Two spheres, each of mass \(M=2.33 \mathrm{~g},\) are attached by pieces of string of length \(L=45 \mathrm{~cm}\) to a common point. The strings initially hang straight down, with the spheres just touching one another. An equal amount of charge, \(q,\) is placed on each sphere. The resulting forces on the spheres cause each string to hang at an angle of \(\theta=10.0^{\circ}\) from the vertical. Determine \(q,\) the amount of charge on each sphere.

In this step, we will analyze the forces acting on one of the spheres. Specifically, there are two forces acting on it: the gravitational force, acting vertically downward, and the electrostatic repulsion, acting horizontally away from the other sphere. The weight of the sphere is given by \(W = Mg\), where \(M\) is the mass of the sphere and \(g\) is the acceleration due to gravity.

The magnitude of the electrostatic repulsion can be calculated using Coulomb's law: \(F = \frac{kq^2}{r^2}\), where \(F\) is the force between the spheres, \(k\) is the electrostatic constant, \(q\) is the charge on each sphere, and \(r\) is the distance between the two spheres.

To find the distance between the spheres, we use the geometry of the problem. The spheres are at the ends of the strings, with angle \(\theta\) from the vertical. The distance between the spheres is the length of the base of an isosceles triangle with vertices at the common point and the two spheres. We can find the length of the base as follows: \(r = 2L \sin(\frac{\theta}{2})\)

The string supporting the sphere can also provide a vertical force to counteract the gravitational force, and a horizontal force to counteract the electrostatic repulsion. Both forces are equal to the tension on the string, \(T\). The vertical equilibrium equation is: \(T\cos(\theta) = Mg\), and the horizontal equilibrium equation is: \(T\sin(\theta) = F\), where \(F = \frac{kq^2}{r^2}\).

We can now solve for the charge \(q\) on each sphere. Divide the horizontal and vertical equilibrium equations: \(\tan(\theta) = \frac{F}{Mg}\). Substitute the expression for \(F\) from Coulomb's law and the expression for \(r\) obtained in Step 3: \(\tan(\theta) = \frac{kq^2}{Mg(2L\sin(\frac{\theta}{2}))^2}\). Rearrange the equation to solve for \(q\) and substitute the given values of \(M = 2.33 \text{ g}\), \(L = 45 \text{ cm}\), and \(\theta = 10.0^\circ\): \(q = \sqrt{\frac{\tan(10^\circ)(2)(2.33\text{ g})(9.81\frac{\text{m}}{\text{s}^2})(45\text{ cm}\times10^{-2}\text{m/cm})^2\sin^2(5^\circ)}{8.99\times10^9\frac{\text{N m}^2}{\text{C}^2}}}\) After calculating, we find that the charge on each sphere is: \(q \approx 1.00 \times 10^{-8} \text{ C}\).