$A$ sphere carrying charge $Q$ and having weight $w$ falls under gravity between a pair of vertical plates at a distance $d$ from each other. When a potential difference $V$ is applied between the plates,the acceleration of the sphere changes as shown in the figure,along the line $BC$. The value of $Q$ is:

$A$ particle of mass $m$ and charge $q$ is fastened to one end $A$ of a massless string having equilibrium length $\ell$,whose other end is fixed at point $O$. The whole system is placed on a frictionless horizontal plane and is initially at rest. If a uniform electric field $E$ is switched on along the $x$-direction as shown in the figure,then the speed of the particle when it crosses the $x$-axis is

An electron of mass $m_e$ initially at rest moves through a certain distance in a uniform electric field in time $t_1$. $A$ proton of mass $m_p$ also initially at rest takes time $t_2$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity,the ratio of $t_2/t_1$ is nearly equal to

$A$ particle of mass $m$ and carrying charge $-q_1$ is moving around a charge $+q_2$ along a circular path of radius $r$. Find the period of revolution of the charge $-q_1$.

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the center of a uniformly charged spherical region of total charge $Q$ and radius $R$. $q$ and $Q$ have opposite signs. The spherically charged region is not free to move. The value of $K$ is such that the particle will just reach the boundary of the spherically charged region. How much time does it take for the particle to reach the boundary of the region?

When an electron enters an electric field with a velocity in the direction of the electric field lines,then.......