How many electrons should be removed from a coin of mass $1.6 \,g$,so that it may float in an electric field of intensity $10^9 \,N/C$ directed upward?

An electron of mass $m_e$ initially at rest moves through a certain distance in a uniform electric field in time $t_1$. $A$ proton of mass $m_p$ also initially at rest takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity,the ratio of $t_2/t_1$ is nearly equal to

The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (Given: mass of an electron $m = 9.1 \times 10^{-31} \ kg$,charge of an electron $e = 1.6 \times 10^{-19} \ C$,and acceleration due to gravity $g = 10 \ m/s^2$.)

$A$ charged particle $q$ is shot towards another charged particle $Q$ which is fixed,with a speed $v$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be

$A$ particle having mass $m$ and charge $q$ is at rest. On applying a uniform electric field $E$ on it,it starts moving. When this particle travels a distance $x$ in the direction of the force,its kinetic energy will be . . . . . . .

An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4 \, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to