In the given figure, find the distance from point $A$ where the electric field is zero (in $cm$).

$A$ thin conducting ring of radius $R$ is given a charge $+Q.$ The electric field at the centre $O$ of the ring due to the charge on the part $AKB$ of the ring is $E.$ The electric field at the centre due to the charge on the part $ACDB$ of the ring is

$A$ wire of length $L = 20 \, cm$ is bent into a semicircular arc. If the two equal halves of the arc are uniformly charged with charges $+Q$ and $-Q$ respectively,where $|Q| = 10^3 \varepsilon_0$ Coulomb and $\varepsilon_0$ is the permittivity of free space,find the net electric field at the centre $O$ of the semicircular arc.

$A$ 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 $-3Q$ are placed at a distance $x$ apart. If the electric field at the location of $Q$ is $E$,then the electric field at the location of $-3Q$ is:

Two equal positive point charges are separated by a distance $2a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which the force experienced by a test charge $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is $................$