Question

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. $F K / \varepsilon_{0}$
• C. $F / K \varepsilon_{0}$
• D. $F / K$

Answer 1

Answer: (D)

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 $)$