Force between two charged spheres hanging from a string

1. Feb 5, 2013

AmagicalFishy

Problem

Two small identical small spheres with mass m are hung from insulating threads of length L, as shown in the ﬁgure. Each sphere has the same charge q1 = q2 = q. The radius of each sphere is very small compared to the distance between them, so that they may be considered as point charges. If the distance between the centres of the two spheres is d and the angle θ is small, show that the equilibrium separation between the two spheres is:

$$(\frac {q^2 L}{2\pi mg \epsilon _0})^{1/3}$$

Hint: When θ is small, sin(θ) ≈ tan(θ)

Relevant equations
$$|\vec{F_{E-field}}| = k\frac{q^2}{d^2}\\ |\vec{F_{Gravity}}| = mg \\ \vec{F_{Tension}} = -k\frac{q^2}{d^2} \hat{i} + mg\hat{j} \\ |\vec{F_{Tension}}| = \sqrt{(k\frac{q^2}{d^2})^2 + (mg)^2}$$

Solution Attempt
So, I can solve this by setting:
$$\sin{\theta} = \frac{d}{2L} \\ \tan{\theta} = \frac{|\vec{F_E}|}{|\vec{F_G}|} = \frac{kq^2}{mgd^2}$$
And just solving for d.

My issue arises when I try to solve it by setting...
$$\sin{\theta} = \frac{|\vec{F_E}|}{|\vec{F_T}|}$$

Then, I get:
$$\frac{|\vec{F_E}|}{|\vec{F_G}|} ≈ \frac{|\vec{F_E}|}{|\vec{F_T}|} \Longrightarrow \frac{1}{|\vec{F_G}|} ≈ \frac{1}{|\vec{F_T}|} \Longrightarrow |\vec{F_G}| ≈ |\vec{F_T}| \Longrightarrow |\vec{F_G}|^2 ≈ |\vec{F_T}|^2 \Longrightarrow m^2g^2 ≈ (k\frac{q^2}{d^2})^2 + (mg)^2$$
Which I know is wrong because, in that case, the (mg)2 cancels out, but I can't figure out what's wrong about approaching it that way. Effectively, it's saying that |FE| is very small (nigh zero), but... isn't that the case anyway since θ is so small? Is there something wrong with my tension force? That's what I imagine the problem is; for some reason, tension's always given me trouble—but I can't isolate anything. It seems, conceptually, everything about solving it entirely with force-magnitudes is correct, but it just doesn't work out mathematically. I don't understand why it doesn't work out mathematically.

Unless:
$$\sqrt{k}q = (\frac {q^2 L}{2\pi mg \epsilon _0})^{1/3}$$

In which case it does work out mathematically, and I just can't simplify things down to x = x to prove it. I've gotten to...
$$\frac{kq^2}{2mg} = L^2$$
... which begs for some clever usage of the Pythagorean theorem or something. I can't figure it out. =\ Help, please.

2. Feb 5, 2013

cepheid

Staff Emeritus
Can I give you a word of advice? You're making this way harder than it needs to be. My advice is: forget about tension! Tension can only pull radially (along the string) not tangentially. On each charge, you're going to have the two forces of gravity and the electrostatic force. Each of these forces will have components parallel to and perpendicular to the strings. The components [STRIKE]perpendicular[/STRIKE] parallel to the strings can be ignored, because they'll just be balanced by the tension. Since that's all the tension does, we can forget about it.

Just equate the component of F_E that is perpendicular to the string with the component of F_g that is perpendicular to the string. I.e. equate the two tangential components. You'll find that one of them tends to want to reduce theta, and the other one tends to want to increase it. At equilibrium, they should balance each other out, meaning that their magnitudes are the same.

If you do it this way, it's maybe a 3 or 4 line calculation.

Last edited: Feb 5, 2013
3. Feb 5, 2013

cepheid

Staff Emeritus
As for this part in red: what the heck are you talking about? The (mg)2 most certainly does not cancel out. PLEASE don't tell me that you just crossed out the mg on both sides of this equation:

(mg)2 ≈ (kq2)/d2 + (mg)2

and then said (kq2)/d2 ≈ 1. Because that would be totally WRONG. That's a PLUS sign on the righthand side there, not a multiplication sign. EDIT: and don't deny it: I know that's what you did, because that how you ended up with the notion that sqrt(k)*q should be equal to d. :tongue:

If you do it correctly, you just end up with (kq2)/d2 ≈ 0, which may be true, but doesn't help you at all.

Last edited: Feb 5, 2013
4. Feb 5, 2013

haruspex

You meant parallel, right?

5. Feb 5, 2013

cepheid

Staff Emeritus
Indeed I did. Good catch, thanks.

6. Feb 6, 2013

AmagicalFishy

Can't you just cancel out (mg)2 by subtracting (mg)2 from both sides and get:
$$0 ≈ (kq^2/d^2)^2 \Longrightarrow 0 ≈ k^2q^4/d^4 \Longrightarrow ...$$

... oh.

It wasn't that I messed up in cancelling out the (mg)2, it was that I messed up in separating coulomb's constant and the charge afterwards (though I did that in a fairly similar way to your description of the (mg)2—which is, admittedly, just as stupid a mistake >.>). Still, you more-than-answered my question; technically, the way I was doing it (aside from that mistake) was mathematically correct... but the way I was doing it also doesn't help me at all.

My first thought upon reading your suggestion starts with taking compTension F of all the force vectors to get the components parallel to the string, subtracting that from the force vectors to get the perpendicular components, and equating the perpendicular components to one another... but I feel like that's the opposite of what you mean when you say "ignore Tension".

Upon trying to do this anyway, I have realized that, while my skills surely do not lie in physics—I'm really, really good at taking what (I think) should be simple instructions/problems and turning them into horrible, error-prone messes of numbers, symbols, and pain.

7. Feb 6, 2013

cepheid

Staff Emeritus
To get components of F_g that are parallel and perpendicular to the string, just resolve the vector into components using trigonometry in the usual way. So you have a vector with magnitude mg that points straight downwards (vertical). The tail of this vector starts at one of the spheres. Now, starting at that tail, draw the component of this vector that is parallel to the string. Starting from the tip of that parallel vector, draw another vector that is perpendicular to the string. Its tail starts at the tip of the parallel vector, and its tip meets the tip of the resultant (total) mg vector. So now you have the parallel and perpendicular component vectors adding together to form the vertical mg vector as the resultant.

What you should have just drawn is a right triangle, one of whose angles is θ (or θ/2 -- I'm not sure if θ is the angle between the strings, or the angle between each string and the vertical). So, the lengths of the two component vectors in the triangle will depend on the length of the resultant (the hypotenuse) by factors of sinθ and cosθ (but I won't say which one is which). Just focus on the perpendicular component and ignore the parallel one.

Now do the same thing all over again for F_E, which is horizontal, in order to figure out its perpendicular component.

8. Feb 6, 2013

AmagicalFishy

Excellent, excellent.

There are a lot of ways to solve this (some way more difficult than others) and I think I fully comprehend all of them (that I can think of). That helped a ton, thank you. :)