The sum of two point positive charges separated by a distance of $1.5 \ m$ in air is $25 \mu C$. If the electrostatic force between the two charges is $0.6 \ N$,then the difference between the two charges is (in $\mu C$)

Three charges $q, Q$ and $+4q$ are placed in a straight line of length $d$ at positions $0, d/3$ and $2d/3$ respectively. In order to make the net force on $q$ be zero, the value of $Q$ should be

Identify the wrong statement in the following. Coulomb's law correctly describes the electric force that

The force between two point charges $q_1$ and $q_2$ placed in a vacuum at a distance $r \ cm$ apart is $F$. The force between them when placed in a medium having a dielectric constant $K=5$ at a distance $r/5 \ cm$ apart will be:

Three charges $+5 q, Q$ and $-2 q$ are kept along a straight line in the same order such that $+5 q$ and $-2 q$ charges are at a distance of $\frac{2 r}{3}$ and $\frac{r}{3}$ from the charge $Q$ respectively. If the net force on the charge $-2 q$ is zero,then $Q$ is

Four electric charges $2 \mu C, Q, 4 \mu C$ and $12 \mu C$ are placed on the $x$-axis at distances $x=0, 1 \ cm, 2 \ cm$ and $4 \ cm$ respectively. If the net force acting on the charge at the origin is zero,then $Q=$ (in $\mu C$)