2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2008

  • 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  Solved  Paper-2007

    A) 40.5 pF

    B) 20.25 pF              

    C) 1.8 pF   

    D) 45 pF

    Correct Answer: A

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner