Two point charges $+3 \mu C$ and $+8 \mu C$ repel each other with a force of $40 \ N$. If a charge of $-5 \mu C$ is added to each of them,then the force between them will become: (in $N$)

Three charges of each magnitude $100 \mu C$ are placed at the corners $A, B$ and $C$ of an equilateral triangle of side $4 \text{ m}$. If the charges at points $A$ and $C$ are positive and the charge at point $B$ is negative,then the magnitude of the total force acting on the charge at $C$ and the angle made by it with $AC$ are:

The repulsive force between two particles of same mass and charge,separated by a certain distance,is equal to the weight of one of them. The distance between them is . . . . . . $\times 10^{-1} \ m$.
Mass of particle $= 1.66 \times 10^{-27} \ kg$
Charge of particle $= 1.6 \times 10^{-19} \ C$
$k = 9 \times 10^9 \ MKS, \ g = 10 \ ms^{-2}$

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

Two balls with charges $+7 \ \mu C$ and $-5 \ \mu C$ attract each other with a force $F$. If $-2 \ \mu C$ charge is added to both,what will be the force between them?

If the distance between two equal point charges is doubled,what would happen to the force between them?