Energy stored in the capacitor is given by:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{V}}}^{{\mathbf{2}}}}$

Capacitance:

$\overline{){\mathbf{C}}{\mathbf{=}}\frac{{\mathbf{\epsilon}}_{\mathbf{0}}\mathbf{A}}{\mathbf{d}}}$

**1)**

Area, A = πr^{2}, but r = 7.0/2 = 3.5 cm

A = π(3.5 × 10^{-2})^{2} = 3.85 × 10^{-3} m^{2}

Capacitance:

A capacitor consists of two 7.0-cm-diameter circular plates separated by 1.0 mm. The plates are charged to 160 V , then the battery is removed.

1) How much energy is stored in the capacitor?

2) How much work must be done to pull the plates apart to where the distance between them is 2.0 mm?