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

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

A thin semi-circular ring ofradius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec E$ at the centre $O$ is

The charge distribution along the semi-circular arc is non-uniform . Charge per unit length $\lambda $ is given as $\lambda  = {\lambda _0}\sin \theta $ , with $\theta $ measured as shown in figure. $\lambda_0$ is a positive constant. The radius of arc is $R$ . The electric field at the center $P$ of semi-circular arc is $E_1$ . The value of $\frac{{{\lambda _0}}}{{{ \in _0}{E_1}R}}$ is

A charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6 × 10^4 $  $Vm^{-1}$. The charge on the drop is $8 × 10^{-18}$ $C$. The radius of the drop is $\left[ {{\rho _{air}} = 1.29\,kg/{m^3};{\rho _{Hg}} = 13.6 \times {{10}^3}kg/{m^3}} \right]$