Question

$N$dielectrics are introduced in series in a capacitor of thickness $D$. Each dielectric have width $d=\frac{D}{N}$ & dielectric constant of $mth$ dielectric is given by$K_m=K\left(1+\frac{m}{N}\right):[N>>{10}^3]$, Area of plates$=A$]. Net capacitance is given by $\frac{\left[K{Є}_{0}A\right]}{\left[\alpha D\ln 2\right]}$. Find the value of$\alpha$.

Answer 1

Step 1: Given data:

No dielectrics $N$are connected in series

Thickness $=D$

Each dielectric has a width $\left(d\right)=\frac{D}{N}$

The dielectric constant of $mth$ dielectric is given by-

$K_m=K\left(1+\frac{m}{N}\right):[N>>{10}^3]$………………..(1)

Area of plates$=A$

Net capacitance, $C=\alpha \left(\frac{K{\epsilon }_{0}A}{D\ln 2}\right)$……………..(2)

Step 2: Use the formula and integrates the function:

Let at a distance of $x$there is a dielectric of width $dx$ then-

$\frac{x}{m}=\frac{d}{N}$………………………(3)

$dc=\frac{{\epsilon }_0{K}_{m}A}{dx}$, where $dc$ is the element capacitance, A is area

The series combination of dielectrics formula,

$\frac{1}{C_{eq}}=\int \frac{1}{dc}$……………..(4)

Therefore it can be written as,

$\frac{1}{dc}=\frac{dx}{K_m{\epsilon }_{0}A}$……………………….(5)

Step 3: Calculate the value of $\alpha$

Substitute the value of$\frac{1}{dc}$in equation (4).

$\frac{1}{C_{eq}}={\int }_0^d\frac{dx}{K_m{\epsilon }_{0}A}$

Now use the equation (2) and (3).

$$\begin{gathered} \frac{1}{C_{eq}}={\int }_0^d\frac{dx}{K{\epsilon }_{0}A\left(1+{\displaystyle \frac{m}{N}}\right)} \\ \frac{1}{C_{eq}}={\int }_0^d\frac{dx}{K{\epsilon }_{0}A\left(1+{\displaystyle \frac{x}{d}}\right)} \\ \frac{1}{C_{eq}}={\int }_0^d\frac{dx}{K{\epsilon }_{0}A\left({\displaystyle \frac{d+x}{d}}\right)} \\ \frac{1}{C_{eq}}=\frac{d}{K{\epsilon }_{0}A}\ln {\left[d+x\right]}_0^d \\ \frac{1}{C_{eq}}=\frac{d}{K{\epsilon }_{0}A}\left(\ln 2d-\ln d\right) \\ \frac{1}{C_{eq}}=\frac{d}{K{\epsilon }_{0}A}\ln 2 \end{gathered}$$

Simplify further,

$C_{eq}=\frac{K{\epsilon }_{0}A}{d\ln 2}$…………………….(6)

Comparing the equation (2) and (6).

$\alpha =1$

Therefore the value of $\alpha =1$.