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JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d’. The space between the plates is now filled with two dielectrics. One of the dielectric has dielectric constant \[{{\kappa }_{1}}=3\] and thickness \[\frac{d}{3}\] while the other one has dielectric constant \[{{\kappa }_{2}}=6\] and thickness \[\frac{2d}{3}\]. Capacitance of the capacitor is now                                                         [AIEEE 2008]

    A)       40.5 pF

    B) 20.25 pF

    C)             1.8 pF

    D) 45 pF

    Correct Answer: A

    Solution :

    [a]  \[{{C}_{0}}=9pF=\frac{{{\varepsilon }_{0}}A}{d}\,\,;\,\,\,\frac{1}{C}=\frac{d/3}{{{\varepsilon }_{0}}A{{\kappa }_{1}}}+\frac{2d/3}{{{\varepsilon }_{0}}A{{\kappa }_{2}}}\]
    \[{{\kappa }_{1}}=3,\,{{\kappa }_{2}}=6\]   
    \[\therefore \]\[C=\frac{9}{2}\frac{{{\varepsilon }_{0}}A}{2}=\frac{9}{2}\left( 9pF \right)=40.5\,\,pF\]


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