A uniformly charged rod of length $4\,m$ and linear charge density $\lambda  = 30\,\mu C/m$ is placed as shown in figure. Calculate the $x-$ component of electric field at point $P$.

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

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 point charges $( + Q)$ and $( - 2Q)$ are fixed on the $X-$axis at positions $a$ and $2a$ from origin respectively. At what positions on the axis, the resultant electric field is zero

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 $\left[ {{\rho _{air}} = 1.29\,kg/{m^3};{\rho _{Hg}} = 13.6 \times {{10}^3}kg/{m^3}} \right]$

A liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5\, k\,Vm^{-1}$ . The density of liquid is $1.26\times10^3\, kg\, m^{-3}$ . The radius of the drop is (neglect buoyancy)