$A$ water drop of mass $10^{-6} \ kg$ carries a charge of $(-10^{-6}) \ C$. What magnitude of electric field should be applied to the drop so that it is in equilibrium with its weight?

Two charges $-q$ each are fixed and separated by a distance $2d$. $A$ third charge $q$ of mass $m$ placed at the midpoint is displaced slightly by $x$ $(x \ll d)$ perpendicular to the line joining the two fixed charges as shown in the figure. Show that the charge $q$ will perform simple harmonic motion with a time period $T = \left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$.

An $\alpha$-particle of mass $6.4 \times 10^{-27} \ kg$ and charge $3.2 \times 10^{-19} \ C$ is situated in a uniform electric field of $1.6 \times 10^{5} \ Vm^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \ m$ path when it starts from rest is

$A$ sphere carrying charge $Q$ and having weight $w$ falls under gravity between a pair of vertical plates at a distance $d$ from each other. When a potential difference $V$ is applied between the plates,the acceleration of the sphere changes as shown in the figure,along the line $BC$. The value of $Q$ is:

An electron falls through a small distance in a uniform electric field of magnitude $2 \times 10^4 \ N C^{-1}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be

$A$ proton and an $\alpha$-particle start from rest in a uniform electric field. The ratio of times taken by them to travel the same distance in the field is