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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    Between the plates of a parallel plate condenser, a plate of thickness \[{{t}_{1}}\]and dielectric constant \[{{k}_{1}}\] is placed. In the rest of the space, there is another plate of thickness \[{{t}_{2}}\] and dielectric constant \[{{k}_{2}}\].  The potential difference across the condenser will be                                    [MP PET 1993]

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

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

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

    D)                                      \[\frac{{{\varepsilon }_{0}}Q}{A}({{k}_{1}}{{t}_{1}}+{{k}_{2}}{{t}_{2}})\]

    Correct Answer: A

    Solution :

                       Potential difference across the condenser \[V={{V}_{1}}+{{V}_{2}}={{E}_{1}}{{t}_{1}}+{{E}_{2}}{{t}_{2}}=\frac{\sigma }{{{K}_{1}}{{\varepsilon }_{0}}}{{t}_{1}}+\frac{\sigma }{{{K}_{2}}{{\varepsilon }_{0}}}{{t}_{2}}\] Þ \[V=\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}} \right)=\frac{Q}{A{{\varepsilon }_{0}}}\left( \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}} \right)\]


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