When a $5 C$ charge is kept in a uniform electric field,a force of $5000 N$ acts on it. Find the potential difference between two points in that field,separated by a distance of $1 cm$. (in $V$)

$A$ hollow conducting shell is charged with $500 \,\mu C$. If a force of $562.5 \,N$ is exerted on the charge, the electric field intensity on its surface is ...... $N/C$.

Two point charges $-5 \mu C$ and $+5 \mu C$ are placed at $A$ and $B$ respectively on a straight line separated by $5 \ cm$. At a point $C$,if the resultant electric field is parallel to the line joining the points $A$ and $B$,then:

Two point charges $-10 \mu C$ and $+5 \mu C$ are situated on the $X$-axis at $x=0$ and $x=\sqrt{2} \ m$. The point along the $X$-axis where the electric field becomes zero is:

Four charges are placed on the corners of a square as shown in the figure,with a side length of $5 \, cm$. If $Q = 1 \, \mu C$,then the electric field intensity at the centre will be:

If the potential at the centre of a uniformly charged hollow sphere of radius $R$ is $V$,then the electric field at a distance $r$ from the centre of the sphere is $(r > R)$.