A vertical electric field of magnitude $4.9 \times 10^{5} N / C$ just prevents a water droplet of a mass $0.1\, g$ from falling. The value of charge on the droplet will be  ........ $\times 10^{-9} \;C$ $\left(\right.$ Given $\left.g =9.8 m / s ^{2}\right)$

$ABC$ is an equilateral triangle. Charges $ + \,q$ are placed at each corner. The electric intensity at $O$ will be

A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20000\, V/m$. If mass of the particle is $9.6 \times {10^{ - 16}}\,kg$, the charge on it and excess number of electrons on the particle are respectively $(g = 10\,m/{s^2})$

A uniformly charged rod of length $4\,m$ and linear charge density $\lambda  = 30\,\mu C/m$ is placed as shown in figure. Calculate the $x-$ component of electric field at point $P$.

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC ^{-1}$. The distance ' $d$ ' between the point charges $A$ and $B$ is..............$m$

A charged particle of mass $5 \times {10^{ - 5}}\,kg$ is held stationary in space by placing it in an electric field of strength ${10^7}\,N{C^{ - 1}}$ directed vertically downwards. The charge on the particle is