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NEET Sample Paper NEET Sample Test Paper-69

  • 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

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

    Correct Answer: A

    Solution :

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


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