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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    Between the plates of parallel plate capacitor of capacity C, two parallel plate of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to 1/5th of the distance between the plates of the original capacitor, then the capacity of the new capacitor is:

    A)  \[\frac{5}{3}C\]                               

    B)  \[\frac{3}{5}C\]

    C)  \[\frac{3C}{10}\]                             

    D)  \[\frac{10C}{3}\]

    Correct Answer: A

    Solution :

                                    Given: Thickness of each plates \[=\frac{1}{5}\]distance between plates \[=\frac{1}{5}\times d\] The capacitance of capacitor is given by \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d-t}\]                                    ?(i) Thickness of two conductor is given by \[t=2\left( \frac{d}{2} \right)=\frac{2d}{5}\] Putting the values of t in Eq. (i) \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d-\frac{2d}{5}}=\frac{5}{3}\frac{{{\varepsilon }_{0}}A}{d}\] \[=\frac{5}{3}C\]                              \[\left( \text{where}\,C=\frac{{{\varepsilon }_{0}}A}{d} \right)\]


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