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

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the centre of a uniformly charged spherical region of total charge $Q$ and radius $R$. Charges $q$ and $Q$ have opposite signs. The spherically charged region is not free to move and kinetic energy $K$ is just sufficient for the charged particle to reach the boundary of the spherical charge. How much time does it take the particle to reach the boundary of the region?

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

The kinetic energy of an electron which is accelerated through a potential of $100 \, V$ is:

$A$ charged oil drop is suspended in a uniform electric field of $3 \times 10^{4} \; V/m$ so that it neither falls nor rises. The charge on the drop will be $..... \times 10^{-18} \; C$. (Take the mass of the drop $= 9.9 \times 10^{-15} \; kg$ and $g = 10 \; m/s^{2}$)

$A$ 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: