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(adsbygoogle = window.adsbygoogle || []).push({});

(1/C_{eq}) = (1/C_{1})+(1/C_{2})+(1/C_{3})

Wouldnt the overall capacitance be lower by having 3 dielectric between the two plates. If we separated the 3 into 3 seprate dielectric capacitors wouldnt the charge across all 3 remain constant Q_{0}=Q_{1}=Q_{2}=Q_{3}.

Therefore the potential for each capacitor depends only on the width of the dielectric since

V_{cap#1-3}=Ed=(Q/Ae_{0})*d

Since A is constant all the same size, e_{0}= 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=Ae_{0}/d, also only depends on the "d".

Basically what im 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

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# Homework Help: Dielectric Compacitors Confirmation

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