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:

Point charges $+8q$ and $-2q$ are placed at $x = 0$ and $x = L$ respectively. At which point on the $X$-axis will the resultant electric field intensity be zero?

$A$ negative charge is placed at the centre of a non-conducting sphere. The direction of the electric field at any point on the surface of the sphere is

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC^{-1}$. The distance $d$ between the point charges $A$ and $B$ is ............ $m$. (in $.0$)

The figure shows some electric field lines corresponding to an electric field. Then:

$A$ point charge of $50 \mu C$ is placed in the $XY$ plane at a location with radius vector $\vec{r}_0 = 2 \hat{i} + 3 \hat{j} \ m$. The electric field strength magnitude at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j} \ m$ is (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2}$). (in $kV \ m^{-1}$)