$A$ positive charge particle of $100 \,mg$ is thrown in the opposite direction to a uniform electric field of strength $1 \times 10^{5} \,NC^{-1}$. If the charge on the particle is $40 \,\mu C$ and the initial velocity is $200 \,ms^{-1}$,how much distance (in $m$) will it travel before coming to rest momentarily?

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 $\,m/s^2$)

$A$ 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

$A$ particle of mass $2 \times 10^{-5} \ kg$ and charge $4 \times 10^{-3} \ C$ starts from rest in a uniform electric field of $5 \ V/m$. What is its kinetic energy after $10 \ s$?

An electron enters between two horizontal plates separated by $2\,mm$ and having a potential difference of $1000\,V$. The force on the electron is:

The intensity of the electric field required to balance a proton of mass $1.7 \times 10^{-27} \ kg$ and charge $1.6 \times 10^{-19} \ C$ is nearly: