Question

The capacitance of a parallel plate condenser is $C_1$(fig. a). A dielectric of dielectric constant ‘K’ is inserted as shown in figure ‘b’ and ‘c’. If $C_2$ and $C_3$ are the capacitances in figures ‘b’ and ‘c’ then :

• A. Both C${}_2$ and C${}_3$>C${}_1$
• B. C${}_3$>C${}_1$ and C${}_2$>C${}_1$
• C. Both C${}_2$ and C${}_3$< C${}_1$
• D. C${}_1$=C${}_2$=C${}_3$

Answer 1

The correct option is A Both C${}_2$ and C${}_3$>C${}_1$

We know that capacitance is given by

$C_1=\frac{{\epsilon }_{0}A}{d.}$
b) In $C_2\frac{k{\epsilon }_{0}A2}{d}and\frac{{\epsilon }_{o}A2}{d}$ are in series
$\Rightarrow \frac{1}{C_2}=\frac{d}{k\epsilon A2}+\frac{d}{2{\epsilon }_{0}A}=\frac{d}{2{\epsilon }_{0}A}(\frac{1}{k}+1)$
$\Rightarrow C_2=\frac{{\epsilon }_{0}A}{d}\left(\frac{2k}{1+k}\right)$ we know k>1
$\therefore \frac{2k}{1+k}>1.$
$\therefore C_2>C_1.$
c) in $C_3\frac{k{\epsilon }_{0}A}{2d}and\frac{k{\epsilon }_{o}A}{d}$ are in parallel
$\therefore C_3=\frac{{\epsilon }_{0}A}{d}(\frac{k}{2}+1)$
$\therefore C_3>C_1.$