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

Electric field at centre $O$ of semicircle of radius $a$ having linear charge density $\lambda$ given is given by

Is electric field scalar or vector ? Why ?

Charges $Q _{1}$ and $Q _{2}$ arc at points $A$ and $B$ of a right angle triangle $OAB$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q _{1} / Q _{2}$ is proportional to

Two point charges $q_{1}$ and $q_{2},$ of magnitude $+10^{-8} \;C$ and $-10^{-8}\; C ,$ respectively, are placed $0.1 \;m$ apart. Calculate the electric fields at points $A, B$ and $C$ shown in Figure

Two point charges $( + Q)$ and $( - 2Q)$ are fixed on the $X-$axis at positions $a$ and $2a$ from origin respectively. At what positions on the axis, the resultant electric field is zero