# Dielectric Capacitance of Parallel Plate Cap

• jesuslovesu
In summary, the problem involves finding the capacitance of a parallel plate capacitor filled with a dielectric material whose relative permittivity varies from 1 to 10. Using the equations C = Q/V and E = σ/ε₀εᵣ, and integrating the electric displacement field, one can determine that the capacitance is given by C = 9Aε₀/dln(10).
jesuslovesu

## Homework Statement

A parallel plate cap of cross sectional area A and thickness d is filled with a dielectric material whose relative permittivity varies from $$\epsilon_r = 1 to \epsilon_r = 10$$
Find the capacitance.

## The Attempt at a Solution

C = Q/V

E = $$\sigma / e0 \epsilon_r$$

I am thinking that since V = $$- \int{}{} E dot dl$$ I need to integrate, I can't quite figure out the relationship between this though.

I kind of would like to just evaluate the from 10 to 1 and get C = 9 A e0 / d but that isn't quite it.
C = 9A e0 / d is close to the answer, but somehow a ln sneaks in there too?

how does the dielectric vary from 1 to 10? linearly?

jesuslovesu said:

## Homework Statement

A parallel plate cap of cross sectional area A and thickness d is filled with a dielectric material whose relative permittivity varies from $$\epsilon_r = 1 to \epsilon_r = 10$$
Find the capacitance.

## The Attempt at a Solution

C = Q/V

E = $$\sigma / e0 \epsilon_r$$

I am thinking that since V = $$- \int{}{} E dot dl$$ I need to integrate
no
$$V=\int \vec E \cdot d\vec \ell=\int_{x=0}^{x=\ell}\frac{D}{\epsilon(x)}dx=D\int_{x=0}^{x=\ell}\frac{1}{\epsilon(x)}dx$$
and if eps varies linearly one does end up with a log term in the capacitance. (here D=\epsilon E is the electric displacemnt field)

I kind of would like to just evaluate the from 10 to 1 and get C = 9 A e0 / d but that isn't quite it.
C = 9A e0 / d is close to the answer, but somehow a ln sneaks in there too?[/QUOTE]

Yes, sorry it's linearly.

Cool thanks, I'll give it a shot now.

## 1. What is the dielectric capacitance of a parallel plate capacitor?

The dielectric capacitance of a parallel plate capacitor is a measure of the ability of the capacitor to store electrical charge. It is influenced by the distance between the plates, the surface area of the plates, and the type of dielectric material between the plates.

## 2. How is the dielectric capacitance of a parallel plate capacitor calculated?

The dielectric capacitance of a parallel plate capacitor can be calculated using the formula C = ε0A/d, where C is the capacitance, ε0 is the permittivity of free space, A is the surface area of the plates, and d is the distance between the plates.

## 3. What factors affect the dielectric capacitance of a parallel plate capacitor?

The dielectric capacitance of a parallel plate capacitor is influenced by the distance between the plates, the surface area of the plates, and the type of dielectric material between the plates. It is also affected by the dielectric constant of the material, which is a measure of its ability to store electrical charge.

## 4. How does the dielectric material affect the capacitance of a parallel plate capacitor?

The type of dielectric material between the plates of a parallel plate capacitor can greatly affect its capacitance. Materials with higher dielectric constants, such as ceramic or mica, can increase the capacitance compared to materials with lower dielectric constants, such as air or vacuum.

## 5. What is the importance of understanding the dielectric capacitance of a parallel plate capacitor?

Understanding the dielectric capacitance of a parallel plate capacitor is important in designing and using electronic circuits. It allows for the precise calculation and control of the amount of charge that can be stored in the capacitor, which is essential for the proper functioning of many electronic devices.

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