Question

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?

Accepted Answer

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. 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. 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 $$ k = \frac{F}{\Delta x} . $$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 ($$ 8.99 	imes 10^9 \, \mathrm{N \cdot m^2 / C^2} $$), $$ q  is $$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). 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 $$ \frac{k_e q^2}{r^2} = k \Delta x .$$

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Concetti chiave

Hooke's Law

Electrostatic Force

Equilibrium Condition