Five point charges each having magnitude $q$ are placed at the corners of a regular hexagon as shown in the figure. The net electric field at the centre $O$ is $\vec E$. To make the net electric field at $O$ equal to $6\vec E$,the charge placed on the remaining sixth corner should be: (in $,q$)

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

$A$ charge $10^{-9} \ C$ is located at the origin of a coordinate system and another charge $Q$ at $(2, 0, 0) \ m$. If the $Y$-component of the electric field at $(3, 1, 1) \ m$ is zero,calculate the value of $Q$.

The number of electrons to be put on a spherical conductor of radius $0.1\,m$ to produce an electric field of $0.036\,N/C$ just above its surface is:

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC^{-1}$. The distance $d$ between the point charges $A$ and $B$ is ............ $m$. (in $.0$)

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices $A, B, C, D, E$ and $F$ respectively of a regular hexagon $ABCDEF$. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is thrice the electric field at centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is