Two point charges $+10 \mu C$ and $+4 \mu C$ are placed $10 \ cm$ apart in air. The work required to be done to bring them $2 \ cm$ closer is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right)$. (in $J$)

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges $Q$ repel each other with a force $F$ when kept apart at a distance $r$. $A$ third spherical conductor having the same radius as that of $B$ but uncharged is brought in contact with $B$,then brought in contact with $C$,and finally removed away from both. The new force of repulsion between $B$ and $C$ is

$A$ charge $q_1$ exerts some force on a second charge $q_2$. If a third charge $q_3$ is brought near,the force of $q_1$ exerted on $q_2$:

Find the value of $q$ so that the net electrostatic force on the $10 \mu C$ charge placed at a vertex of the square is zero. The charges are arranged as shown in the figure.

The force between two point charges $A$ and $B$ is $F$. If $75\%$ of the charge of $A$ is transferred to $B$,then the new force between $A$ and $B$ is:

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