Question

A force $F$ acts between sodium and chlorine ions of salt (sodium chloride), when ions are $1cm$ apart in air. The permittivity of air and dielectric constant of water are ${\epsilon }_0$ and $K$ respectively. When a piece of salt is placed in water, electrical force acting between sodium and chlorine ions $1cm$ apart is

• A. $\frac{F}{K}$
• B. $\frac{FK}{{\epsilon }_0}$
• C. $\frac{F}{K{\epsilon }_0}$
• D. $\frac{F{\epsilon }_0}{K}$

Answer 1

The correct option is A $\frac{F}{K}$

When placed $1cm$ apart in air, the force between $N{a}^{+}$ and $C{l}^{-}$ ions = $F$

Force between two charges placed in vaccum/air is $F=\frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}$
Force between two charges placed in dielectric of dielectric constant $K$ is $F_k=\frac{q_1{q}_2}{4\pi K{ϵ}_0{r}^2}$

So, when placed $1cm$ apart in water, the force between $N{a}^{+}$ and $C{l}^{-}$ ions is $F_k=\frac{F}{K}$