VMMC VMMC Medical Solved Paper-2003

  • question_answer
     A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constants \[{{K}_{1}}\], and \[{{K}_{2}}\], its capacity becomes Q. The ratio of \[{{C}_{1}}\] to C is:

    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 :

    Capacity of parallel plate condenser is \[C=\frac{{{\varepsilon }_{0}}A}{d}\]                      ?(1) When the space is equally filled with parallel layers of materials of dielectric constants \[{{K}_{1}}\]and \[{{K}_{2}},\]the capacity becomes \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{\left( \frac{d}{2{{K}_{1}}}+\frac{d}{2{{K}_{2}}} \right)}=\frac{2{{K}_{1}}{{K}_{2}}{{\varepsilon }_{0}}A}{({{K}_{1}}+{{K}_{2}})d}\] Thus, from Eqs. (1) and (2), \[\frac{{{C}_{1}}}{C}=\frac{2{{K}_{1}}{{K}_{2}}{{\varepsilon }_{0}}A}{({{K}_{1}}+{{K}_{2}})d}.\frac{d}{{{\varepsilon }_{0}}A}\]                 \[=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\]


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