# Calculate capacitance with a dielectric

In summary, the final capacitance of an isolated charged parallel-plate capacitor with plate separation d=2 mm and area A=10 cm2, after a polystyrene dielectric slab with thickness equal to half the plate separation and only half the area of the plates is inserted, is given by the equation C = ε0A/2d + ε0A2/2d + κε0A/2d, where κ is the dielectric constant of air, and gives a final capacitance of C = 2.2135e-10 Farads."

## Homework Statement

An isolated charged parallel-plate capacitor with the plate separation d=2 mm and area of A=10 cm2 is initially empty. One slides a polystyrene dielectric slab with thickness equal to half the plate separation and with only half the area of the plates into the region between the plates. What is the final capacitance?

## Homework Equations

C = κ((ε0A)/d)

Since no specification is made, I assume there is air in-between, thus use K of 1.00059 (given in a table)

Since the dielectric slab is only half the area, I convert it to 5 cm2, or 0.0025 m2

## The Attempt at a Solution

(8.854e-12 X 0.0025) / 0.0001m

= 2.2135e -10

I can't tell 100% what you've actually done. I'd have to say this isn't correct considering it looks like you've neglected the rest of the capacitor after the dielectric was inserted.

What it looks like to me is that you'll need to break down the capacitor into what looks like 3 partsThe part that is the same size as the dielectric, but is the air that is above it.
The dielectric part itself
The rest of the plate which isn't covered by the dielectric at all

C = εA/d2 {area is halved} + εA2/2d {area and d are halved} + kε2A/2d {area and d are halved, but has the dieletric}

It doesn't say to only focus on the area which is occupied on the dielectric

Liquidxlax said:
I can't tell 100% what you've actually done. I'd have to say this isn't correct considering it looks like you've neglected the rest of the capacitor after the dielectric was inserted.

What it looks like to me is that you'll need to break down the capacitor into what looks like 3 parts

The part that is the same size as the dielectric, but is the air that is above it.
The dielectric part itself
The rest of the plate which isn't covered by the dielectric at all

C = εA/d2 {area is halved} + εA2/2d {area and d are halved} + kε2A/2d {area and d are halved, but has the dieletric}

It doesn't say to only focus on the area which is occupied on the dielectric

If I try this, it gives me:

C = ε0A/2d + ε0A/d + κε0A/d (since the 2nd and third expression the 2's cancel out)

But then the unit give out something weird: C2/Nm2 X m2/m = C2m/N

I must be doing something wrong, because it's not any unit I can use. Although if the units were re-arranged, they'd give a Farad, considering I have Nm which gives J, it could theoretically give C / V, or C / J/C

Liquidxlax said:
What it looks like to me is that you'll need to break down the capacitor into what looks like 3 parts

The part that is the same size as the dielectric, but is the air that is above it.
The dielectric part itself
The rest of the plate which isn't covered by the dielectric at all

C = εA/d2 {area is halved} + εA2/2d {area and d are halved} + kε2A/2d {area and d are halved, but has the dielectric}

It doesn't say to only focus on the area which is occupied on the dielectric

Yes, you do need to break the capacitor into what looks like three parts.

However, you can't simply add all three capacitances arithmetically. You must consider which parts act as if they're in parallel, and which parts act as if they're in series.

SammyS said:
Yes, you do need to break the capacitor into what looks like three parts.

However, you can't simply add all three capacitances arithmetically. You must consider which parts act as if they're in parallel, and which parts act as if they're in series.

Wouldn't the whole system be in parallel? Considering the two plates are parallel, and then the dielectric is slid in-between the two

Wouldn't the whole system be in parallel? Considering the two plates are parallel, and then the dielectric is slid in-between the two
In a word: No.

The half with the dielectric is as if you have a capacitor, area, A/2, plate separation, d/2, with the dielectric, in series with a similar size capacitor filled with air.

By the way, considering significant digits and all, we usually take the dielectric constant or air as being 1 .

## 1. How do you calculate capacitance with a dielectric?

Capacitance with a dielectric can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the capacitor plates, and d is the distance between the plates.

## 2. What is the role of a dielectric in capacitance?

A dielectric material is used in capacitors to increase the capacitance by reducing the electric field between the plates. It does this by polarizing in the presence of an electric field and creating an opposing field, effectively increasing the distance between the plates.

## 3. Can you calculate capacitance with multiple dielectric materials?

Yes, if there are multiple dielectric materials, the effective permittivity (ε) is calculated by taking the sum of the individual permittivity values weighted by their respective areas. The formula for this is ε = (ε1A1 + ε2A2 + ...)/Atotal.

## 4. How does the dielectric constant affect capacitance?

The dielectric constant, also known as the relative permittivity, is a measure of how much a dielectric material can increase the capacitance. A higher dielectric constant means a higher capacitance, as the material has a greater ability to polarize and reduce the electric field between the plates.

## 5. What are some common dielectric materials used in capacitors?

Some common dielectric materials used in capacitors include air, paper, plastic, and ceramic materials. Each material has a different dielectric constant and is chosen based on the specific requirements of the capacitor, such as size, capacitance, and operating voltage.

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