A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength 81 7 10^5 Vm ⁻¹ . When the field is switched off, the drop is observed to fall with terminal velocity 2 10⁻³ ~ms ⁻¹ . Given g=9.8 ~ms ⁻² , viscosity of the air =1.8 10⁻⁵ Ns m ⁻² and the density of oil =900 ~kg ~m ⁻³ , the magnitude of q is

Step-by-step solution

aligned & q E=m g & 6 r v=m g & 4 3 r^3 g=m g & r= ( 3 m g 4 g )^ 1 / 3 aligned Substituting the value of r in Eq. (ii) we get, 6 v ( 3 m g 4 g )^ 1 / 3 =m g or (6 v)^3 ( 3 m g 4 g )=(m g)^3 Again substituting m g=q E we get, (q E)^2= ( 3 4 g )(6 v)^3 or q E= ( 3 4 g )^ 1 / 2 (6 v)^ 3 / 2 q= 1 E ( 3 4 g )^ 1 / 2 (6 v)^ 3 / 2 Substituting the values we get , aligned q= & 7 81 10^5 3 4 900 9.8 216 ^3 & (1.8 10⁻⁵ 2 10⁻³ )^3 & =8.0 10⁻¹⁹ C aligned correct option is ( d ) .