An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be:

An oil drop having a mass $4.8 \times 10^{-13} \,kg$ and charge $2.4 \times 10^{-18} \,C$ stands still between two charged horizontal plates separated by a distance of $1 \,cm$. If now the polarity of the plates is changed,the instantaneous acceleration of the drop is: $(g = 10 \,m/s^2)$ (in $\,m/s^2$)

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 $g = 10 \ m/s^2$.)

An electron with kinetic energy $K_{1}$ enters between parallel plates of a capacitor at an angle $\alpha$ with the plates. It leaves the plates at an angle $\beta$ with the plates. If the kinetic energy of the electron when leaving is $K_{2}$,then the ratio of kinetic energies $K_{1}:K_{2}$ is ....... .

In a certain region,a uniform electric field $E$ and a magnetic field $B$ are present in opposite directions. At the instant $t = 0$,a particle of mass $m$ carrying a charge $q$ is given a velocity $v_0$ at an angle $\theta$ with the $y$-axis,in the $yz$-plane. The time after which the speed of the particle would be minimum is equal to

$A$ 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