I solve for the three smallest consecutive values of theta (in the equation below) without knowing $n_{e}$?

$$\left|\frac{1}{2n_{e}}\right| = \left|{cos}^{3}{\theta}\right|$$

So, how can I can I solve for the three smallest consecutive values of theta without knowing what $n_{e}$ is?

Just cycle through the valid values for q, solve for y in each case (if possible), then translate that back to theta.

In trying to find out what $n_{e}$ is I did not quite follow what you meant by "valid values for q?"

Is there a way to at least determine whether the charge $q$ is positive or negative, or in other words is there a way to justify whether,

$$q \geq 0$$

or

$$q < 0$$

Yea, I am stuck any help is appreciated.

Thanks,

-PFStudent

P.S.: Sorry for the late reply.

 

[learningphysics]

I think you can make some assumptions. Like the angle is between 0 and 90 degrees. Also ne can only take integer values and the maximum magnitude of the charge is 5e according to the question.

The higher the cosine, the smaller the angle... take ne = 1, 2 and 3. and solve for the angle in each case.

 

[Doc Al]

I note from earlier that,

-----------------------------------------------

$$q \leq 5e$$

and defined on the interval,

q $\equiv$ $(-\infty, 5e]$

Comments below...

In trying to find out what $n_{e}$ is I did not quite follow what you meant by "valid values for q?"

The way I read the problem, you are given that the unknown charges are negative (-q), where q can only be e, 2e, 3e, 4e, or 5e.

 

[Kushal]

i worked through [art of it, i don't know if its correct. You should first be caluculating the forces from the 3 charges on e2.
F1 = (k*e^2)/R^2
F3 = F4 = (k*e*-q)/(R/cos theta)^2

Now considering only the horizontal components, the vertical ones from F3 and F4 will cancel out.

F1 = 2 F3 * cos theta
This equation is for the electron 2 to remain in position.

When I've solved this equation i got:

cos theta = cube root(e/2q)

i ignored the negative of q coz i think it should be cancelling out somewhere.

i used the principle that as theta decreases (from 90 degrees to 0 degrees for acute angles) the value of cos-1 [cube root(e/2q)] increases.

i hope you can solve the problem from here.

 

[PFStudent]

Hey,

The way I read the problem, you are given that the unknown charges are negative (-q), where q can only be e, 2e, 3e, 4e, or 5e.

Hmmm...Ok, so by the problem stating a charge $-q$, it is implied with reference to the following equation,

-----------------------------------------------

$$q = n_{e}e, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}n_{e} = \pm1, \pm 2, \pm 3,...,$$

e $\equiv$ elementary charge

$$e = 1.60217646{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}10^{-19}{\textcolor[rgb]{1.00,1.00,1.00}{.}}C$$

-----------------------------------------------

That, the sign of the charge, that is $n_{e}$ has been established or in other words $n_{e}$ is negative and when the problem gave $-q$ it was with the sign taken from $n_{e}$ and moved in front of $q$. So, when the problem mentions for "physically possible values of $q$ $\leq$ 5e," it actually means,

$$|q| \leq 5e$$

Thus, re-interpreting the problem,

$-5e < q < 0$, Where the coefficient of e must be an integer.

$q = [-5e, 0]$, Where the coefficient of e must be an integer.

Therefore,

$$q = n_{e}e, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}n_{e} = \pm1, \pm 2, \pm 3,...,$$

$$q = n_{e}e$$

$n_{e} = [-5, 0]$, Where $n_{e}$ must be an integer.

Let, $n_{e} = -N$, Where $N = [0, 5]$, $N$ must be an integer.

$n_{e} = -N$, $N = [0, 5]$, $N$ must be an integer.

$$q = \left(-N\right)e$$

$$q = -Ne$$

$$n_{e} = -N$$

Going back to where I left off,

$$\left|\frac{1}{2n_{e}}\right| = \left|{cos}^{3}{\theta}\right|$$

$$\left|\frac{1}{2\left(-N\right)}\right| = \left|{cos}^{3}{\theta}\right|$$

$$\left|\frac{1}{2N}\right| = \left|{cos}^{3}{\theta}\right|$$

Note that,

$$\frac{1}{2N} > 0$$

Then,

$$\left|\frac{1}{2N}\right| = \frac{1}{2N}$$

So now,

$$\left|\frac{1}{2N}\right| = \left|{cos}^{3}{\theta}\right|$$

Becomes,

$$\frac{1}{2N} = \left|{cos}^{3}{\theta}\right|$$

I think you can make some assumptions. Like the angle is between 0 and 90 degrees.

I agree. The reason theta must be restricted in the interval $\left[0, \frac{\pi}{2}\right]$ is because of the following.
If theta is allowed to be smaller than zero the following would happen.
As theta approaches zero degrees particles 3 and 4 would come closer to each other and because they are like-charge would produce a mutual repulsion force approaching infinity as theta approaches zero.
If theta is allowed to be greater than 90 degrees than the following would happen.
As theta approaches 90 degrees particles 3 and 4 move farther away from one another in the y-direction. If theta reaches or exceeds 90 degrees particles 3 and 4 are removed from the y-axis and therefore distort the given state of the problem.

Then by reasoning values for theta that are real the following must be true,

$$\theta = \left[0, \frac{\pi}{2}\right]$$

So since $\theta$ is now restricted to Quadrant I, then,

$$cos\theta \geq 0$$

Therefore,

$$\left|{cos}{\theta}\right| = {cos}{\theta}$$

$$\left|{cos}^{3}{\theta}\right| = {cos}^{3}{\theta}$$

So now,

$$\frac{1}{2N} = \left|{cos}^{3}{\theta}\right|$$

Becomes,

$$\frac{1}{2N} = {cos}^{3}{\theta}$$

Through some algebra,

$${cos}{\theta} = \frac{1}{\sqrt[3]{2N}}$$

-----------------------------------------------
$$arccos(x) = y$$

D: $x = \left[-1, 1\right]$
R: $y = \left[0, \pi\right]$
-----------------------------------------------

$${\theta} = arccos\left(\frac{1}{\sqrt[3]{2N}}\right)$$

Therefore,

$${\theta}\left(N\right) = arccos\left(\frac{1}{\sqrt[3]{2N}}\right)$$

Sig. Fig. $\equiv$ 3 (I guess…)

$${\theta}\left(0\right) = Undefined$$

$${\theta}\left(1\right) = 37.4{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 0.653{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

$${\theta}\left(2\right) = 50.9{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 0.889{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

$${\theta}\left(3\right) = 56.6{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 0.988{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

$${\theta}\left(4\right) = 60.0{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 1.04{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

$${\theta}\left(5\right) = 62.3{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 1.088{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

i used the principle that as theta decreases (from 90 degrees to 0 degrees for acute angles) the value of cos-1 [cube root(e/2q)] increases.

I should have realized that earlier, would have made this problem a little easier. Thanks.

Going back to what the problem originally asked,

“For physically possible values of $q \leq 5e$, what are (a) the smallest, (b) second smallest, and (c) third smallest values of $\theta$ for which electron 2 is held in place?”

(a)
$${\theta}\left(1\right) = 37.4{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 0.653{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

(b)
$${\theta}\left(2\right) = 50.9{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 0.889{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

(c)
$${\theta}\left(3\right) = 56.6{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees, 0.988{\textcolor[rgb]{1.00,1.00,1.00}{.}}Radians$$

Thanks,

-PFStudent

P.S.: Thanks for the help: Doc Al, learningphysics, and Kushal.

 

[BlackWyvern]

That looks insanely hard.
Is it possible to use 1/d^2 and solve for where the force is zero?