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)
$\frac{F}{2}$
$F$
$2F$
$4F$
The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times {10^{ - 11}}\,m,$ will be (Charge on electron $=$ $1.6 \times 10^{-19}$ $C$, mass of electron = $ 9.1 \times 10^{-31}$ $kg$, mass of proton = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$
Two free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so placed that the entire system is in equilibrium. Then the third charge is :-
Force between two point charges $q_1$ and $q_2$ placed in vacuum at ' $r$ ' $\mathrm{cm}$ apart is $F$. Force between them when placed in a medium having dielectric $\mathrm{K}=5$ at $\mathrm{r} / 5$ $\mathrm{cm}$ apart will be:
A charged particle having some mass is resting in equilibrium at a height $H$ above the centre of a uniformly charged non-conducting horizontal ring of radius $R$. The force of gravity acts downwards. The equilibrium of the particle will be stable $R$
A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium, if $q$ is equal to