A parallel-plate capacitor having plate area $$20 cm^2$$ and separation between the plates 1.00 mm is connected to a battery of $$12.0 V$$. The plates are pulled apart to increase the separation to $$2.0 mm$$. Calculate the stored energy in the electric field before and after the process.
area $$=a = 20 cm^2 = 2 \times 10^{-2} m^2$$
$$d = $$ separation $$= 1mm = 10^{-3} m$$
$$Ci = \dfrac{\varepsilon_0 \times 2 \times 10^{-3}}{10^{-3}} = 2 \varepsilon_0$$ $$Cf = \dfrac{\varepsilon_0 \times 2 \times 10^{-3}}{2 \times 10^{-3}} = \varepsilon_0$$
Before the process
$$E_i = (1/2) \times Ci \times v^2 = (1/2) \times 2 \times 8.85 \times 10^{-12} \times 144 = 12.7 \times 10^{-10} J$$
After the force
$$E_i = (1/2) \times Cf \times v^2 = (1/2) \times 8.85 \times 10^{-12} \times 144 = 6.35 \times 10^{-10}J$$