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

In a region of space,suppose there exists a uniform electric field $\vec{E} = 10 \hat{i} \text{ V/m}$. If a positive charge moves with a velocity $\vec{v} = -2 \hat{j} \text{ m/s}$,its potential energy:

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 $\,m/s^2$)

Two identical positive charges $Q$ each are fixed at a distance of $2a$ apart from each other. Another point charge $q_0$ with mass $m$ is placed at the midpoint between the two fixed charges. For a small displacement $x$ along the line joining the fixed charges,the charge $q_0$ executes $SHM$. The time period of oscillation of charge $q_0$ will be:

What is the magnitude of the upward electric field (in $V/m$) required to keep a coin of mass $10^{-6} \ kg$ and charge $10^{-6} \ C$ in equilibrium? (Take $g = 10 \ m/s^2$)

$A$ small sphere of mass $m = 0.5 \, kg$ carrying a positive charge $q = 110 \, \mu C$ is connected to a light,flexible,and inextensible string of length $r = 60 \, cm$ and whirled in a vertical circle. If a vertically upwards electric field of strength $E = 10^5 \, N/C$ exists in the space,what is the minimum velocity of the sphere required at the highest point so that it may just complete the circle? $(g = 10 \, m/s^2)$