A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength 
Vm –1 . When the field is switched off, the drop is observed to fall with terminal velocity 2 × 10 –3 m s –1 . Given g = 9.8 m s –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:
Text Solution
Verified by ExpertsThe correct answer is:
D
In equilibrium,
mg = qE
In absence of electric field,
mg = 6π η rv
qE = 6πqrv
m = 
πr 3 d. = 
= 
After substituting value we get,
q = 8 × 10 –19
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