The dimensional formula of electric field is . . . . . . .

$A$ 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 $kV \ m^{-1}$)

$A$ uniform electric field of $500 \ Vm^{-1}$ is directed at $30^{\circ}$ with the positive $X$-axis as shown in the figure. The potential difference $(V_B - V_A)$ if $OA = 3 \ m$ and $OB = 5 \ m$ is

$A$ cube of side $a$ has point charges $+Q$ located at each of its vertices except at the origin where the charge is $-Q$. The electric field at the centre of the cube is

Electric field at a distance $r$ from an infinitely long uniformly charged straight conductor,having linear charge density $\lambda$ is $E_1$. Another uniformly charged conductor having same linear charge density $\lambda$ is bent into a semicircle of radius $r$. The electric field at its centre is $E_2$. Then

At a certain distance from a point charge,the electric field is $500\,V/m$ and the potential is $3000\,V$. What is this distance in meters?