$A$ liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5 \times 10^3 \, Vm^{-1}$. The density of the liquid is $1.26 \times 10^3 \, kg \, m^{-3}$. The radius of the drop is (neglect buoyancy):

$A$ particle of mass $2 \ g$ and charge $6 \ \mu C$ is accelerated from rest through a potential difference of $60 \ V$. The speed acquired by the particle is (in $ms^{-1}$)

$A$ block of mass $m$ and charge $q$ is connected to a point $O$ with an inextensible string. This system is on a horizontal table. An electric field $E$ is applied perpendicular to the string and in the plane of the horizontal table. The tension in the string when it becomes parallel to the electric field is

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20\ V$ (i.e.,plate $2$ is at a higher potential). The plates are separated by $d = 0.1\ m$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $1$. What is its speed when it hits plate $2$? $(e = 1.6 \times 10^{-19}\ C, m_e = 9.11 \times 10^{-31}\ kg)$

$A$ 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?

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