Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A force F acts between sodium and chlorine ions  of salt (sodium chloride) when put 1 cm apart in the air. The permittivity of air and dielectric constant of water are \[{{\varepsilon }_{0}}\] and K respectively. When a piece of salt is put in water electrical force acting between sodium and chlorine ion 1 cm apart is

    A) \[F\]                                     

    B) \[FK\text{/}{{\text{ }\!\!\varepsilon\!\!\text{ }}_{\text{0}}}\]

    C) \[\text{F/K}{{\text{ }\!\!\varepsilon\!\!\text{ }}_{\text{0}}}\]                                   

    D) \[\text{F/K}\]

    Correct Answer: D

    Solution :

    Force in air \[{{F}_{a}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] Force in dielectric medium \[{{F}_{m}}=\frac{1}{4\pi {{\varepsilon }_{0}}K}\,\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] or            \[{{F}_{m}}=\frac{1}{K}\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{q}_{1}}{{q}_{2}}}{{{e}^{2}}} \right)\]                 \[=\frac{F}{K}\]  (\[\because \] r is constant in both conditions )


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