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 $ACB$ ?

Charges $Q _{1}$ and $Q _{2}$ arc at points $A$ and $B$ of a right angle triangle $OAB$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q _{1} / Q _{2}$ is proportional to

The electric field intensity at a point in vacuum is equal to

Two uniform spherical charge regions $S_1$ and $S_2$ having positive and negative charges overlap each other as shown in the figure. Point $O_1$ and $O_2$ are their centres and points $A, B, C$ and $D$ are on the line joining centres $O_1$ and $O_2$. Electric field from $C$ to $D$

For a uniformly charged ring of radius $R$, the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is