question_answer

A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constants \[{{K}_{\text{1}}}\] and \[{{K}_{\text{2}}}\] its capacity becomes \[{{C}_{\text{1}}}\]. The ratio of \[{{C}_{\text{1}}}\] to C is  [MP PMT 2001]

A)            \[{{K}_{1}}+{{K}_{2}}\]

B)                                      \[\frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}-{{K}_{2}}}\]

C)            \[\frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}}\]     

D)            \[\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\]

Correct Answer: D

Solution :

                   \[{{C}_{A}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A}{d/2},\,{{C}_{B}}=\frac{{{K}_{2}}{{\varepsilon }_{0}}A}{d/2}\]                    \ \[{{C}_{eq}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\]                    \[=\frac{{{C}_{A}}{{C}_{B}}}{{{C}_{A}}+{{C}_{B}}}=\left( \frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right)\,\frac{{{\varepsilon }_{0}}A}{d}\]                     \[\left( \because \,\,c=\frac{{{\varepsilon }_{0}}A}{d} \right)\]