$A$ charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6 \times 10^4 \ Vm^{-1}$. The charge on the drop is $8 \times 10^{-18} \ C$. The radius of the drop is $\left[ \rho_{air} = 1.29 \ kg/m^3, \rho_{Hg} = 13.6 \times 10^3 \ kg/m^3, g = 9.8 \ m/s^2 \right]$.

$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

Two long parallel plates $A$ and $B$ are separated by a distance of $4 \ cm$ with an electric field of $45.5 \ Vm^{-1}$ between the plates directed normally from plate $A$ to plate $B$,as shown in the figure. An electron is projected from plate $A$ with velocity $v$ at an angle of $30^{\circ}$ with the surface of plate $A$. The maximum value of $v$ so that the electron does not hit plate $B$ is (Assume gravity-free space,charge of electron $= 1.6 \times 10^{-19} \ C$ and mass of electron $= 9.1 \times 10^{-31} \ kg$): (in $km \ s^{-1}$)

The figure shows the tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge-to-mass ratio?

Acceleration of a charged particle of charge $q$ and mass $m$ moving in a uniform electric field of strength $E$ is

$A$ particle of charge $1\ \mu C$ and mass $1\ g$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ s$. Then its final speed cannot be.......$m/s$.