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 $,q$)

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:

$A$ circular wire loop of radius $10 \ cm$ carries a total charge of $10^{-5} \ C$ distributed uniformly over its length. $A$ small length of $3.14 \times 10^{-6} \ m$ of wire is cut off. The magnitude of the electric field at the centre due to the remaining wire is:
(Assume $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \ SI$ units) (in $N \ C^{-1}$)

Two point charges $Q$ and $-3 Q$ are placed at some distance apart. If the electric field due to $-3 Q$ at the location of $Q$ is $E \hat{i}$,then the electric field due to $Q$ at the location of $-3 Q$ is:

$A$ thin non-conducting ring of radius $r$ has a linear charge density $\lambda = \lambda_0 \cos \phi$,where $\lambda_0$ is a constant and $\phi$ is the azimuthal angle. The magnitude of the electric field strength at the centre of the ring is

Four point charges are placed at the vertices of a regular hexagon as shown in the figure below. What will be the net electric field at the centre?