# Problem Restatement Two identical spheres each with charge \( q = 6.80\,\mathrm{nC} \) hang in static equilibrium from massless filaments of length \( L \). The angle between the two filaments is \( \theta = 4.50^\circ \). Find the length \( L \) in meters. --- ## Step 1: Draw the Free-Body Diagram Each sphere is subject to: - Tension \( T \) (along the filament) - Gravitational force \( mg \) (downward) - Electrostatic repulsion \( F_e \) (horizontally, away from the other sphere) The spheres hang symmetrically, so each one makes an angle \( \alpha = \theta/2 \) with the vertical. --- ## Step 2: Write Force Equilibrium Equations ### Vertical (y-direction): \[ T \cos\alpha = mg \] ### Horizontal (x-direction): \[ T \sin\alpha = F_e \] --- ## Step 3: Express Electrostatic Force The spheres are separated by distance \( d \): \[ d = 2L \sin\alpha \] Coulomb's Law gives: \[ F_e = \frac{1}{4\pi\epsilon_} \frac{q^2}{d^2} \] where \( \epsilon_ = 8.854 \times 10^{-12}\,\mathrm{C^2/(N\cdot m^2)} \). --- ## Step 4: Eliminate Tension \( T \) Divide the second equation by the first: \[ \frac{T \sin\alpha}{T \cos\alpha} = \frac{F_e}{mg} \implies \tan\alpha = \frac{F_e}{mg} \] --- ## Step 5: Substitute \( F_e \) and \( d \) \[ \tan\alpha = \frac{1}{mg} \cdot \frac{1}{4\pi\epsilon_} \frac{q^2}{d^2} \] But \( d = 2L\sin\alpha \), so \( d^2 = 4L^2\sin^2\alpha \): \[ \tan\alpha = \frac{1}{mg} \cdot \frac{q^2}{4\pi\epsilon_} \cdot \frac{1}{4L^2\sin^2\alpha} \] \[ \tan\alpha = \frac{q^2}{16\pi\epsilon_ m g L^2 \sin^2\alpha} \] --- ## Step 6: Solve for \( L \) \[ L^2 = \frac{q^2}{16\pi\epsilon_ m g \sin^2\alpha \tan\alpha} \] \[ L = \sqrt{ \frac{q^2}{16\pi\epsilon_ m g \sin^2\alpha \tan\alpha} } \] --- ## Step 7: Plug in the Numbers Given: - \( q = 6.80\,\mathrm{nC} = 6.80 \times 10^{-9}\,\mathrm{C} \) - \( \theta = 4.50^\circ \implies \alpha = 2.25^\circ \) - Assume mass of each sphere \( m \) is **not given** (you must specify \( m \) to get a numerical answer!). - \( g = 9.80\,\mathrm{m/s^2} \) - \( \epsilon_ = 8.854 \times 10^{-12}\,\mathrm{C^2/(N\cdot m^2)} \) Calculate trigonometric terms: \[ \alpha = 2.25^\circ = .03927\,\mathrm{rad} \] \[ \sin\alpha = \sin(2.25^\circ) = .03927 \] \[ \tan\alpha = \tan(2.25^\circ) = .03932 \] Now, substitute: \[ L = \sqrt{ \frac{(6.80 \times 10^{-9})^2}{16\pi \times 8.854 \times 10^{-12} \times m \times 9.80 \times (.03927)^2 \times .03932} } \] --- ## Step 8: Final Formula and Discussion Since the mass \( m \) is **not specified**, the **final answer** is: \[ \boxed{ L = \sqrt{ \frac{(6.80 \times 10^{-9})^2} {16\pi \times 8.854 \times 10^{-12} \times m \times 9.80 \times (.03927)^2 \times .03932} } } \] - If a value of \( m \) is given, substitute it in kg and compute \( L \). - **You must know the mass of each sphere to get a numerical answer.** --- ## Summary - **Key method:** Use force equilibrium and geometry to relate \( L \), \( q \), \( m \), \( \theta \). - **Final answer:** See box above; plug in \( m \) when known.