Question

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

• A. $8.0\times {10}^{-19}C$
• B. $4.8\times {10}^{-19}C$
• C. $1.6\times {10}^{-19}C$
• D. $3.2\times {10}^{-19}C$

Answer 1

The correct option is A $8.0\times {10}^{-19}C$

During equilibrium in presence of electric field qE = mg
$\Rightarrow qE=\frac{4}{3}\pi r^3\rho g$ ...(i)
When the drop descends with constant velocity
$mg=6\pi \eta rV$
$\Rightarrow \frac{4}{3}\pi r^3\rho g=6\pi \eta rV$
Putting the values, we obtain
$r=\frac{3}{7}\times {10}^{-5}m$
Putting back in equation (i)
$Q\approx 8\times {10}^{-19}C$