$A$ 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 charged spheres separated at a distance $d$ exert a force $F$ on each other. If they are immersed in a liquid of dielectric constant $2$,then what is the force (if all conditions are same)?

Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. $A$ third identical uncharged conducting sphere $C$ is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. The new force of repulsion between spheres $A$ and $B$ (radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

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:

As shown in the figure,charges $+q$ each are placed at the four vertices of a square of side length $a$. The Coulomb force acting on the charge placed at vertex $D$ is . . . . . . .

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\vec{F}$ between the two is