In the given diagram,find the distance of the neutral point from the particle of charge $e$ in $cm$.

Two point charges $A$ and $B$ with magnitudes $+4q$ and $-4q$ are placed along a line separated by a distance $r$. The force acting between them is $F$. If $25\%$ of the charge from point $A$ is transferred to point $B$,the force between the charges now becomes:

Two positive point charges are separated by a distance of $4 \ m$ in air. If the sum of the two charges is $36 \mu C$ and the electrostatic force between them is $0.18 \ N$,then the bigger charge is (in $\mu C$)

There are two charges $+1 \ \mu C$ and $+5 \ \mu C$. The ratio of the forces acting on them will be

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

Two identical pendulums $A$ and $B$ are suspended from the same point. The bobs are given positive charges,with $A$ having more charge than $B$. They diverge and reach equilibrium,with $A$ and $B$ making angles $\theta_1$ and $\theta_2$ with the vertical,respectively. Then: