A tiny spherical oil drop carrying a net charge $$q$$ is balanced in still air, with a vertical uniform electric field of strength $$\frac{81}{7} \pi \times 10^5 \mathrm{~V} / \mathrm{m}$$. When the field is switched OFF, the drop is observed to fall with terminal velocity $$2 \times 10^{-3} \mathrm{~ms}^{-1}$$. Here $$g=9.8 \mathrm{~m} / \mathrm{s}^2$$, viscosity of air is $$1.8 \times 10^{-5} \mathrm{Ns} / \mathrm{m}^2$$ and density of oil is $$900 \mathrm{~kg} \mathrm{~m}^{-3}$$. The magnitude of $$q$$ is

Four charges $$+q_1+2 q_1+q$$ and $$-2 q$$ are placed at the corners of a square $$A B C D$$ respectively. The force on a unit positive charge kept at the centre $$O$$ is

An electric dipole with dipole moment $$4 \times 10^{-19} \mathrm{C}-\mathrm{m}$$ is aligned at $$30^{\circ}$$ with the direction of a uniform electric field of magnitude $$5 \times 10^4 \mathrm{NC}^{-1}$$, the magnitude of the torque acting on the dipole is

A charged particle of mass $$m$$ and charge $$q$$ is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after $$t$$ seconds is