Finding electric field through capacitor

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In a parallel-plate capacitor with two dielectrics, the electric field through each dielectric can be calculated using the formula E = V/d. For Dielectric A, with a width of 0.001m and relative permittivity of 2, the electric field is 4000 V/m. For Dielectric B, with a width of 0.003m and relative permittivity of 4, the electric field is 2000 V/m. The discussion emphasizes the importance of the electric displacement field D, which relates to the electric field E and is calculated using D = eE. Understanding these relationships is crucial for analyzing capacitors with multiple dielectrics.
geft
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A parallel-plate capacitor with two dielectrics is placed in a circuit with 10V applied.

Dielectric A has 0.001m width and relative permittivity 2. Dielectric B has 0.003m width and relative permittivity 4. The plates have area of 0.01m^2. What is the electric field through each dielectric? (Ans: 4000 V/m and 2000 V/m)

I know the formula used is E = V/d, but how do I divide the voltage between these dielectrics?
 
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hi geft! :smile:
geft said:
I know the formula used is E = V/d, but how do I divide the voltage between these dielectrics?

why would the width matter?

start with the electric displacement field D

what is its magnitude, and how is E related to it? :wink:
 
D = eE

E = D/e = V/d

...but I wouldn't be able to get the displacement field without the electric field, right?
 
displacement field was invented for the capacitor …

D = -Q/A :wink:
 
Didn't know that formula, thanks! :)
 
oooh, i forgot to mention …

that's exactly why the electric displacement field D is measured in units of coulombs per square metre, C/m2 :biggrin:
 
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