A particle ' $A$ ' has charge ' $+q$ ' and a particle ' $B$ ' has charge ' $+4 q$ '. Each has same mass ' $m$ '. When they are allowed to fall from rest through the same potential, the ratio of their speeds will become (particle A to particle B)

$A$ sphere ' $A$ ' of radius ' $R$ ' has a charge ' $Q$ ' on it. The field at point B outside the sphere is ' $E$ '. Now another sphere of radius ' $2 R$ ' having a charge ' $-2 Q$ ' is placed at B. The total field at the point midway between A and B due to both the spheres is

The point charges $+\mathrm{q},-\mathrm{q},-\mathrm{q},+\mathrm{q},+\mathrm{Q}$ and -q are placed at the vertices of a regular hexagon ABCDEF as shown in figure. The electric field at the centre of hexagon ' $O$ ' due to the five charges at $A, B, C, D$ and $F$ is thrice the electric field at centre ' $O$ ' due to charge +Q at E alone. The value of Q is

A small particle carrying a negative charge of $1.6 \times 10^{-19} \mathrm{C}$ is suspended in equilibrium between two horizontal metal plates 8 cm apart having a potential difference of 980 V across them. The mass of the particle is $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right]$