JIPMER Jipmer Medical Solved Paper-2012

  • question_answer
    A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

    A)  \[(K+3)\frac{C}{4}\]

    B)                         \[(K+2)\frac{C}{4}\]

    C)  \[(K+1)\frac{C}{4}\]                      

    D)  \[\frac{KC}{4}\]

    Correct Answer: A

    Solution :

    Capacitance of a capacitor with dielectric as air \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}}{d}\left( \frac{3A}{4} \right)=\frac{3{{\varepsilon }_{0}}A}{4d}\] Similarly, the capacitance of a capacitor with dielectric constant K \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}K}{d}\left( \frac{A}{4} \right)=\frac{{{\varepsilon }_{0}}AK}{4d}\] \[{{C}_{1}}\] and \[{{C}_{2}}\] are in parallel \[{{C}_{net}}={{C}_{1}}+{{C}_{2}}\] \[=\frac{3{{\varepsilon }_{0}}A}{4d}+\frac{{{\varepsilon }_{0}}AK}{4d}=\frac{C}{A}(K+3).\]


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