Question

A charged oil drop is suspended in uniform field of $3\times {10}^{4}V/m$ so that it neither falls nor rises. The charge on the drop will be: (take the mass of the charge $=9.9\times {10}^{-15}$ kg and $g=10m/s^2$)

• A. $3.3\times {10}^{-18}C$
• B. $3.2\times {10}^{-18}C$
• C. $1.6\times {10}^{-18}C$
• D. $4.8\times {10}^{-18}C$

Answer 1

The correct option is A $3.3\times {10}^{-18}C$

Under such and equilibrium, the electric and gravitational forces balance each other.

Thus $qE=mg$

$⟹q=\frac{mg}{E}=\frac{(9.9\times {10}^{-15})\times \left(10\right)}{3\times {10}^4}=3.3\times {10}^{-18}C$