$A$ 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 '$q$' is [ $\epsilon_0$ = permittivity of free space ]

$A$ 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 $ms^{-1}$)

$A$ particle with charge $q$ approaches a fixed particle with charge $Q$ with an initial velocity $v$. It comes to a stop at a closest distance $r$ from $Q$. If the particle $q$ is given an initial velocity $2v$,the closest distance will be:

The figure shows the tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge-to-mass ratio?

$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be: