Question

An oil drop has a charge $-9.6\times {10}^{-19}C$ and mass $1.6\times {10}^{-15}gm$. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the following are correct. $(g=10m{s}^2)$ (Assume that the magnitude of resistance force is same in both the cases)

• A. The electric field is directed upward
• B. The electric field is directed downward
• C. The intensity of electric field is $\frac{1}{3}\times {10}^2{NC}^{-1}$
• D. The intensity of electric field is $\frac{1}{6}\times {10}^5{NC}^{-1}$

Answer 1

Answer: (A), (D)

The correct options are
A The electric field is directed upward
D The intensity of electric field is $\frac{1}{6}\times {10}^5{NC}^{-1}$

(c) At equilibrium, electric force on drop balances weight of drop.

$qE=mg$

$E=mg/q$

$E=\frac{-1.6\ast {10}^{-15}\ast 10}{-9.6\ast {10}^{-19}}$

$E=\frac{1}{6}\ast {10}^5$

Upward