On moving a charge of $20 \; C$ by $2 \; cm$,$2 \; J$ of work is done. The potential difference between the points is (in $V$):

$A$ regular hexagon of side $6 \ cm$ has a charge of $2 \ \mu C$ at each of its vertices. What is the potential at the centre of the hexagon? $\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right]$

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:

The figure shows a solid hemisphere with a charge of $5 \ nC$ distributed uniformly throughout its volume. The hemisphere lies on a plane and point $P$ is located on this plane,along a radial line from the center of curvature at a distance of $15 \ cm$. The electric potential at point $P$ due to the hemisphere is ..... $V$.

Two electric charges of $9 \mu C$ and $-3 \mu C$ are placed $0.16 \ m$ apart in air. There will be a point $P$ on the line joining the two charges and between them where the electric potential is zero. The distance of $P$ from the $9 \mu C$ charge is: (in $m$)

Assertion: Electrons move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.