Question

Two Styrofoam balls with the same mass \(m=9.0 \times 10^{-8} \mathrm{kg}\) and the same positive charge \(Q\) are suspended from the same point by insulating threads of length \(L=0.98 \mathrm{m} .\) The separation of the balls is \(d=0.020 \mathrm{m} .\) What is the charge \(Q ?\)

Short answer

Answer: The charge Q on each Styrofoam ball is approximately \(3.1 \times 10^{-8}\ \text{C}\).

Step-by-step solution

Determine the acting forces

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 (\(T\)).

Apply Newton's second law of motion

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 \(T\sin\theta = F_e\).

Apply Coulomb's law

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.

Use the geometry of the system to find angle θ

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 \(x\): \(d = 2x\).

Combine the equations to find the charge Q

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 \(3.1 \times 10^{-8}\ \text{C}\).