You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.0 g mass from it. This stretches the spring to a length of 5.0 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm. What is the magnitude of the charge (in nC) on each bead?

Verified step by step guidance

1. Step 1: Identify the forces acting on the spring in each scenario. In the first case, the spring is stretched due to the gravitational force exerted by the 1.0 g mass. In the second case, the spring is stretched due to the electrostatic force between the two charged beads.
2. Step 2: Use Hooke's Law to relate the force and the spring's extension. Hooke's Law is given by \( F = k \Delta x \), where \( F \) is the force, \( k \) is the spring constant, and \( \Delta x \) is the change in length of the spring from its unstretched length.
3. Step 3: Calculate the spring constant \( k \) using the first scenario. The force due to gravity is \( F = mg \), where \( m \) is the mass (1.0 g = 0.001 kg) and \( g \) is the acceleration due to gravity (9.8 m/s²). The extension \( \Delta x \) is the difference between the stretched length (5.0 cm = 0.05 m) and the unstretched length (4.0 cm = 0.04 m). Substitute these values into \( k = \frac{F}{\Delta x} \).
4. Step 4: Use the spring constant \( k \) calculated in Step 3 to analyze the second scenario. The force stretching the spring in this case is the electrostatic force between the two charged beads, given by Coulomb's Law: \( F = \frac{k_e q^2}{r^2} \), where \( k_e \) is Coulomb's constant (\( 8.99 \times 10^9 \, \mathrm{N \cdot m^2 / C^2} \)), \( q \) is the charge on each bead, and \( r \) is the distance between the beads (equal to the stretched length of the spring, 4.5 cm = 0.045 m).
5. Step 5: Equate the electrostatic force \( F \) from Coulomb's Law to the spring force \( F = k \Delta x \), where \( \Delta x \) is the extension of the spring in the second scenario (4.5 cm - 4.0 cm = 0.005 m). Solve for \( q \) (the charge on each bead) by rearranging the equation \( \frac{k_e q^2}{r^2} = k \Delta x \).

Key Concepts

Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its extension or compression from its natural length, expressed as F = kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. This principle is essential for understanding how the spring behaves when weights are added or when forces are applied.

Electrostatic Force

The electrostatic force is the force between charged objects, described by Coulomb's Law, which states that the force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This concept is crucial for analyzing the interaction between the charged beads and how it affects the spring's extension.

Equilibrium Condition

The equilibrium condition occurs when the net force acting on an object is zero, meaning all forces balance out. In this scenario, the gravitational force acting on the mass and the spring force must equal the electrostatic force between the charged beads, allowing us to set up equations to solve for the unknown charge on the beads.