# How much will this capacitor hold?

I am making a homemade capacitor for a science project. I have a 5 volt power source that I need to charge the capacitor with. If I cut out two 10"x10" slips of foil and 1 10"x10" piece of paper, and I put the paper in between, what would the capacitance be? Please keep in mind I need to have it charge to 5 volts. If this model doesn't work, can you please direct me to a better model. I cannot just buy a capacitor, though.

• You forgot to mention the topology. – Ignacio Vazquez-Abrams Nov 3 '13 at 22:19
• What do you mean? I don't know much about capacitors. – Justin Chang Nov 3 '13 at 22:21
• Is it flat? Folded? Rolled up? – Ignacio Vazquez-Abrams Nov 3 '13 at 22:22
• hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html#c1 - probably about 40nF if i got my maths right and the assembly isn't rolled up. I assumed the paper was 0.4mm - just a guess – Andy aka Nov 3 '13 at 22:53
• By the way, if you use two sheets of paper and then roll up the result, you get up to twice the capacitance because most of the foil will be forming a capacitor to the other foil on both sides. Also, the tighter you wrap it or otherwise compress it, the higher the capacitance. This is essentially a air gap capacitor. At 5 V you don't need to worry about insulation distance, so the closer the two plates are, the higher the capacitance. – Olin Lathrop Nov 3 '13 at 23:07

Lets assume you are making the simplest type of plate capacitor which consists of two conductors (your foils) separated by an insulator (your paper). (see http://en.wikipedia.org/wiki/Capacitor and http://en.wikipedia.org/wiki/Permittivity)

You will need to know (or look up) the permittivity of the insulation material and to measure the thickness with reasonable accuracy but at best you could only estimate the value of the capacitor that you would make. Printing paper is about 0.1 mm (but can vary considerably depending on its 'weight' - g/m^2). Paper has a permittivity of about 3.3 times that of a vacuum (ε0 = 8.8541878176.. × 10^−12 F/m) i.e. 2.8182 x 10^-11 F/m.

Putting these figures into you calculation (and converting inches to m etc.) will give

   C   = 2.8x 10^-11 * 0.066 (area in m^2) / 10^-4(thickness in m)   Farads


giving a capacitor of around 0.018uF

(I'm certain somebody will correct me if I have my arithmetic wrong)

The actual capacitance will certainly be less than this due to faults in construction (air pockets increasing plate distance etc.).

Personally I wouldn't use paper for the insulator/dielectric as moisture content can vary considerably and produces a 'leaky' capacitor. The last time I did this with my students we used printer transparency sheets.