The magnitude of an electric field which can just suspend a deuteron of mass $3.2 \times 10^{-27} \ kg$ freely in air is

$A$ proton is $1840$ times heavier than an electron. When it is accelerated through a potential difference of $1 \ kV$,what will be its kinetic energy in $keV$?

$A$ small sphere carrying a charge $q$ is hanging in between two parallel plates by a string of length $L$. The time period of the pendulum is $T_0$. When the parallel plates are charged,the time period changes to $T$. The ratio $T/T_0$ is equal to

$A$ charge of magnitude $2e$ and mass $4m$ is moving in an electric field $E$. The acceleration imparted to the above charge is

$A$ charged particle of mass $m$ and charge $q$ is at rest. It is accelerated in a uniform electric field of intensity $E$ for time $t$. The kinetic energy of the particle after time $t$ is

$A$ 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: