Re: Leyden jar capacitors as source for electrostatic experiments

Leyden Jar Capacitors

Without the water you can calculate the capacitance of the bottles to be 

C=k(eA/d)  

C = Capacitance
K = the dielectric constant
e = the permittivity of free space, constant, e = 8.854*10^12 C^2/(N*m^2).
A = the surface area of the capacitor, constant. 
d = the distance between the wire and the surface of the bottle, constant. 

Keeping everything the same, e (a universal constant that won’t change), 
A, and d, and adding water to make a Leyden Jar capacitor as you’re doing, 
you notice that:

k(air) = 1.00059 
k(water) = 80 

By adding water you’ve essentially increased your capacitance 80 times.

So what does this do for your experiment?

This lets you store more charge in the capacitor at less voltage as 
charge, voltage and capacitance are all related by the equation:

C=Q/V.

Let’s say for example that the static charge on your television is the 
same every time you turn it on.  It draws a lot of charge to the glass and 
then saturates.  This means that while you may have the same amount of 
charge every time you do the experiment you have still cut your voltage by 
a factor of 80 by adding the water.

So when before you may have had 80,000V from the television you now have 
only 1,000V.  (I have no idea how much voltage is actually on the 
television, mind you, this is just a guess.)  From reading a bit about 
your experiment it says that you need 4,500V to make it barely turn and 
7,000V to make it turn rapidly.  So you’re probably around the 4,500V mark 
with the Leyden Jar set up.

What do the Leyden Jar capacitors let you do?

You can use them to store more charge on the capacitors than you could 
with the empty bottles.  Feasibly, you could store more and more charge 
until you get quite a bunch of energy in these things.  It follows that 
your voltage will have to increase with the increase in charge as your 
capacitance is a constant. 

Reference:  Serway, "Physics for scientists & Engineers, 3rd Ed."