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

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting,massless strings of same length. At equilibrium,the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $\rho_l = 800 \ kg \ m^{-3}$ and dielectric constant $K = 21$. If the angle between the strings remains the same after the immersion,then
$(A)$ electric force between the spheres remains unchanged
$(B)$ electric force between the spheres reduces
$(C)$ mass density of the spheres is $840 \ kg \ m^{-3}$
$(D)$ the tension in the strings holding the spheres remains unchanged

$A$ charge $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$. The force of repulsion between them will be maximum when:

Two charges of magnitude $5\, nC$ and $-2\, nC$ are placed at points $(2\, cm, 0, 0)$ and $(x\, cm, 0, 0)$ respectively in a region of space where there is no other external field. If the electrostatic potential energy of the system is $-0.5\,\mu J$,find the value of $x$ in $cm$.

Two particles with charges $+3.72 \mu C$ and $+1.86 \mu C$ are at some distance apart. If $20 \%$ of the charge is transferred from the first particle to the second particle,then the electrostatic force between them is

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is