Dielectric Compacitors Confirmation

[Trenthan]

Dielectric compacitors can contain 1 or multiple dielectrics, if we say have a capacitor with 3 dielectrics between the two plates. We can model it as 3 capacitors in series. Since we know that Capacitors in series the equivalent is

(1/C_eq) = (1/C₁)+(1/C₂)+(1/C₃)

Wouldnt the overall capacitance be lower by having 3 dielectric between the two plates. If we separated the 3 into 3 seprate dielectric capacitors wouldn't the charge across all 3 remain constant Q₀=Q₁=Q₂=Q₃.

Therefore the potential for each capacitor depends only on the width of the dielectric since
V_cap#1-3=Ed=(Q/Ae₀)*d

Since A is constant all the same size, e₀= 8.85*10^-12, and Q is constant since they are in series. The potential depends only on "d" the width of the dielectric, thus capacitance C=Ae₀/d, also only depends on the "d".

Basically what I am trying to say is if we have 3 individual capacitors, than arrange them in series the capacitance islower. Therefore if we want to build a capacitor with a high capacitance we only use a single dielectric since multiples decrease the capacitance of the capacitor due to the series arrangement of dielectric in the capacitor.

Hope this is clear thanks Trent

 

[Trenthan]

ok i believe i am right or on the right track i found this website which helped a lot hopefully it is correct

http://dev.physicslab.org/Document....ctrostatics_DielectricsBeyondFundamentals.xml

if someone could just check site is reliable and correct than I've solved my own puzzle, its got some good example too** might help others

Cheers Trent