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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2010

  • question_answer
    Capacitance of a parallel plate capacitor becomes 4/3 times its original value, if a dielectric slab of thickness \[t=\frac{d}{2}\] is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is

    A)  8     

    B)  4      

    C)  6       

    D)  2

    Correct Answer: C

    Solution :

     Capacitance,\[{{C}_{air}}=\frac{{{\varepsilon }_{0}}A}{d}\]with dielectric slab \[C'=\frac{{{\varepsilon }_{0}}A}{\left( d-t+\frac{t}{k} \right)}\] Given,      \[C'=\frac{4}{3}C\] \[C'=\frac{{{\varepsilon }_{0}}A}{\left( d-t+\frac{t}{K} \right)}=\frac{4}{3}\times \frac{{{\varepsilon }_{0}}A}{d}\] \[\Rightarrow \] \[K=2\]


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