Two point charges $+9e$ and $+e$ are placed at a distance of $16\, cm$ from each other. Where should a third charge $q$ be placed between them so that the system remains in equilibrium?

$A$ particle of mass $1 \, mg$ and charge $q$ is placed at the midpoint of two stationary particles,each carrying the same charge $q$,kept at a distance of $2 \, m$. If the free charged particle is displaced from its equilibrium position by a small distance $x$ $(x \ll 1 \, m)$,it executes $SHM$. Its angular frequency of oscillation will be $.... \times 10^{8} \, rad/s$ if $q^{2} = 10 \, C^{2}$.

Two point charges of $+2 \, \mu \text{C}$ and $+6 \, \mu \text{C}$ repel each other with a force of $12 \, \text{N}$. If a charge of $-4 \, \mu \text{C}$ is added to each,the force will be:

Force of attraction between two point charges $Q$ and $-Q$ separated by $d \, \text{m}$ is $F_e$. When these charges are placed on two identical spheres of radius $R = 0.3 \, d$ whose centres are $d \, \text{m}$ apart,the force of attraction between them is

Two point charges $+e$ and $+4e$ are kept at a distance $d$ units apart. $A$ third point charge $+q$ is placed between the two charges at a distance $x$ units from charge $+e$ so as to be in equilibrium. The value of $x$ is:

Two charges $q_1$ and $q_2$ are placed in vacuum at a distance $d$ and the force acting between them is $F$. If a medium of dielectric constant $4$ is introduced around them,the force now will be