If a particle of mass $10 \ mg$ and charge $2 \ \mu C$ at rest is subjected to a uniform electric field of potential difference $160 \ V$,then the velocity acquired by the particle is (in $ms^{-1}$)

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the centre of a uniformly charged spherical region of total charge $Q$ and radius $R$. Charges $q$ and $Q$ have opposite signs. The spherically charged region is not free to move and kinetic energy $K$ is just sufficient for the charged particle to reach the boundary of the spherical charge. How much time does it take the particle to reach the boundary of the region?

The uniform electric field intensity between the two plates of a parallel plate capacitor is $1 \times 10^3 \ Vm^{-1}$ acting vertically upwards as shown in the figure. The plates are sufficiently long and have a separation of $2 \ cm$. $A$ particle of negative charge $1 \ \mu C$ and mass $2 \ g$ is projected at an angle $45^{\circ}$ with the electric field from the lower plate with a velocity '$u$'. The maximum velocity acquired by the particle,if it does not hit the upper plate,is (in $ms^{-1}$)

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.

$A$ 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 $10^{-3} \ kg$ and charge $1.0 \ C$,is initially at rest. At time $t = 0$,the particle comes under the influence of an electric field $\vec{E}(t) = E_0 \sin(\omega t) \hat{i}$,where $E_0 = 1.0 \ N \ C^{-1}$ and $\omega = 10^3 \ rad \ s^{-1}$. Consider the effect of only the electrical force on the particle. Then the maximum speed,in $m \ s^{-1}$,attained by the particle at subsequent times is . . . . . .