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A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta=\frac{ d }{ N }$. The dielectric constant of the $m ^{\text {th }}$ layer is $K _{ m }= K \left(1+\frac{ m }{ N }\right)$. For a very large $N \left(>10^3\right)$, the capacitance $C$ is $\alpha\left(\frac{ K \varepsilon_0 A }{ d \;ln 2}\right)$. The value of $\alpha$ will be. . . . . . . .
[ $\epsilon_0$ is the permittivity of free space]
$1$
$3$
$5$
$6$
Solution
$\delta=d x=\frac{d}{N} \& \frac{m}{N}=\frac{x}{d}$
$K_m=K\left(1+\frac{m}{N}\right)$
$\Rightarrow K_m=K\left(1+\frac{x}{d}\right)$
$C^{\prime}=\frac{K_m A \epsilon_0}{d x}$
$\frac{1}{C_{e q}}=\int_0^d \frac{d x}{K_m A \epsilon_0}=\frac{1}{K A \epsilon_0} \int_0^d \frac{d x}{\left(1+\frac{x}{d}\right)}$
$\Rightarrow \frac{1}{C_{e q}}=\frac{d}{K A \epsilon_0}\left[\ln \left(1+\frac{x}{d}\right)\right]_0^d$
$\Rightarrow \frac{1}{ C _{ eq }}=\frac{ d }{ KA \epsilon_0}[\ln 2-\ell n (1)]$
$\Rightarrow C _{ eq }=\frac{ KA \epsilon_0}{ d \ell n 2} \Rightarrow \alpha=1$
