CET Karnataka Medical CET - Karnataka Medical Solved Paper-2007

  • 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)  \[\left( K+3 \right)\frac{C}{4}\]

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

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

    D)  \[\frac{KC}{4}\]

    Correct Answer: A

    Solution :

    The condenser with air as the dielectric has capacitance \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}}{d}\left( \frac{3A}{4} \right)=\frac{3{{\varepsilon }_{0}}A}{4d}\] similarly, the condenser with K as the dielectric constant has capacitance \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}K}{d}\left( \frac{A}{4} \right)=\frac{{{\varepsilon }_{0}}AK}{4d}\] Since, \[{{C}_{1}}\]and \[{{C}_{2}}\] are in parallel \[{{C}_{net}}={{C}_{1}}+{{C}_{2}}\] \[=\frac{3{{\varepsilon }_{0}}A}{4d}+\frac{{{\varepsilon }_{0}}AK}{4d}\] \[=\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{3}{4}+\frac{K}{4} \right]\] \[=\frac{C}{4}(K+3)\]


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