Capacitors -> Storing 500 Joules in a capacitor with a simple 9 volt battery
You don't. 

500 J of energy at 9 volts implies a capacitance C such that$$C = \frac {2 E}{V^2} = \frac{1000}{81} = 12.3 \text{ F}$$ Now, the formula for capacitance of two plates of area A and separation d is$$C = \frac{k\times \epsilon_0 \times A}{d}$$ where $$\epsilon_0 = 8.845\times 10^{-12} $$ and k is the relative permittivty of the dielectric. Assuming the spacing between plates is .01 inch (.000254 meters) which is a reasonable thickness for paper, and the relative permittivity of the dielectric is 2, this can be rearranged to give $$A =\frac{ C\times d}{k\times \epsilon_0} = \frac{12.3\times 2.54\times 10^{-4}}{{2\times8.845\times 10^{-12} }} = 1.76\times 10^8\text{ m}^2$$ or about 176 square kilometers. This says that, for instance, if you were planning to use aluminum foil with a paper dielectric, you'd need about 300 square kilometers of foil (paper has a relative permittivity of about 2.3). A 200-square-foot roll of aluminum foil (18.4 square meters) currently costs about 10 dollars. This suggests that you will need to spend about 160 million dollars if you plan to buy your foil at the local supermarket.

You'll also need about 300 square kilometers of paper, but I'll leave the pricing of that up to you. 

You might also give some thought to exactly how much this will weigh. (Hint: you're looking at about 76 thousand cubic meters.)

You can, of course, use something thinner, like Saran wrap, which is about 1/20 the thickness I've specified. This will cut your foil requirements by a factor of 20, but I'm not sure 8 million bucks for foil is what you'd call a practical number.