$A$ particle of mass $M$ and charge $q$, initially at rest, is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be

An electron,initially at rest,is accelerated through a potential difference of $200 \ V$,so that it acquires a velocity of $8.4 \times 10^6 \ m/s$. The value of $e/m$ for the electron will be:

$A$ cathode ray tube contains a pair of parallel metal plates $1.0 \, cm$ apart and $3.0 \, cm$ long. $A$ narrow horizontal beam of electrons with a velocity $3 \times 10^7 \, m/s$ passes down the tube midway between the plates. When a potential difference of $550 \, V$ is maintained across the plates,it is found that the electron beam is deflected such that it just strikes the end of one of the plates. The specific charge of the electron in $C/kg$ is:

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 $\,m/s^2$)

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:

Calculate the work done by the electrostatic force on an electron moving in a circular orbit around the nucleus.