The maximum electric field that can be held in air without producing ionization of air is $10^7\,V/m$. The maximum potential,therefore,to which a conducting sphere of radius $0.10\,m$ can be charged in air is:

$A$ particle has a mass $400$ times that of an electron and a charge double that of an electron. It is accelerated by a potential difference of $5 \ V$. If the particle was initially at rest,what will be its final kinetic energy in $eV$?

Two tiny spheres carrying charges $1.8 \mu C$ and $2.8 \mu C$ are located $40 \ cm$ apart. The potential at the mid-point of the line joining the two charges is:

Consider two concentric conducting spheres of radii $R$ and $2R$ respectively. The inner sphere is given a charge $+Q$. The outer sphere is grounded. The potential at $r = \frac{3R}{2}$ is

Two charges $q_1$ and $q_2$ are separated by a distance of $30\ cm$. $A$ third charge $q_3$,initially at $C$ as shown in the figure,is moved along the circular path of radius $40\ cm$ from $C$ to $D$. If the difference in potential energy due to the movement of $q_3$ from $C$ to $D$ is given by $\frac{q_3 K}{4 \pi \epsilon_0}$,find the value of $K$.

The work done to take an electron from rest where potential is $-60\, V$ to another point where potential is $-20\, V$ is given by.....$eV$.