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.
$qE+qE\theta_0^2$
$qE+\frac{qE\theta_0^2}{4}$
$qE+\frac{qE\theta_0^2}{3}$
$qE+\frac{qE\theta_0^2}{6}$
For a uniformly charged ring of radius $R$, the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is
If the net electric field at point $\mathrm{P}$ along $\mathrm{Y}$ axis is zero, then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$, where $\mathrm{x}=$. . . . . .
What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,newton/coulomb$ $[1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N{m^2}/{C^2}]$
The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,\frac{V}{m}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10\,\frac{m}{{{s^2}}}$)
Write equation of electric field by system of $\mathrm{'n'}$ charges.