# 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) $\; \frac{F\varepsilon _{0}}{K}\:$

A Avinash

As we learned

When dielectric insert between the charges -

$\dpi{100} F_{med}=\frac{F_{air}}{K}=\frac{1q_{1}q_{2}}{4\pi \varepsilon _{0}kr^{2}}$

- 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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