A parallel-plate capacitor with a plate area of 20 cm^2 and a...

The force between the plates of a parallel-plate capacitor connected to a constant voltage source $V$ is given by:

$F = \dfrac{\epsilon_0 A V^2}{2 x^2}$

The work done by the external agent in pulling the plates apart from a separation $x_1$ to $x_2$ is:

$W = \int_{x_1}^{x_2} F \, dx = \int_{x_1}^{x_2} \dfrac{\epsilon_0 A V^2}{2 x^2} \, dx$

W = \dfrac{\epsilon_0 A V^2}{2} \left[ -\dfrac{1}{x} \right]_{x_1}^{x_2} = \dfrac{\epsilon_0 A V^2}{2} \left(\dfrac{1}{x_1} - \dfrac{1}{x_2} \right)

Given values are:

$A = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$

$V = 12.0 \text{ V}$

$x_1 = 1.00 \text{ mm} = 10^{-3} \text{ m}$

$x_2 = 2.0 \text{ mm} = 2 \times 10^{-3} \text{ m}$

$\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

Substituting the values into the work equation:

$W =$\dfrac{8.85 \times 10^{-12} \times 20 \times 10^{-4} \times (12.0)^2}{2} \left( \dfrac{1}{10^{-3}} - \dfrac{1}{2 \times 10^{-3}} \right)$<br><br>$W = 8.85 \times 10^{-15} \times 144 \times (1000 - 500)$<br><br>$W = 8.85 \times 10^{-15} \times 144 \times 500$<br><br>$W = 6.37 \times 10^{-10} \text{ J}$<br><br>Answer:$6.37 \times 10^{-10} \text{ J}