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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Charge and Coulombs Law

  • question_answer
    A force \[F\] acts between sodium and chlorine ions of salt (sodium chloride) when put \[1\,cm\] apart in 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 ions \[1\,cm\] apart is                                     [MP PET 1995]

    A)                    \[\frac{F}{K}\]              

    B)                    \[\frac{FK}{{{\varepsilon }_{0}}}\]           

    C)                    \[\frac{F}{K{{\varepsilon }_{0}}}\]                          

    D)                    \[\frac{F{{\varepsilon }_{0}}}{K}\]

    Correct Answer: A

    Solution :

                 When put 1 cm apart in air, the force between Na and Cl ions = F. When put in water, the force between Na and Cl ions \[=\frac{F}{K}\]           


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