Two metal plates having a potential difference of $800 \; V$ are $0.02 \; m$ apart horizontally. $A$ particle of mass $1.96 \times 10^{-15} \; kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge,then the charge on the particle is:

An electron is accelerated through a potential difference of $1000 \ V$. Its velocity is nearly

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

$A$ cathode ray tube contains a pair of parallel metal plates $1.0 \, cm$ apart and $3.0 \, cm$ long. $A$ narrow horizontal beam of electrons with a velocity $3 \times 10^7 \, m/s$ passes down the tube midway between the plates. When a potential difference of $550 \, V$ is maintained across the plates,it is found that the electron beam is deflected such that it just strikes the end of one of the plates. The specific charge of the electron in $C/kg$ is:

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).