Electric field strength due to a point charge of $5\,\mu C$ at a distance of $80\, cm$ from the charge is

Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta $ with the $x-$ axis, then the angle $\theta$ is

A thin disc of radius $b = 2a$  has a concentric hole of radius $'a'$ in it (see figure). It carries uniform surface charge $'\sigma '$ on it. If the electric field on its axis at height $'h'$ $(h << a)$ from its centre is given as $'Ch'$ then value of $'C'$ is

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 ${\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}$ ]

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

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.......$N/C$ $(g = 10\,m{s^{ - 1}})$