The $e/m$ ratio for an electron is $1.8 \times 10^{11} \ C \ kg^{-1}$. What will be its velocity when accelerated through a potential difference of $9 \ V$?

$A$ uniform electric field $\vec{E}$ exists between the plates of a charged condenser. $A$ charged particle enters the space between the plates perpendicular to $\vec{E}$. The path of the particle between the plates is a

In the following figure,two parallel metallic plates are maintained at different potentials. If an electron is released midway between the plates,it will move:

The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 \times 10^3 \ Vm^{-1}$ acting vertically upwards as shown in the figure. The plates are sufficiently long and have a separation of $2 \ cm$. $A$ particle of negative charge $1 \ \mu C$ and mass $2 \ g$ is projected at an angle $45^{\circ}$ with the electric field from the lower plate with a velocity '$u$'. The maximum velocity acquired by the particle,if it does not hit the upper plate,is (in $ms^{-1}$)

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

$A$ $3 \text{ C}$ charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$. The work done in moving the charge is . . . . . . $J$.