NEET Sample Paper NEET Sample Test Paper-77

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

    A)  \[(4+k)\frac{C}{4}\]                 

    B)  \[(3+k)\frac{C}{4}\]

    C)  \[\frac{kC}{2}\]                       

    D)  \[\frac{kC}{4}\]  

    Correct Answer: B

    Solution :

    Capacitor with air as the dielectric has capacitance \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}}{d}\left( \frac{3\,\,A}{4} \right)\,\,=\,\,\frac{3{{\varepsilon }_{0}}A}{4d}\] Similarly the capacitor with k as dielectric constant has capacitance \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}k}{d}\left( \frac{A}{4} \right)\,\,=\,\,\frac{{{\varepsilon }_{0}}Ak}{4d}\] \[\because \,\,\,\,{{C}_{1}}\,\,and\,\,{{C}_{2}}\] are in parallel \[{{C}_{net}}=\,\,{{C}_{1}}\,\,+\,\,{{C}_{2}}\] \[=\,\,\,\,\frac{3{{\varepsilon }_{0}}4}{4d}+\frac{{{\varepsilon }_{0}}Ak}{4d}=\frac{{{\varepsilon }_{0}}A}{d}\,\left[ \frac{3}{4}+\frac{k}{4} \right]\] \[=\,\,\,\frac{C}{4}(3+k)\]


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