In the given figure electric field at center $O$ due to section $AB$ of uniformly charged ring is $\overrightarrow E$. What will be electric field at $O$ due to section $ACB$ ?

The given diagram shows two semi infinite line of charges having equal (in magnitude) linear charge density but with opposite sign. The electric field at any point on $x$ axis for $(x > 0)$ is along the unit vector

A thin semi-circular ring ofradius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec E$ at the centre $O$ 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…………..$m$

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \mathrm{~N} / \mathrm{C}$ normally. A charged particle of mass $2 \mathrm{~g}$ being suspended through a silk thread of length $20 \mathrm{~cm}$ and remain stayed at a distance of $10 \mathrm{~cm}$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{\mathrm{x}}} \ \mu \mathrm{C}$ where $\mathrm{x}=$ ____________.  use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]