$ABC$ is a right-angled triangle in which $AB = 3 \ cm$,$BC = 4 \ cm$ and the right angle is at $B$. Three charges $+15 \ \mu C$,$+12 \ \mu C$ and $-20 \ \mu C$ are placed respectively at $A$,$B$ and $C$. The force acting on the charge at $B$ is (in $N$)

Three charges, each of $+1 \mu C$, are placed at the corners of an equilateral triangle. If the repulsive force between any two charges is $F$, then the net force on any one charge will be:

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is

$A$ point charge $Q$ is placed at the center of the line joining two equal point charges $+q$ and $+q$. The value of $Q$ if the system of the charges is in equilibrium,is

Two point charges $q_1$ and $q_2$ are $l$ distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is:

Two identical conducting spheres $A$ and $B$ having charges $+q$ and $-q$ respectively are kept at a distance $d$ apart and experience a Coulombian force $F$ between them. If $50\%$ of the charge is transferred from sphere $B$ to $A$,then the new Coulombian force between them is . . . . . . .