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

A thin disc of radius $b = 2a$  has a concentric hole of radius $'a'$ in it (see figure). It carries uniform surface charge $'\sigma '$ on it. If the electric field on its axis at height $'h'$ $(h << a)$ from its centre is given as $'Ch'$ then value of $'C'$ is

Two charges $+Q$ and $-2 Q$ are located at points $A$ and $B$ on a horizontal line as shown below.The electric field is zero at a point which is located at a finite distance

The acceleration of an electron in an electric field of magnitude $50\, V/cm$, if $e/m$ value of the electron is $1.76 \times {10^{11}}\,C/kg$, is

Find out electric filed intensity at point $A(1,0,2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2,1)$

Write equation of electric field by point charge. How does it depend on distance ?