- #1

Yuravian

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## Homework Statement

In Fig. 21-42 (I attached an MSPaint rendition of it), two tiny conducting balls of identical mass

*m*and Identical charge

*q*hang from non-conducting threads of length

*L*. Assume [tex]\theta[/tex] is so small that [tex]tan \theta[/tex] can be replaced by its approximate equal, [tex] sin \theta[/tex]. (a) Show that [tex]x=(\frac{q^2L}{2\pi\epsilon_{0}mg})^{1/3}[/tex] gives the equilibrium separation of the balls. (b) If

*L*=120cm,

*m*=10 g, and

*x*=5.0 cm, what is |

*q*|?

## Homework Equations

Well, I think Coulomb's law is clearly involved because of the [tex]\frac{1}{2\pi\epsilon_{0}}[/tex] bit, which is equal to 2*

*k*.

So then, restating Coulomb's law: [tex]F=\frac{1}{4\pi\epsilon_{0}}*\frac{\left|q_1\right|\left|q_2\right|}{r^2}[/tex]

## The Attempt at a Solution

The problem here is I don't know how to start the problem. It appears simple enough but I can't seem to get an answer out of it. Here is what I was thinking: since [tex]q_1[/tex] and [tex]q_2[/tex] are both the same, the top bit of coulomb's law becomes [tex]q^2[/tex] since any number squared is positive (no need for abs. value). splitting out coulomb's law from the full equation within parenthesis gives [tex]x=(\frac{1}{4\pi\epsilon_{0}}*\frac{q^2}{r^2}*\frac{2Lr^2}{mg})^{1/3}[/tex]. The problem with this is that I had to work in [tex]\frac{r^2}{r^2}[/tex], when I'm pretty sure there's an equivalency in there somewhere.

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