Confusion about a capacitor with 2 different dielectrics

In summary, the conversation discusses a problem involving a parallel-plate capacitor with two slabs of dielectric between the plates. The goal is to find the capacitance of the capacitor. The proposed solution involves treating the two slabs as separate conductors and using the equation V = Q/(4πε0.r), but this is incorrect. A more appropriate equation is V = ∫0d E(z)dz, where E is the electric field and d is the plate separation. Another potential issue is the lack of specificity in the problem statement regarding the arrangement of the dielectric slabs. An easier approach involves using the equivalent capacitance formula for two capacitors in series or parallel.
  • #1
Hieu
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0
1. The problem statement, all variables, and given/known data
A parallel-plate capacitor has the space between the plates filled with to slabs of dielectric, one with constant K1, & one constant K2. each slab has thickness d⁄2, where d is plate separation, show the capacitance is?
cramster-equation-2012241714286346397246825332121459.gif

Homework Equations


C=Q/V
E=Q/(ε0.A)
mrqBvc

V=Q/(4πε0.r)

The Attempt at a Solution


I consider the conductor as 2 different conductors with different slabs of the dielectric. I think that is not a bad idea to find the answer, but my result is different from the answer, maybe it is wrong somewhere. Can you guys help me? Here is my solution.
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  • #2
Hello. Welcome to PF!

The mistake is using the equation ##V = \large \frac{Q}{4\pi\epsilon_0 r}##. This is not applicable to parallel plates.
 
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  • #3
You can express V as a line integral over the E field. Try using: $$V=\int_0^d E(z)dz=\int_0^{\frac {d} {2}} E_1(z)dz +\int_{\frac {d} {2}}^{d} E_2(z)dz $$where ##E=\frac {\sigma} {K\epsilon_0}##, d is plate separation and ##\sigma## is the charge density.

Peace
Fred
 
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  • #4
Hi Hieu,

Your problem statement should be more specific about the geometric arrangement of the dielectric slabs between the plates. Do they divide the space horizontally or vertically? Given the expected answer we can deduce that the arrangement is two horizontal slabs dividing the space vertically (case B below), but it might be not so obvious to a new student approaching this type of problem for the first time.
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  • #5
Fred Wright said:
You can express V as a line integral...

There is a much easier way for an old duffer like me that involves knowing how to calculate the equivalent capacitance for two capacitors in series or parallel.
 
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FAQ: Confusion about a capacitor with 2 different dielectrics

What is a capacitor?

A capacitor is a device that stores electrical energy by accumulating an electric charge between two conducting plates separated by an insulating material called a dielectric.

What is a dielectric?

A dielectric is an insulating material that is placed between the two conducting plates of a capacitor. It helps to increase the capacitance of the capacitor by reducing the electric field between the plates.

What happens when a capacitor has two different dielectrics?

When a capacitor has two different dielectrics, the electric field between the plates will be different for each dielectric. This will result in a different amount of charge being stored on each plate, leading to a difference in capacitance for each dielectric.

How does the dielectric constant affect a capacitor?

The dielectric constant, also known as the relative permittivity, is a measure of how well a material can store electrical energy in an electric field. The higher the dielectric constant, the more charge can be stored on the plates of a capacitor, resulting in a higher capacitance.

Why is it important to consider the dielectric in a capacitor?

The dielectric in a capacitor plays a crucial role in determining its overall capacitance. By choosing a dielectric with a higher dielectric constant, a capacitor can store more charge and have a higher capacitance, making it more efficient in storing electrical energy.

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