$A$ charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20,000\, V/m$. If the mass of the particle is $9.6 \times 10^{-16}\, kg$,the charge on it and the excess number of electrons on the particle are respectively $(g = 10\, m/s^2)$.

An electron having charge '$e$' and mass '$m$' is moving in a uniform electric field $E$. Its acceleration will be

$A$ positively charged particle moving along the horizontal axis with a certain velocity enters a uniform electric field directed along the vertical axis. Its

An electron is accelerated through a potential difference of $1000 \ V$. Its velocity is nearly

As shown in the figure below,a point charge $q$ moves from point $P$ to a point $S$ traversing a path $PQRS$ in a uniform electric field $\vec{E}$. The electric field is directed along a direction parallel to the $x$-axis. The coordinates of $P$,$Q$,$R$,and $S$ are $(a, b, 0)$,$(2a, 0, 0)$,$(a, -b, 0)$,and $(0, 0, 0)$ respectively. What is the work done by the field in the process?

Two charges $-q$ each are fixed and separated by a distance $2d$. $A$ third charge $q$ of mass $m$ placed at the midpoint is displaced slightly by $x$ $(x \ll d)$ perpendicular to the line joining the two fixed charges as shown in the figure. Show that the charge $q$ will perform simple harmonic motion with a time period $T = \left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$.