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  • I have a doubt, kindly clarify. - Electrostatics - JEE Main-3

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

  • Option 1)

    \frac{F}{K}\;

  • Option 2)

    \; \frac{FK}{\varepsilon _{0}}\:

  • Option 3)

    \: \frac{F}{K\varepsilon _{0}}\:\;

  • Option 4)

    \; \frac{F\varepsilon _{0}}{K}\:

 

Answers (1)

best_answer

As we learned

When dielectric insert between the charges -

F_{med}=\frac{F_{air}}{K}=\frac{1q_{1}q_{2}}{4\pi \varepsilon _{0}kr^{2}}

- wherein

 

 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}

 


Option 1)

\frac{F}{K}\;

Option 2)

\; \frac{FK}{\varepsilon _{0}}\:

Option 3)

\: \frac{F}{K\varepsilon _{0}}\:\;

Option 4)

\; \frac{F\varepsilon _{0}}{K}\:

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Avinash

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