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
$mge$
$\frac{{mg}}{e}$
$\frac{e}{{mg}}$
$\frac{{{e^2}}}{{{m^2}}}g$
The acceleration of an electron in an electric field of magnitude $50\, V/cm$, if $e/m$ value of the electron is $1.76 \times {10^{11}}\,C/kg$, is
An oil drop carries six electronic charges, has a mass of $1.6 \times 10^{-12} g$ and falls with a terminal velocity in air. The magnitude of vertical electrical electric field required to make the drop move upward with the same speed as was formely moving is ........$kN/C$
What will be the magnitude of electric field at point $O$ as shown in figure ? Each side of the figure is $I$ and perpendicular to each other.
Explain electric field and also electric field by point charge.
As shown in the figure, a particle A of mass $2\,m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q.$ The system is placed in an electric field $\vec E$ . The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E $ . After the system settles into equilibrium, one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find maximum tension in the rod.