An electric field due to a positively charged long straight wire at a distance $r$ from it is proportional to $r^{-1}$ in magnitude. Two electrons are orbiting such a long straight wire in circular orbits of radii $1 \mathring{A}$ and $2 \mathring{A}$. The ratio of their respective time periods is

An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E$. The direction of the electric field is now reversed,keeping its magnitude the same. $A$ proton is allowed to fall from rest in it through the same vertical distance $h$. The time of fall of the electron,in comparison to the time of fall of the proton,is:

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

In a parallel plate air capacitor,a cathode beam comprising $n = 10^6$ electrons is emitted with a velocity $v_0 = 10^8\ m/s$ into the space between the plates. The potential difference between the plates is $\phi = 400\ V$,the separation between the plates is $d = 2\ cm$ and the area of each plate is $l^2 = 100\ cm^2$ (where $l$ is the length of the plate). The deflection of the electron beam is ........... $mm$.

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)

$A$ particle of mass $m$ and carrying charge $-q_1$ is moving around a charge $+q_2$ along a circular path of radius $r$. Find the period of revolution of the charge $-q_1$.