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 positively charged thin metal ring of radius $R$ is fixed in the $xy - $ plane with its centre at the $O$. A negatively charged particle $P$ is released from rest at the point $(0,\,0,\,{z_0})$, where ${z_0} > 0$. Then the motion of $P$ is

Four equal positive charges are fixed at the vertices of a square of side $L$. $Z$-axis is perpendicular to the plane of the square. The point $z = 0$ is the point where the diagonals of the square intersect each other. The plot of electric field due to the four charges, as one moves on the $z-$ axis.

Two equal negative charges $-\, q$ each are fixed at the points $(0, a)$ and $(0, -a)$ on the $Y$ -axis. A positive charge $Q$ is released from rest at the point $(2a, 0)$ on the $X$ -axis. The charge $Q$ will :-

A vertical electric field of magnitude $4.9 \times 10^{5} N / C$ just prevents a water droplet of a mass $0.1\, g$ from falling. The value of charge on the droplet will be  ........ $\times 10^{-9} \;C$ $\left(\right.$ Given $\left.g =9.8 m / s ^{2}\right)$

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a=\frac{\sqrt{3}}{2} L$.