$A$ particle of charge $1\ \mu C$ and mass $1\ g$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ s$. Then its final speed cannot be.......$m/s$.

An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^4 \, N/C)$. The velocity of the electron when it reaches plate $B$ will be nearly equal to

$A$ 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:

$A$ body having a specific charge of $8\,\mu \text{C/g}$ is resting on a frictionless plane at a distance of $10\,\text{cm}$ from a wall. It starts moving towards the wall when a uniform electric field of $100\,\text{V/m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic,then the time period of the motion will be (in seconds):

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)

$A$ 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?