0
$\begingroup$

I am having an issue with the question S1. My solution is slightly different to the one on the paper and was just wondering if anyone could tell me where I am going wrong.

Q: A parallel plate capacitor a composite dielectric. A thin sheet of dielectric of permittivty $\epsilon_1$ and a thickness $t_1$ is placed on top of a second thin dielectric sheet of permittyivty $\epsilon_2$ and thickness $t_2$. On top and bottom are parallel conducting plates of area S. How that the value of the capacitance is given by.

$$C=\frac{S}{\frac{t_1}{\epsilon _{1\:}}+\frac{t_2}{\epsilon _2}}[1]$$

My solution is as follows:

Outline of equations used for solution

$$D=\epsilon E\:\left[2\right]$$

where $\epsilon=\epsilon_0 \epsilon_1$

$$V=\frac{D}{\epsilon }t \space [3]$$

$$Q=CV \space [4]$$

$$D=\frac{Q}{A} \space [5]$$

My workings

$$V_T=V_1+V_2 \space [6]$$

$$V_T=\frac{D}{\epsilon _0\epsilon _{1\:}}t_1+\frac{D}{\epsilon \:_0\epsilon \:_{2\:}}t_2\:\left[7\right]$$

$$V_T=\frac{D}{\epsilon _0}\left(\frac{t_1}{\epsilon 1}+\frac{t_2}{\epsilon _2}\right)\:\left[8\right]$$

$$V_T=\frac{Q}{A\epsilon _0}\left(\frac{t_1}{\epsilon 1}+\frac{t_2}{\epsilon _2}\right)\:\left[9\right]$$

$$\frac{\left(V_T\epsilon _{0\:}A\right)}{Q}=\left(\frac{t_1}{\epsilon 1}+\frac{t_2}{\epsilon _2}\right)\:\left[10\right]$$

$$\frac{Q}{V_T\epsilon _0A}=\frac{1}{\left(\frac{t_1}{\epsilon \:1}+\frac{t_2}{\epsilon \:_2}\right)}\:\left[11\right]$$

$$C=\frac{A\epsilon _{0\:}}{\left(\frac{t_1}{\epsilon \:1}+\frac{t_2}{\epsilon \:_2}\right)}\:\left[12\right]$$

where

$A$=$S$

As you can see I have an extra epsilon but I cant see how the epsilon has be canceled out.

$\endgroup$
  • $\begingroup$ I think it's just a case of different notation. You are using $\epsilon_1$ to mean the $relative$ permittivity (or dielectric constant) of dielectric 1. This is non-standard notation. The question-setter is using $\epsilon_1$ to mean the absolute permittivity of dielectric 1, that is $\epsilon_0\times relative\ permittivity$. $\endgroup$ – Philip Wood Jan 22 '18 at 12:02
0
$\begingroup$

Just nomenclature. Your $\epsilon_1$ is relative permittivity while their $\epsilon_1$ is permittivity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy