Calculating Capacitance: A Derivation

In summary, the conversation discusses the calculation of the capacitance of a parallel plate capacitor with a dielectric layer of relative permittivity εr and thickness d/2 fitted against one of the plates. The steps taken include using Gauss's law and the equation D = ε0(1+εr)E to determine the potential between the plates. While there is uncertainty about the accuracy of the solution, it is possible to approach the problem by considering the capacitor as two capacitors connected in series.
  • #1
raggle
8
0

Homework Statement



A parallel plate capacitor consists of two plates, each of area A, separated by a small distance d. in this gap, a dielectric of relative permittivity εr and thickness d/2 is fitted tight against one of the plates, leaving an air gap of thickness d/2 between it and the other plate. Calculate the capacitance of the capacitor

Homework Equations



Gauss's law ∫D.dS = ρ

D = ε0(1+εr)E

C = [itex]Q/V[/itex]

The Attempt at a Solution



First I said the plates have a charge density σ. By using Gauss's law in the dielectric I got D = σ, and then the second equation gives

E = D/ε0(1+εr) = σ/ε0(1+εr)

Then (this is where I'm worried I start going wrong) I use this to figure out the potential between the plates, and I split the integral up into two integrals, one inside the dielectric with E = σ/ε0(1+εr) and another outside the dielectric with E = σ/2ε0. Altogether this ends up giving:

V = -([itex]\int_{0}^{d/2} \frac{\sigma dl}{2\epsilon_0 (1+\epsilon)} + \int_{d/2}^{d} \frac{\sigma dl}{2\epsilon_0}[/itex])

and going through the integrals gives

V = [itex]\frac{(\epsilon - 1)d\sigma}{4\epsilon_0}[/itex]

Finally, putting Q = Aσ, I ended up with

C = [itex]4A \epsilon_0/(\epsilon -1)d[/itex]

Could someone tell me if I made a mistake somewhere? I'm quite bad at calculating capacitance.
Also is it possible to do this problem by thinking of the capacitor as two capacitors connected in series? Because when I try the problem that way I end up getting a d in the numerator, so I think I've slipped up somewhere.

Thanks!
 
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  • #2
raggle said:
D = ε0(1+εr)E

This should be D = ε0εrE


raggle said:
Then (this is where I'm worried I start going wrong) I use this to figure out the potential between the plates, and I split the integral up into two integrals, one inside the dielectric with E = σ/ε0(1+εr) and another outside the dielectric with E = σ/2ε0. Altogether this ends up giving:

Where are those 2's coming from?
 

1. What is a capacitor with dielectric?

A capacitor with dielectric is an electronic component that stores electrical energy by using an insulating material, known as a dielectric, between two conductive plates. The dielectric material increases the capacitance of the capacitor, allowing it to store more charge.

2. How does a capacitor with dielectric work?

A capacitor with dielectric works by creating an electric field between its two conductive plates. When a voltage is applied, the electrons in the conductive plates are attracted to each other, causing a build-up of charge. The dielectric material between the plates prevents the plates from touching and allows the capacitor to store the charge.

3. What materials are commonly used as dielectrics in capacitors?

Common materials used as dielectrics in capacitors include ceramic, plastic, paper, and electrolytic materials. The choice of dielectric material depends on the desired capacitance and other properties such as stability, temperature sensitivity, and cost.

4. What are the advantages of using a capacitor with dielectric?

The main advantage of using a capacitor with dielectric is that it can store more charge compared to a capacitor without a dielectric. This allows for higher capacitance values and better energy storage. Additionally, dielectrics can also improve the stability, temperature tolerance, and reliability of the capacitor.

5. How do you calculate the capacitance of a capacitor with dielectric?

The capacitance of a capacitor with dielectric can be calculated using the formula C = εA/d, where C is the capacitance in farads, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. The permittivity of the dielectric can be found in tables or specified by the manufacturer.

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