Fully filled capacitors with parallel dielectrics problem

In summary, the conversation discusses the calculation of the electric field and surface density of induced charge in a capacitor with two dielectrics. The first dielectric has a surface area of S/3 and a dielectric constant of er1, while the second dielectric has a surface area of 2S/3 and a dielectric constant of er2. The question is to calculate the electric field and surface density of charge. The calculation involves applying Gauss' theorem, but the resulting equations do not hold true. The reason for this discrepancy is that the charge density on the plates is not uniform due to the different dielectric constants. Further explanation is needed to understand this concept.
  • #1
Amaelle
310
54
consider that we have two dielectrics inside a capacitor as shown in the picture, let0s consider also that Q is the charge of the capacitor and d the distance between the two plates , the first dielectric occupy a surface of S/3 with a dielectric constant of er1 and the second a surface of 2S/3 with a dielectric constant of er2, the question is calculate the electric field inside the capacitor and the surface density of the induced charge (see the uploaded image)
During the calculation I faced the following doubts: I applied gauss theorem to the flux density in both areas of the dielectrics and I got:
QEAJg.gif


σ is the surface density of the free charges =Q/S,

8YAxA.gif


doing the same with the second surface I got that
y48Zg.gif


and because my two dielectric are in parallel so E1=E2

this gives me
MHZ6c.gif


BUT this not true! so where did my logic fails??

Many thanks in advance!
dielectric.png
dielectric.png
dielectric.png
 

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  • y48Zg.gif
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  • #2
Amaelle said:
BUT this not true! so where did my logic fails??
Charge density on the plates will not be uniform.
 
  • Like
Likes Amaelle
  • #3
Thanks for your prompt reply, it makes sens but can you elborate more?
 
  • #4
I just go why the charge density is different from dielectric to dielectric
many thanks !
 

1. What is a fully filled capacitor with parallel dielectrics problem?

A fully filled capacitor with parallel dielectrics problem refers to a scenario where a capacitor is filled with two or more dielectric materials that are arranged in parallel. The problem arises when trying to calculate the total capacitance of the system.

2. How do you calculate the total capacitance of a fully filled capacitor with parallel dielectrics?

To calculate the total capacitance, you can use the equation C = C1 + C2 + ... + Cn, where C1, C2, and Cn are the capacitance values of each individual dielectric material. Alternatively, you can use the formula C = ε0 * A / d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

3. How does the dielectric constant affect the total capacitance of a fully filled capacitor?

The dielectric constant, also known as the relative permittivity, is a measure of how well a material can store electrical energy. It affects the total capacitance by increasing it, as it reduces the electric field between the plates and allows for more charge to be stored.

4. Can dielectric materials with different dielectric constants be used in a fully filled capacitor with parallel dielectrics?

Yes, dielectric materials with different dielectric constants can be used in a fully filled capacitor with parallel dielectrics. The total capacitance will be a combination of the individual capacitance values of each material, and it will also depend on the arrangement and thickness of each material.

5. How can a fully filled capacitor with parallel dielectrics be used in practical applications?

Fully filled capacitors with parallel dielectrics can be used in various applications, such as in electronic circuits, power supplies, and energy storage systems. They can also be used to increase the capacitance of a system without increasing the physical size of the capacitor.

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