A particle $A$ has charge $ + q$ and a particle $B$ has charge $ + \,4q$ with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference, the ratio of their speed $\frac{{{v_A}}}{{{v_B}}}$ will become

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

The charge $q$ is fired towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distance of approach would be

Three particles, each having a charge of $10\,\mu C$ are placed at the corners of an equilateral triangle of side $10\,cm$. The electrostatic potential energy of the system is…..$J$ (Given $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N – {m^2}/{C^2}$)

If $OP = 1\,\,cm$ and $OS = 2\,\, cm$, work done by electric field in shifting a point charge $\frac {4\sqrt 2}{27}\,\, μC$ from point $P$ to $S$ in given figure is