Discussion Forum

 A parallel plate capacitor having plates of area S and plate separation d, has capacitance  C₁   in air. When two dielectrics of different relative permittivities (ε₁= 2 and ε₂  =4) are introduced between the two plates as shown in the figure, the capacitance becomes C₂. The ratio   $\frac{C_{2}}{C_{1}}$  is

Answer:

Explanation:

$C_{1}=\frac{\epsilon_{0}s}{d},C=\frac{2\epsilon_{0}\frac{s}{2}}{\frac{d}{2}}=\frac{2\epsilon_{0}s}{d}$

$C'=\frac{4\epsilon_{0}\frac{s}{2}}{\frac{d}{2}}=\frac{4\epsilon_{0}s}{d}$

and    $C"=\frac{2\epsilon_{0}\frac{s}{2}}{d}=\frac{\epsilon_{0}s}{d}$

  $C_{2}=\frac{CC'}{C+C'}+C"=\frac{4}{3}\frac{\epsilon_{0}s}{d}+\frac{\epsilon_{0}s}{d}$

=   $\frac{7}{3}\frac{\epsilon_{0}s}{d}\frac{C_{2}}{C_{1}}=\frac{7}{3}$