🚀 go-pugleaf

Joseph Bellina describes an experiment which seems to confirm familiar
textbook formulas involving capacitors. This always happens (I will try to
explain below) when capacitors used have the same dielectric material. I
predict that the results of his experiments will be differnt when
capacitors made from different dielectric material are used instead.

Here is an explanation why charges on capacitors in series are identical
when dielectric materials are the same. To simplify assume that capacitors
have parallel plates. Consider C1 and C2 in series with a source of
constant voltage V. The leakage current i=V/(R1+R2), where R1 and R2 are
leakage resistances. The voltage on C1 is V1=V*R1/(R1+R2) and therefore
Q1=C1*V1. Similarly, the voltage on C2 is V2=V*R2/(R1+R2) and therefore
Q2=C2*V2. Thus the ratio Q1/Q2=(C1*R1)/(C2*R2). Use parallel plate formula
to show that (C1/C2)=e1/e2, where e1 and e2 stand for dielectric constants.
Also note that R1/R2=r1/r2, where r1 and r2 stand for resistivity because
(R=r*length/area).

As you can see, the ratio of charges, Q1/Q2 is equal to (e1*r1)/(e2*r2).
That ratio is unity when the dielectric materials are the same, EVEN WHEN
C1 AND C2 ARE VERY DIFFERENT ! In general, however, Q1 and Q2 are expected
to be different. Try a cheap paper capacitor and an expensive mica capacitor
to show how different Q1 and Q2 can be. By the way, the equal charge hypothesis
was "experimentally verified" by J.J. Thomson (who discovered electrons)
using two Leyden jars, capacitors made from the same material, glass.

         Ludwik Kowalski             kowalski@apollo.montclair.edu