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

What is the magnitude of a point charge which produces an electric field of $2\, N/coulomb$ at a distance of $60\, cm$ $(1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N - {m^2}/{C^2})$

The distance between the two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$ At what point on the line joining the two, the intensity will be zero

An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20000\, V/m$. If mass of the particle is $9.6 \times {10^{ - 16}}\,kg$, the charge on it and excess number of electrons on the particle are respectively $(g = 10\,m/{s^2})$

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q.$ The distance of the point $P$ from the centre of the rod is $a=\frac{\sqrt{3}}{2} L$.