$A$ 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:

The ratio of the relative strengths of the gravitational force and the electromagnetic force between two charged particles is

Three charges $q$,$Q$,and $4q$ are placed in a straight line of length $l$ at points distant $0$,$\frac{l}{2}$,and $l$ respectively from one end. In order to make the net force on $q$ zero,the charge $Q$ must be equal to

$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes:

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 $= 1.6 \times 10^{-19} \ C$
$k = 9 \times 10^9 \ MKS, \ g = 10 \ ms^{-2}$

Three charges $4q, Q$ and $q$ are placed at positions $0, \frac{l}{2}$ and $l$ respectively along a straight line. If the resultant force on $q$ is zero,then $Q$ is equal to