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

At a certain distance from a point charge,the electric field intensity is $500 \ V/m$ and the electric potential is $3000 \ V$. What is this distance in $m$?

Which of the following statement$(s)$ is/are correct?
$(A)$ If the electric field due to a point charge varies as $r^{-2.5}$ instead of $r^{-2}$,then the Gauss law will still be valid.
$(B)$ The Gauss law can be used to calculate the field distribution around an electric dipole.
$(C)$ If the electric field between two point charges is zero somewhere,then the sign of the two charges is the same.
$(D)$ The work done by the external force in moving a unit positive charge from point $A$ at potential $V_A$ to point $B$ at potential $V_B$ is $(V_B - V_A)$.

$A$ charge $q$ is placed in the middle of a line joining two equal and like point charges $Q$. For what value of $q$ will the system be in equilibrium?

Five balls marked $a$ to $e$ are suspended using separate threads. Pairs $(b, c)$ and $(d, e)$ show electrostatic repulsion,while pairs $(a, b)$,$(c, e)$,and $(a, e)$ show electrostatic attraction. The ball marked $a$ must be:

Two small conducting spheres of equal radius have charges of $10\ \mu C$ and $-20\ \mu C$ respectively and are placed at a distance $R$ from each other,experiencing a force $F_1$. If they are brought in contact and then separated to the same distance,the new force between them is $F_2$. Find the ratio $F_1 : F_2$.