A bullet of mass $2\, gm$ is having a charge of $2\,\mu C$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10\,m/s$

If $4 \times {10^{20}}eV$ energy is required to move a charge of $0.25$ coulomb between two points. Then what will be the potential difference between them......$V$

A simple pendulum with a bob of mass $m = 1\ kg$ , charge $q = 5\mu C$ and string length $l = 1\ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 × 10^6\ V/m$ at its bottom most point $A$ , as shown in figure. It is given a speed $u$ such that the particle leave the circular path at its topmost point $C$ . Find the speed $u$ . (Take $g = 10\ m/s^2$ )

An $\alpha $-particle is accelerated through a potential difference of $200\,V$. The increase in its kinetic energy is.......$eV$

${\rm{ }}1\,ne\,V{\rm{ }} = {\rm{ }}......\,J.$ (Fill in the gap)

Charge $q_{2}$ is at the centre of a circular path with radius $r$. Work done in carrying charge $q_{1}$, once around this equipotential path, would be