Question

Suppose $n$ conducting plates are placed face to face, and the distance between two successive plates is $d$. Each plate is half of the area of the previous one. If area of first plate is $A$. If the area of the first plate is $A$, the equivalent capacitance of the system is given as $\frac{{\epsilon }_{0}A}{d(x^n-2)}$. Find $x$ :

Answer 1

There are $(n-1)$ capacitors and they are in series.

If $A_1=A,$ then $A_2=A/2,A_3=A/4=A/2^2,A_4=A/8=A/2^3,.......,A_n=A/2^{n-1}.$

$\frac{1}{C_{eq}}=\frac{1}{\frac{A{ϵ}_0}{2d}}+\frac{1}{\frac{A{ϵ}_0}{2^{2}d}}+\frac{1}{\frac{A{ϵ}_0}{2^{3}d}}+......+\frac{1}{\frac{A{ϵ}_0}{2^{n-1}d}}$

$=\frac{d}{A{ϵ}_0}[2+2^2+2^3+....+2^{n-1}]$

$=\frac{d}{A{ϵ}_0}\left[\frac{2(2^{n-1}-1)}{2-1}\right]$ ( this is GP series , sum $S_{n-1}=\frac{a(2^{n-1}-1)}{(r-1)},$ first term $a=2,$ common ratio $r=2$ )

$\frac{1}{C_{eq}}=\frac{d}{A{ϵ}_0}[2^n-2]$

$C_{eq}=\frac{A{ϵ}_0}{d[2^n-2]}$

Thus, $x=2$