# Two charged spheres repel (attached to strings)

1. Sep 3, 2006

It's been awhile since I've had physics I, so this problem is giving me a headache.

Q) Two very small conducting spheres, each of a mass $1.0 \times 10^{-4}\,\,(kg)$, are suspended at a common point by very thin nonconducting threads of a length $0.2 \,\,(m)$. A charge $Q$ is placed on each sphere. The electric force of repulsion separates the spheres, and an equilibrium is reached when the suspending threads make an angle of $10 \,\, (deg)$. Assuming a gravitational force of $9.80 \,\, (N/kg)$ and a negligible mass for the threads, find $Q$.

My Work)
We first will deal with:
$$\vec F_{12} = \frac{\hat R_{12} k q_1 q_2}{R^2_{12}}$$
$$\sum \vec F_i = m \vec a$$
Since equilibrium is reached, $\vec a = \vec 0$. Thus,

$$\sum F_i = \vec T_2 + \vec F_G + \vec F_{12} = \vec 0$$

We now find the forces.

$$\vec F_G = -\hat y (9.8 \times 10^{-4})$$

Setting up the coordinate system we assume the orgin as at the point of interesection of the two threads. Thus, a vector that points to sphere-two is:
$$\vec S_2 = \hat x(0.2 \sin 5^{\circ}) - \hat y(0.2 \cos 5^{\circ})$$

The vector quantities for coloumbs law are as follows:
$$\vec R_{12} = \hat x (2(0.2\sin 5^{\circ}))$$
$$R = 0.4 \sin 5^{\circ}$$
$$\hat R = \hat x$$

Thus, since the spheres have an equal charge
$$\vec F_{12}=\frac{\hat x k Q^2}{0.16 \sin^2 5^{\circ}}$$

Now I know the tension has to exert a force that holds the sphere in place, so gravity and the electric repulsion keep it from moving away. So do I just say that $\vec T_2 = -\vec S_2$?

I'm not really sure what to do. Is what I'm doing even correct?