The intensity of the electric field required to keep a water drop of radius ${10^{ - 5}}\, cm$ just suspended in air when charged with one electron is approximately

A ring of radius $R$ is charged uniformly with a charge $+\,Q$ . The electric field at a point on its axis at a distance $r$ from any point on the ring will be

Two charges $ + 5\,\mu C$ and $ + 10\,\mu C$ are placed $20\, cm$ apart. The net electric field at the mid-Point between the two charges is

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

The electric field due to a charge at a distance of $3\, m$ from it is $500\, N/coulomb$. The magnitude of the charge is.......$\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N - {m^2}}}{{coulom{b^2}}}} \right]$

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 $a=\frac{\sqrt{3}}{2} L$.