$A$ positively charged ball hangs from a silk thread. We put a positive test charge $q_0$ at a point and measure $F/q_0$. Then, it can be predicted that the electric field strength $E$ is:

Two point charges $(+Q)$ and $(-2Q)$ are fixed on the $X$-axis at positions $x = a$ and $x = 2a$ from the origin, respectively. At what position on the axis is the resultant electric field zero?

The charges $2q, -q, -q$ are located at the vertices of an equilateral triangle. At the center of the triangle:

Two charges $Q$ and $4Q$ are separated by a distance of $6 \text{ cm}$. The distance of the point from $4Q$ at which the net electric field is zero is: (in $\text{ cm}$)

$A$ negative charge $-q$ is placed at the midpoint between two fixed equal positive charges $Q$,separated by a distance $2d$. If the negative charge is given a small displacement $x$ $(x \ll d)$ perpendicular to the line joining the positive charges,how will the net force $(F)$ acting on it approximately depend on $x$?

Two point charges $q_{1}$ and $q_{2},$ of magnitude $+10^{-8} \; C$ and $-10^{-8} \; C,$ respectively,are placed $0.1 \; m$ apart. Calculate the electric fields at points $A, B$ and $C$ shown in the figure.