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

The figure shows the paths of three charged particles in a uniform electric field. Which particle has the highest charge-to-mass ratio?

An electron of mass $m$ and charge $e$ is accelerated from rest through a potential difference $V$ in vacuum. The final speed of the electron will be

$A$ uniform electric field,$\vec{E} = -400 \sqrt{3} \hat{y} \text{ NC}^{-1}$ is applied in a region. $A$ charged particle of mass $m$ carrying positive charge $q$ is projected in this region with an initial speed of $u = 2 \sqrt{10} \times 10^6 \text{ ms}^{-1}$. This particle is aimed to hit a target $T$,which is $5 \text{ m}$ away from its entry point into the field as shown schematically in the figure. Take $\frac{q}{m} = 10^{10} \text{ Ckg}^{-1}$. Then-
$(A)$ the particle will hit $T$ if projected at an angle $45^{\circ}$ from the horizontal
$(B)$ the particle will hit $T$ if projected either at an angle $30^{\circ}$ or $60^{\circ}$ from the horizontal
$(C)$ time taken by the particle to hit $T$ could be $\sqrt{\frac{5}{6}} \mu\text{s}$ as well as $\sqrt{\frac{5}{2}} \mu\text{s}$
$(D)$ time taken by the particle to hit $T$ is $\sqrt{\frac{5}{3}} \mu\text{s}$

$A$ 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

In a Millikan oil drop experiment,a charged drop of mass $1.8 \times 10^{-14} \ kg$ is stationary between its plates. The distance between the plates is $0.90 \ cm$ and the potential difference is $2.0 \ kV$. The number of electrons on the drop is: