A parallel-plate capacitor with a plate area of 20 cm ^2 and a separation of 1.00 mm between the plates is connected to a 12.0 V battery. The plates are subsequently pulled apart to increase the separation to 2.0 mm . Using the expression for the force between the plates, find the work done by the person pulling the plates apart.

Step-by-step solution

The force between the plates of a parallel-plate capacitor connected to a constant voltage source V is given by: F = ₀ A V^2 2 x^2 The work done by the external agent in pulling the plates apart from a separation x₁ to x₂ is: W = _ x₁ ^ x₂ F , dx = _ x₁ ^ x₂ ₀ A V^2 2 x^2 , dx W = ₀ A V^2 2 [ - 1 x ]_ x₁ ^ x₂ = ₀ A V^2 2 ( 1 x₁ - 1 x₂ ) Given values are: A = 20 cm ^2 = 20 10⁻⁴ m ^2 V = 12.0 V x₁ = 1.00 mm = 10⁻³ m x₂ = 2.0 mm = 2 10⁻³ m ₀ = 8.85 10⁻¹² F/m Substituting the values into the work equation: W = 8.85 10⁻