A parallel-plate capacitor has a plate area of 100 cm ^2 and a separation of 1.0 cm between its plates. It is connected across a battery with an emf of 24 volts. Determine the force of attraction between the plates.

Step-by-step solution

The force of attraction between the plates of a parallel-plate capacitor is given by: F = 1 2 ₀ E^2 A Since the electric field E = V d , the formula becomes: F = ₀ A V^2 2 d^2 Given values are: A = 100 cm ^2 = 10⁻² m ^2 d = 1.0 cm = 10⁻² m V = 24 V ₀ = 8.85 10⁻¹² F/m Substituting these values into the formula: F = 8.85 10⁻¹² 10⁻² (24)^2 2 (10⁻²)^2 F = 8.85 10⁻¹⁴ 576 2 10⁻⁴ F = 8.85 288 10⁻¹⁰ N F = 2548.8 10⁻¹⁰ N F 2.5 10⁻⁷ N