How many electrons should be removed from a coin of mas $1.6 \,g$, so that it may float in an electric field of intensity $10^9 \,N / C$ directed upward?

An oil drop of radius $2\, mm$ with a density $3\, g$ $cm ^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5}\, V\, m ^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? (consider $\left. g =9.81\, m / s ^{2}\right)$

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 $\left( {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,N - {m^2}/{c^2}} \right)$

$ABC$ is an equilateral triangle. Charges $ + \,q$ are placed at each corner. The electric intensity at $O$ will be

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 $P$.

Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^4 \mathrm{~N} / \mathrm{C}$ normally. A charged particle of mass $2 \mathrm{~g}$ being suspended through a silk thread of length $20 \mathrm{~cm}$ and remain stayed at a distance of $10 \mathrm{~cm}$ from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{\mathrm{x}}} \ \mu \mathrm{C}$ where $\mathrm{x}=$ ____________.  use $g=10 \mathrm{~m} / \mathrm{s}^2$ ]