The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31}\,kg$ and $1.6 \times$ $10^{-19}\,C$.)

Find ratio of electric field at point $A$ and $B.$ Infinitely long uniformly charged wire with  linear charge density $\lambda$ is kept along $z-$ axis

Two charges $\pm 10\; \mu C$ are placed $5.0\; mm$ apart. Determine the electric field at $(a)$ a point $P$ on the axis of the dipole $15 cm$ away from its centre $O$ on the side of the positive charge, as shown in Figure $(a),$ and $(b)$ a point $Q , 15\; cm$ away from $O$ on a line passing through $O$ and normal to the axis of the dipole, as shown in Figure.

Two point charges $Q$ and $-3Q$ are placed at some distance apart. If the electric field at the location of $Q$ is $E$ then at the locality of $ - 3Q$, it is

An infinite number of electric charges each equal to $5\, nC$ (magnitude) are placed along $X$-axis at $x = 1$ $cm$, $x = 2$ $cm$ , $x = 4$ $cm$ $x = 8$ $cm$ ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at $x = 0$ is $\left( {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,N - {m^2}/{c^2}} \right)$

Two charges $q$ and $3 q$ are separated by a distance ' $r$ ' in air. At a distance $x$ from charge $q$, the resultant electric field is zero. The value of $x$ is :