$A$ uniform electric field $\vec{E}$ exists between the plates of a charged condenser. $A$ charged particle enters the space between the plates perpendicular to $\vec{E}$. The path of the particle between the plates is a

$A$ drop of $10^{-6} \ kg$ water carries $10^{-6} \ C$ charge. What electric field should be applied to balance its weight? (Assume $g = 10 \ m/s^2$)

Two large vertical and parallel metal plates having a separation of $1 \ cm$ are connected to a $DC$ voltage source of potential difference $X$. $A$ proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical $JUST$ after release. Then $X$ is nearly

$A$ particle of mass $1 \, g$ and charge $-0.1 \, \mu C$ is projected from the ground with a velocity $10\sqrt{2} \, m/s$ at an angle of $45^o$ with the horizontal in a region having a uniform electric field $1 \, kV/cm$ in the horizontal direction. Acceleration due to gravity is $10 \, m/s^2$ in the vertical downward direction. Select the $INCORRECT$ statement.

$A$ particle of mass $0.5 \ g$ and charge $10 \ \mu C$ is subjected to a uniform electric field of $8 \ NC^{-1}$. If the particle is initially at rest,the velocity of the particle after a time of $5 \ s$ is: (in $ms^{-1}$)

$A$ 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