Physics

Physics

Problem

An oil drop, carrying six electronic charges and having a mass of 1.6×10−121.6 \times 10^{-12} g, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy. (1) 105 NC−110^5 \text{ NC}^{-1} (2) 104 NC−110^4 \text{ NC}^{-1} (3) 3.3×104 NC−13.3 \times 10^4 \text{ NC}^{-1} (4) 3.3×105 NC−13.3 \times 10^5 \text{ NC}^{-1}
Intuitive explanation

The key insight is that terminal velocity occurs when forces balance—gravity pulling down equals drag pulling up. To reverse the drop's motion at the same speed, you need the electric force to not just overcome gravity, but to create an equal and opposite net force. Since drag reverses direction when motion reverses, the electric force must be twice the weight to produce the same upward acceleration that gravity alone produced downward.

Intuitive explanation

The key insight is that terminal velocity occurs when forces balance—gravity pulling down equals drag pulling up. To reverse the drop's motion at the same speed, you need the electric force to not just overcome gravity, but to create an equal and opposite net force. Since drag reverses direction when motion reverses, the electric force must be twice the weight to produce the same upward acceleration that gravity alone produced downward.

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