Two charges $e$ and $3e$ are placed at a distance $r$ from each other. The distance of the point where the electric field intensity will be zero is .........

Five point charges each having magnitude $q$ are placed at the corners of a regular hexagon as shown in the figure. The net electric field at the centre $O$ is $\vec E$. To make the net electric field at $O$ equal to $6\vec E$,the charge placed on the remaining sixth corner should be: (in $,q$)

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \ N/C$ normally. $A$ charged particle of mass $2 \ g$ is suspended by a silk thread of length $20 \ cm$ and remains at a distance of $10 \ cm$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} \ \mu C$ where $x=$ . . . . . . . (Use $g=10 \ m/s^2$)

Two point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. The net electric field due to these two point charges will be zero at a point on the $x$-axis at:

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices of a regular hexagon $ABCDEF$ as shown in the figure. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is twice the electric field at centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is

Two particles $A$ and $B$ having charges $20\, \mu C$ and $-5\, \mu C$ respectively are held fixed with a separation of $5\, cm$. At what position should a third charged particle be placed so that it does not experience a net electric force?