Question

The figure shows a capacitor having three layers between its plates. Layer $x$ is vacuum, $y$ is conductor and $z$ is a dielectric. Which of the following change(s) will result in increase in capacitance?

• A. Replace $x$ by conductor
• B. Replace $y$ by dielectric
• C. Replace $y$ by condctor
• D. Replace $x$ by dielectric

Answer 1

Answer: (A), (C), (D)

Replace $x$ by dielectric

$C_1=\frac{A{\epsilon }_0{k}_1}{d},C_2=\frac{A{k}_2{\epsilon }_0}{d},C_2=\frac{A{k}_3{\epsilon }_0}{d}$

As all are in series, the dielectric constant of $x,y$ and $z$ are $k=1,k=\infty ,k=k$ respectively. so the equivalent capacitance will be,

$\frac{1}{C_0}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$=\frac{d}{A{k}_1{\epsilon }_0}+\frac{d}{A{k}_2{\epsilon }_0}+\frac{d}{A{k}_3{\epsilon }_0}$

$=\frac{d}{A{\epsilon }_0}[\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}]$

$C_0=\frac{A{\epsilon }_0}{d}\frac{1}{(1+0+\frac{1}{k})}$

$C_0=\frac{A{\epsilon }_0}{d}\frac{1}{1+\frac{1}{k}}=\frac{A{\epsilon }_0}{d}\left(\frac{k}{1+k}\right)$

$C_0=\left(\frac{k}{1+k}\right)\frac{A{\epsilon }_0}{d}$

(A) $C_1=\frac{A{\epsilon }_0}{d}\frac{1}{(\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{k})}$

$C_1=\frac{A{\epsilon }_0}{d}\left[k\right]$

$\because k>1$

$\Rightarrow \frac{1}{k}<1$

$\Rightarrow \frac{1}{k}+1<2$

$\Rightarrow \frac{1+k}{k}<2$

$\frac{k}{k+1}>\frac{1}{2}$

$\therefore C_1>C_0$

(B) $C_2=\frac{A{\epsilon }_0}{d}\frac{1}{(1+\frac{1}{k}+\frac{1}{k})}$

$=\frac{A{\epsilon }_0}{d}\frac{1}{(1+\frac{2}{k})}$

$=\frac{A{\epsilon }_0}{d}\left[\frac{k}{k+2}\right]\because \frac{2}{k}<2$

$\frac{2}{k}+1<3$

$\frac{2+k}{k}<3$

$\frac{k}{k+2}>\frac{1}{3}$

$\therefore C_2<C_0$

(c) $C_3=\frac{A{\epsilon }_0}{d};C_3>C_0$

(D) $C_4=\frac{A{\epsilon }_0}{d}\frac{1}{(\frac{1}{k}+\frac{1}{\infty }+\frac{1}{k})}$

$=\frac{A{\epsilon }_0}{d}\left[\frac{k}{2}\right]$

$\because k>1$

$\frac{k}{2}>\frac{1}{2}$

$\therefore C_4>C_0$

Therefore A, C & D are correct options.