Is This Electromagnetic Theory Solution Correct?

[Usiia]

Just to say that I'm getting an "unsolvable" cubic equation in $\tan \theta$. Does the question expect an exact solution?

It is a cubic relation.

 

[Usiia]

Answer provided that I just found at the back of the book $$ \tan^3 \theta ( 1+ \tan^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

 

[Usiia]

Answer provided that I just found at the back of the book $$ \tan^3 \theta ( 1+ \tan^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

Of course this means nothing to me. I am supposing that this is a related rates problem and that I need more than what I think I have so far in terms of the tangent.

 

[PeroK]

Of course this means nothing to me. I am supposing that this is a related rates problem and that I need more than what I think I have so far in terms of the tangent.

First, generate the force balancing equations for $T, F$ and $mg$.

 

[Usiia]

First, generate the force balancing equations for $T, F$ and $mg$.

Need to go to work now, but will come back to this with this approach.

Thank you so much!

 

[PeroK]

Answer provided that I just found at the back of the book $$ \tan^3 \theta ( 1+ \tan^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

I get $$ \tan \theta (\sin^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$which is the same as $$\frac{ \tan^3 \theta}{1+ \tan^2 \theta} = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

 

[Usiia]

I get $$ \tan \theta (\sin^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$which is the same as $$\frac{ \tan^3 \theta}{1+ \tan^2 \theta} = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

You are right, I missed the "/" it is the second.

 

[Usiia]

I get $$ \tan \theta (\sin^2 \theta ) = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$which is the same as $$\frac{ \tan^3 \theta}{1+ \tan^2 \theta} = \dfrac{q^2}{8 \pi \epsilon_0 mgl^2} $$

I was able to reach your calculations. If you're feeling generous, give this a look?

1. $$ F_c = \dfrac{1}{4 \pi \epsilon_0 } \dfrac{2q^2}{r^2} $$
2. $$ \tau_x = F_c $$ (stability)
3. $$ \tan \theta = \dfrac{\tau_x }{mg} $$
4. $$ r = 2 l \sin \theta $$
5. $$ q = 2q_{1}^{2} $$ (charge is twice the other of equal mass)

Substituting 4 into 1 yields,

$$ F_c = \dfrac{1}{4 \pi \epsilon_0 } \dfrac{2q^2}{(l \sin \theta)^2} $$
$$ F_c = \dfrac{q^2}{8 \pi \epsilon_0 l^2 \sin^2 \theta}$$And Substituting 2 into 3 yields,

$$ \tan \theta = \dfrac{F_c}{mg} $$

Due to the stability condition. After simplification and arithmetic,

$$ \dfrac{\dfrac{q^2}{8 \pi \epsilon_0 l^2 \sin^2 \theta} }{mg} = \tan \theta $$

$$ \dfrac{q^2}{8 \pi \epsilon_0 mg l^2 \sin^2 \theta} = \tan \theta $$

$$ \dfrac{q^2}{8 \pi \epsilon_0 mg l^2} = \tan \theta \sin^2 \theta$$

 

[PeroK]

$$ r = l \sin \theta $$

This can't be right. $r = 2l \sin \theta$.

$$ F_c = \dfrac{2q^2}{4 \pi \epsilon_0 } \dfrac{2q^2}{(l \sin \theta)^2} $$

You have an extra factor of $2q^2$ there. The next line is correct, but doesn't follow from that:

$$ F_c = \dfrac{q^2}{8 \pi \epsilon_0 l^2 \sin^2 \theta}$$

You got the right answer in the end, but there are still some mistakes.

 

[Usiia]

I edited it, equations needed pruning and it missed my eye. Number 4 needed to be as you mentioned.

 

[PeroK]

I edited it, equations needed pruning and it missed my eye. Number 4 needed to be as you mentioned.

Okay, but it's still wrong in some of your equations:

$$ F_c = \dfrac{1}{4 \pi \epsilon_0 } \dfrac{2q^2}{(l \sin \theta)^2} $$

 

[Usiia]

@PeroK Thank you so much for your continued help. I haven't done this in many many years and it's slowly coming back :)

 

[Usiia]

You are correct, I will try to take this into a latex editor and look it over. It's very difficult making sure it's right on the forum just using preview.

I am trying to copy this from my paper, but not uploading any more scratch work by hand.

 

[Usiia]

I believe this should be correct completely with the substitutions.

1. $$ F_c = \dfrac{1}{4 \pi \epsilon_0 } \dfrac{2q^2}{r^2} $$
2. $$ \tau_x = F_c $$ (stability)
3. $$ \tan \theta = \dfrac{\tau_x }{mg} $$
4. $$ r = 2 l \sin \theta $$
5. $$ q_1q_2 = 2q_{1}^{2} $$ (charge is twice the other of equal mass)

Substituting 4 into 1 yields,

$$ F_c = \dfrac{1}{4 \pi \epsilon_0 } \dfrac{2q^2}{(2 l \sin \theta)^2} $$
$$ F_c = \dfrac{q^2}{8 \pi \epsilon_0 l^2 \sin^2 \theta}$$And Substituting 2 into 3 yields,

$$ \tan \theta = \dfrac{F_c}{mg} $$

Due to the stability condition. After simplification and arithmetic,

$$ \dfrac{\dfrac{q^2}{8 \pi \epsilon_0 l^2 \sin^2 \theta} }{mg} = \tan \theta $$

$$ \dfrac{q^2}{8 \pi \epsilon_0 mg l^2 \sin^2 \theta} = \tan \theta $$

$$ \dfrac{q^2}{8 \pi \epsilon_0 mg l^2} = \tan \theta \sin^2 \theta$$