For the given figure the direction of electric field at $A$ will be

A charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be.......$N/C$ $(g = 10\,m{s^{ - 1}})$

Give reason : ''Small and light pieces of paper are attracted by comb run through dry hair.''

Two point charges $q_1\,(\sqrt {10}\,\,\mu C)$ and $q_2\,(-25\,\,\mu C)$ are placed on the $x-$ axis at $x = 1\,m$ and $x = 4\,m$ respectively. The electric field (in $V/m$ ) at a point $y = 3\,m$ on $y-$ axis is, [ take ${\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}$ ]

The magnitude of electric field intensity $E$ is such that, an electron placed in it would experience an electrical force equal to its weight is given by

Two point charges $( + Q)$ and $( - 2Q)$ are fixed on the $X-$axis at positions $a$ and $2a$ from origin respectively. At what positions on the axis, the resultant electric field is zero