91Ó°ÊÓ

The problem can be better understood by considering a diagram of the two balls. Label the two balls A and B, and label the point from which the threads are suspended as O. The forces acting on balls A and B are: 1. Gravitational force (\(F_g\)), which is acting straight down, 2. Electrical force (\(F_e\)), which is a repulsive force acting horizontally, 3. Tension in the thread (\(T\)).

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, i.e., \(\vec{F} = m\vec{a}\). Since the balls are in equilibrium, the net forces acting in the vertical and horizontal directions are equal to zero. For the vertical component, we write \(T\cos\theta = F_g\), and for the horizontal component, we write \(T\sin\theta = F_e\).

Coulomb's law gives the electrical force \(F_e\) of charged particles: \(F_e = \frac{kQ^2}{d^2}\), where \(k = 8.99 \times 10^9 \mathrm{Nm^2C^{-2}}\) is the electrostatic constant.

The balls are in equilibrium and hanging at an angle. Let's denote the height of each ball below the suspension point as \(h\), and the horizontal distance of each ball from the suspension point as \(x\). We can write two trigonometric relations: 1. \(\sin\theta = \frac{x}{L}\), and 2. \(\cos\theta = \frac{h}{L}\). The distance between the two balls, \(d\), is twice the value of \(x\): \(d = 2x\).

Using the equilibrium equations, we now have: \begin{align*} T\cos\theta = mg, \\ T\sin\theta = \frac{kQ^2}{d^2}. \end{align*} Divide the second equation by the first equation to eliminate the tension T: \begin{align*} \tan\theta = \frac{kQ^2}{mgd^2}. \end{align*} Solve for \(Q\) using the given values for \(m\), \(L\), and \(d\): \begin{align*} Q = \sqrt{\frac{mgd^2\tan\theta}{k}}. \end{align*} Substitute the values to find \(Q\): \begin{align*} Q &= \sqrt{\frac{(9.0\times 10^{-8}\ \text{kg})(9.81\ \text{m/s}^2)(0.020\ \text{m})^2\tan \left( \frac{d}{2L}\right)}{8.99 \times 10^9\ \text{Nm}^2\text{C}^{-2}}} \\ Q &\approx 3.1 \times 10^{-8} \text{ C}. \end{align*} The charge \(Q\) on the Styrofoam balls is approximately \(3.1 \times 10^{-8}\ \text{C}\).