What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,newton/coulomb$ $[1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N{m^2}/{C^2}]$

Four equal positive charges are fixed at the vertices of a square of side $L$. $Z$-axis is perpendicular to the plane of the square. The point $z = 0$ is the point where the diagonals of the square intersect each other. The plot of electric field due to the four charges, as one moves on the $z-$ axis.

Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta $ with the $x-$ axis, then the angle $\theta$ 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 a third charged particle should be placed so that it does not experience a net electric force?

The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31}\,kg$ and $1.6 \times$ $10^{-19}\,C$.)

A wire of length $L\, (=20\, cm)$, is bent into a semicircular arc. If the two equal halves of the arc were each to be uniformly charged with charges $ \pm Q\,,\,\left[ {\left| Q \right| = {{10}^3}{\varepsilon _0}} \right]$ Coulomb where $\varepsilon _0$ is the permittivity (in $SI\, units$) of free space] the net electric field at the centre $O$ of the semicircular arc would be