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 electric field intensity at a point in vacuum is equal to

$A$ charge of $1 \mu C$ each is placed on five corners of a regular hexagon of side $1 \ m$. The electric field at its centre is . . . . . . $N$/$C$.

The point charges $Q$ and $-2Q$ are placed at a distance $r$ apart. If the electric field at the location of $Q$ (point $A$) due to $-2Q$ is $\vec E$,then the electric field at the location of $-2Q$ (point $B$) due to $Q$ will be:

An infinite number of electric charges,each with a magnitude of $5 \, nC$,are placed along the $X$-axis at $x = 1 \, cm$,$x = 2 \, cm$,$x = 4 \, cm$,$x = 8 \, cm$,and so on. If the consecutive charges have opposite signs,find the electric field in $N/C$ at $x = 0$. (Given: $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, N \cdot m^2/C^2$)

$A$ thin,positively charged metal ring of radius $R$ is fixed in the $X-Y$ plane with its center at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, Z_0)$,where $Z_0 > 0$. Describe the motion of the particle.