When air is replaced by a dielectric medium of constant $K$,the maximum force of attraction between two charges separated by distance '$d$' is . . . . . . .

The law governing the force between electric charges is known as

Three point charges $q$,$-2q$,and $2q$ are placed on the $x$-axis at distances $x=0$,$x=\frac{3}{4}R$,and $x=R$ respectively from the origin as shown. If $q = 2 \times 10^{-6} \, C$ and $R = 2 \, cm$,the magnitude of the net force experienced by the charge $-2q$ is .......... $N$.

Two charges $+10 \mu C$ and $-10 \mu C$ are separated by $10 \text{ cm}$. The magnitude of the force acting on another charge $5 \mu C$ placed at the midpoint of the line joining the two charges will be: [Use $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$] (in $\text{ N}$)

Two equal point charges exert a force $F$ on each other when they are placed distance $d$ apart in air. When they are placed distance $D$ apart in a medium of dielectric constant $K$, they exert the same force. The distance $D$ is equal to

Two charges $q$ and $-3q$ are placed fixed on the $x$-axis separated by a distance $d$. Where should a third charge $2q$ be placed such that it will not experience any force?