Three equal charges $+q$ each are placed at the three vertices of an equilateral triangle. The electric field at the centroid of the triangle is . . . . . . . ('$r$' is the length of the side of the triangle).

What is the electric field at a distance $x = \sqrt{8} R$ from the center on the axis of a charged ring with charge $Q$ and radius $R$?

Four point charges $-q, +q, +q$ and $-q$ are placed on the $y$-axis at $y = -2d, y = -d, y = +d$ and $y = +2d$,respectively. The magnitude of the electric field $E$ at a point on the $x$-axis at $x = D$,with $D >> d$,will vary as:

In a region,the electric field varies as $E = 2x^2 - 4$,where $x$ is the distance in $m$ from the origin along the $x$-axis. $A$ positive charge of $1 \,\mu C$ is released with minimum velocity from infinity to cross the origin. Then:

Two point charges $a$ and $b$,whose magnitudes are the same,are positioned at a certain distance from each other with $a$ at the origin. $A$ graph is drawn between the electric field strength $E$ at points between $a$ and $b$ and the distance $x$ from $a$. $E$ is taken as positive if it is along the line joining $a$ to $b$. From the graph,it can be decided that:

Two point charges $(+Q)$ and $(-2Q)$ are placed at distances $a$ and $2a$ respectively from the origin on the $X$-axis. At what distance from the origin is the electric field zero?