# Self Study - Foundations of Electromagnetic Theory Problem 2-1 (two charged particles hanging on strings)

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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
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,

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

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

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

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Homework Helper
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$$r = l \sin \theta$$
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
Usiia
I edited it, equations needed pruning and it missed my eye. Number 4 needed to be as you mentioned.

Homework Helper
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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,

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

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

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

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BvU and PeroK