Question

If a cell of emf $V$ is connected as shown in figure and a dielectric of dielectric constant $k$ is inserted between the first and second plates, the charge on the $n$th plate is given as $\frac{{\epsilon }_{0}A}{2d(x^{n-1}-2+\frac{1}{k})}V$. Find $x$

Answer 1

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

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

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

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

$=\frac{d}{A{ϵ}_0}[\frac{2}{k}+2(2+2^2+....+2^{n-2}\left)\right]$

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

$\frac{1}{C_{eq}}=\frac{2d}{A{ϵ}_0}[\frac{1}{k}+(2^{n-1}-2\left)\right]$

$C_{eq}=\frac{A{ϵ}_0}{2d[\frac{1}{k}+(2^{n-1}-2\left)\right]}$

$Q_{eq}=C_{eq}V=\frac{A{ϵ}_0}{2d[\frac{1}{k}+(2^{n-1}-2\left)\right]}V$

As they are in series so charge on each capacitor is equal to the equivalent charge.

Charge on n th plate is $Q_n=Q_{eq}=\frac{A{ϵ}_0}{2d[\frac{1}{k}+(2^{n-1}-2\left)\right]}V$

Thus, $x=2$