Question

A charged oil drop is prevented from falling under gravity by a vertical electric field
between two horizontal metal plates charged to a potential difference 6920 V, the
distance between the plates being 1.3 cm. When the field is cut off the drop falls in air
with a uniform velocity of 1.9 × 10−4

Answer 1

$$r=\sqrt{\frac{9\eta v}{2g\rho}}$$

$$r=\sqrt{\frac{9(1.81\cdot10^{-5})(1.9\cdot10^{-4})}{2(9.8)(900)}}=1.32\cdot10^{-6}\ m$$