- #1
AmagicalFishy
- 50
- 1
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:
[tex](\frac {q^2 L}{2\pi mg \epsilon _0})^{1/3}[/tex]
Hint: When θ is small, sin(θ) ≈ tan(θ)
Relevant equations
[tex]|\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}[/tex]
Solution Attempt
So, I can solve this by setting:
[tex]\sin{\theta} = \frac{d}{2L} \\
\tan{\theta} = \frac{|\vec{F_E}|}{|\vec{F_G}|} = \frac{kq^2}{mgd^2}[/tex]
And just solving for d.
My issue arises when I try to solve it by setting...
[tex]\sin{\theta} = \frac{|\vec{F_E}|}{|\vec{F_T}|}[/tex]
Then, I get:
[tex]\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[/tex]
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:
[tex]\sqrt{k}q = (\frac {q^2 L}{2\pi mg \epsilon _0})^{1/3}[/tex]
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...
[tex]\frac{kq^2}{2mg} = L^2[/tex]
... which begs for some clever usage of the Pythagorean theorem or something. I can't figure it out. =\ Help, please.
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:
[tex](\frac {q^2 L}{2\pi mg \epsilon _0})^{1/3}[/tex]
Hint: When θ is small, sin(θ) ≈ tan(θ)
Relevant equations
[tex]|\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}[/tex]
Solution Attempt
So, I can solve this by setting:
[tex]\sin{\theta} = \frac{d}{2L} \\
\tan{\theta} = \frac{|\vec{F_E}|}{|\vec{F_G}|} = \frac{kq^2}{mgd^2}[/tex]
And just solving for d.
My issue arises when I try to solve it by setting...
[tex]\sin{\theta} = \frac{|\vec{F_E}|}{|\vec{F_T}|}[/tex]
Then, I get:
[tex]\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[/tex]
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:
[tex]\sqrt{k}q = (\frac {q^2 L}{2\pi mg \epsilon _0})^{1/3}[/tex]
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...
[tex]\frac{kq^2}{2mg} = L^2[/tex]
... which begs for some clever usage of the Pythagorean theorem or something. I can't figure it out. =\ Help, please.