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A large electroscope is made with "leaves" that are $78$-cm-long wires with tiny $24$-g spheres at the ends. When charged, nearly all the charge resides on the spheres.

If the wires each make a $26^{\circ}$ angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

We know that the formulas to use are:

$$F = mg \quad \text{ and } \quad F = \dfrac{kq_1q_2}{r^2}$$

Working out the force equations, we obtain:

$$\Sigma F_x = F_T \cos(\theta) - mg = 0 \rightarrow F_T = \dfrac{mg}{\cos(\theta)}$$ $$\Sigma F_y = F_T \sin(\theta) - F_E = 0 \rightarrow F_E = F_T\sin(\theta) = mg\tan(\theta)$$


$$F_E = k\dfrac{(\frac{Q}{2})^2}{d^2} = mg\tan(\theta) \rightarrow Q = 2d\sqrt{\dfrac{mg\tan(\theta)}{k}}$$

By substitution

$$Q = 2(7.8 \times 10^{-1} \mbox{m})\sqrt{\dfrac{(24 \times 10^{-3} \mbox{kg})(9.8 \mbox{m}/\mbox{s}^2)(\tan(26^{\circ}))}{9.0 \times 10^{9} \mbox{N}\cdot\mbox{m}^2/\mbox{C}^2}}$$

I got $5.6 \times 10^{-6}$, but it's incorrect.

Any suggestions or advices here?

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I think the problem is in the force equilibrium condition. The wire is able to balance any force whatsoever (so long it does not break), so the equilibrium condition I would seek is tha the force components on the perpendicular to the wire add up to zero.

We have the gravity component in this direction: $mg \sin \theta$, the distance between the spheres is, by trigonometry, not $d$ (the length of the wire), but $d \sin \theta$ so the force between them is $$ \frac{k (\frac{Q}{2})^2}{(d \sin \theta)^2}$$

this needs to be projected on the perpendicular to the wire so multiply by $\cos \theta$, the equilibriuk condition is then

$$\frac{k (\frac{Q}{2})^2}{(d \sin \theta)^2} \cos \theta = mg \sin \theta $$

from here the answer is your answer times $\sin \theta$, which is close to

$Q = 2.45 \times 10^{-6}$

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Never mind. Your answer is incorrect by the way because the trig is incorrect. I found the correct answer, which is approximately $4.9 \times 10^{-6}$. – NasuSama Sep 8 '13 at 21:48

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