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 liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5\, k\,Vm^{-1}$ . The density of liquid is $1.26\times10^3\, kg\, m^{-3}$ . The radius of the drop is (neglect buoyancy)

Obtain the equation of electric field at a point by system of $\mathrm{'n'}$ point charges.

A circular ring carries a uniformly distributed positive charge. The electric field $(E) $ and potential $ (V) $ varies with distance $(r)$ from the centre of the ring along its axis as

A thin conducting ring of radius $R$ is given a charge $+Q.$ The electric field at the centre $O$ of the ring due to the charge on the part $AKB$ of the ring is $E.$ The electric field at the centre due to the charge on the part $ACDB$ of the ring is 

Two point charge $-q$ and $+q/2$ are situated at the origin and at the point $(a, 0, 0)$ respectively. The point along the $X$ - axis where the electric field vanishes is