A charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6 × 10^4 $  $Vm^{-1}$. The charge on the drop is $8 × 10^{-18}$ $C$. The radius of the drop is $\left[ {{\rho _{air}} = 1.29\,kg/{m^3};{\rho _{Hg}} = 13.6 \times {{10}^3}kg/{m^3}} \right]$

Two charges $ + 5\,\mu C$ and $ + 10\,\mu C$ are placed $20\, cm$ apart. The net electric field at the mid-Point between the two charges is

Two charges $+Q$ and $-2 Q$ are located at points $A$ and $B$ on a horizontal line as shown below.The electric field is zero at a point which is located at a finite distance

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

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.

A positively charged ball hangs from a silk thread. We put a positive test charge ${q_0}$ at a point and measure $F/{q_0}$, then it can be predicted that the electric field strength $E$