$A$ water drop of mass $10^{-6} \ kg$ carries a charge of $(-10^{-6}) \ C$. What magnitude of electric field should be applied to the drop so that it is in equilibrium with its weight?

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,V/m$. At equilibrium,the angle that the pendulum makes with the vertical is (take $g = 10\,m/s^2$)

$A$ simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the final tension in the string changes. The final tension in the string, when the pendulum attains an equilibrium position, is . . . . . . . ($g$: acceleration due to gravity)

$A$ proton, a deuteron, and an $\alpha$-particle having the same momentum enter a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

$A$ problem of practical interest is to make a beam of electrons turn at a $90^{\circ}$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 \times 10^{-17} \ J$ enters through a small hole in the bottom plate. The strength of the electric field that is needed if the electron is to emerge from an exit hole $1.0 \ cm$ away from the entrance hole,traveling at right angles to its original direction,is $y \times 10^5 \ N/C$. The value of $y$ is

An electron and a proton are placed in the same electric field. The ratio of their accelerations is ......