Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A parallel plate capacitor is filled up by a substance of\[{{t}_{1}}\] thickness of dielectric constant \[{{K}_{1}}\] and other substance of \[{{t}_{2}}\] thickness of dielectric constant \[{{K}_{2.}}\] The capacity of this capacitor will be

    A) \[{{\varepsilon }_{0}}A/\left( \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}} \right)\]          

    B) \[{{\varepsilon }_{0}}A/\left( \frac{{{K}_{1}}}{{{K}_{2}}}+\frac{{{K}_{2}}}{{{t}_{2}}} \right)\]

    C) \[{{\varepsilon }_{0}}A/\left( \frac{{{t}_{1}}}{{{K}_{2}}}+\frac{{{t}_{2}}}{{{K}_{1}}} \right)\]          

    D) \[{{\varepsilon }_{0}}A/\left( \frac{{{K}_{2}}}{{{t}_{1}}}+\frac{{{K}_{1}}}{{{t}_{2}}} \right)\]

    Correct Answer: A

    Solution :

    For two mediums of different thickness \[{{t}_{1}},{{t}_{2}}\]capacitance                 \[C=\frac{{{\varepsilon }_{0}}A}{\left( \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}} \right)}\]


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